LaTeX source for Answers to Bayesian Statistics: An Introduction (4th edn)

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\begin{document}

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\begin{flushleft}
{\Huge{\textbf{Bayesian Statistics:}}}\\
\vspace{10 mm}
{\Large{\textbf{An Introduction}}}\\
\vspace{20 mm}
{\huge{\textbf{PETER M. LEE}}}\\
\vspace{3 mm}
{\large{\textit{\textbf{\mbox{\noindent
Formerly Provost of Wentworth College,}}}}}\\
{\large{\textit{\textbf{\mbox{\noindent
University of York, England}}}}}\\
\vspace{15mm}
{\Large{\textit{\textbf{Third Edition}}}}\\
\vspace{55mm}
A member of the Hodder Headline Group \\
LONDON $\bullet$ SYDNEY $\bullet$ AUCKLAND \\

\vspace{3 mm}
Copublished in North, Central and South America by \\
John Wiley \& Sons Inc., New York $\bullet$ Toronto
\end{flushleft}
\vfill

\mainmatter

\appendix

\setcounter{chapter}{3}

\allowdisplaybreaks

\pagestyle{plain}

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq Considering trumps and non-trumps separately, required
probability is
$2\binom{3}{3}\binom{23}{10}\left/\binom{26}{13}\right.=\frac{11}{50}.$
Probability of a $2:1$ split is 39/50, and the conditional probability
that the king is the odd one is 1/3, so probability one player has the
king and the other the remaining two is 13/50.

\nextq \!\!(a) If $\Pr(A)=0$ or $\Pr(A)=1$.

\blankline (b) Take $A$ as odd red die'', $B$ as odd blue die'' and
$C$ as odd sum''.

\blankline (c) Take $C$ as $\{HHH,\, THH,\, THT,\, TTH\}$.

\nextq \!\!(a)\quad$\Pr(\text{homozygous})=1/3;\quad \Pr(\text{heterozygous})=2/3.$

$\Pr(BB\,|\,\text{7 black}) = \frac{(1/3)(1)}{(1/3)(1)+(2/3)(1/2^7)} = \frac{64}{65}.$

\blankline (c)\quad Similarly we see by induction (wih case $n=0$ given
by part (a)) that $\Pr(BB\,|\,\text{first$n$black})=2^{n-1}/(2^{n-1}+1)$
since
\begin{align*}
\Pr(BB\,|\,\text{first $n+1$ black})
& = \frac{\{2^{n-1}/(2^{n-1}+1)\}(1)}
{\{2^{n-1}/(2^{n-1}+1)\}(1)+\{1/(2^{n-1}+1)\}(1/2)} \\
& = \frac{2^n}{2^n+1}.
\end{align*}

\nextq Use $\Pr(GG)=\Pr(M)\Pr(GG|M)+\Pr(D)\Pr(GG|D)$ to get
$\Pr(M)=\frac{\Pr(GG)-p^2}{p(1-p)}$

\nextq $p(3)=p(4)=1/36$, $p(5)=p(6)=2/36$, $p(7)=\dots=p(14)=3/36$,
$p(15)=p(16)=2/36$, $p(17)=p(18)=1/36$.  As for the distribution
function,
$F(x)=\left\{\begin{array}{ll} 0 & \text{for x < 3;} \\ 1/36 & \text{for 3\leqslant x < 4;} \\ 2/36 & \text{for 4\leqslant x < 5;} \\ 4/36 & \text{for 5\leqslant x < 6;} \\ 6/36 & \text{for 6\leqslant x < 7;} \\ (3[x]-12)/36 & \text{for n\leqslant x < n+1 where 7\leqslant n < 15;} \\ 32/36 & \text{for 15\leqslant x < 16;} \\ 34/36 & \text{for 16\leqslant x < 17;} \\ 35/36 & \text{for 17\leqslant x < 18;} \\ 1 & \text{for x\geqslant18} \end{array}\right.$
where $[x]$ denotes the integer part of $x$.

\nextq $\Pr(k=0)=(1-\pi)^n=(1-\lambda/n)^n\to\text{e}^{-\lambda}$.  More
generally
\begin{align*}
p(k) & = \binom{n}{k}\pi^k(1-\pi)^{n-k} \\
& = \frac{\lambda^k}{k!}\left(1-\frac{\lambda}{n}\right)^n
\frac{n}{n}\frac{n-1}{n}\dots\frac{n-k+1}{n}
\left(1-\frac{\lambda}{n}\right)^{-k} \\
& \to\frac{\lambda^k}{k!}\exp(-\lambda).
\end{align*}

\nextq \!\!(a) $\Pr\bigl(\,\random k=0\bigr)=\Pr(\random m=\random n=0)= \Pr(\random m=0)\,\Pr(\random n=0)= \text{e}^{-\lambda}\text{e}^{-\mu}=\text{e}^{-(\lambda+\mu)}$,
\begin{align*}
\Pr(\,\random k=1\bigr) & = \Pr\bigl(\{\random m=1,\,\random n=0\}\ \text{or}\
\{\random m=0,\,\random n=1\}\bigr) \\
& = \lambda\text{e}^{-(\lambda+\mu)}+
\mu\text{e}^{-(\lambda+\mu)}=
(\lambda+\mu)\text{e}^{-(\lambda+\mu)}.
\end{align*}

\blankline (b) More generally
\begin{align*}
\Pr\bigl(\,\random k=k\bigr)
& = \sum_{m=0}^k \Pr(\random m=m,\,\random n=k-m)=
\sum_{m=0}^k \Pr(\random m=m)\,\Pr(\random n=k-m) \\
& = \sum_{m=0}^k \frac{\lambda^k}{k!}\exp(-\lambda)
\frac{\mu^{k-m}}{(k-m)!}\exp(-\mu) \\
& = \frac{1}{k!}\exp(-(\lambda+\mu))
\sum_{m=0}^{k} \binom{k}{m}\lambda^m\mu^{k-m}=
\frac{(\lambda+\mu)^k}{k!}\exp(-(\lambda+\mu)).
\end{align*}

\blankline (c) By definition of conditional probability
\begin{align*}
\Pr\bigl(\random m=m\,|\,\random k=k\bigr)
& =  \frac{\Pr\bigl(\random m=m,\,\random k=k\bigr)}
{\Pr\bigl(\random k=k\bigr)}=
\frac{\Pr\bigl(\random m=m,\phantom{\random k}\random n=k-m\bigr)}
{\Pr\bigl(\,\random k=k\bigr)} \\
& = \frac{\frac{\lambda^m}{m!}\exp(-\lambda))\,
\frac{\mu^{k-m}}{(k-m)!}\exp(-\mu)}
{\frac{(\lambda+\mu)^k}{k!}\exp(-(\lambda+\mu))} \\
& = \binom{k}{m}\left(\frac{\lambda}{\lambda+\mu}\right)^m
\left(1-\frac{\lambda}{\lambda+\mu}\right)^{k-m}.
\end{align*}

\nextq Let $y=x^2$ where $x\sim\N(0,1)$.  Then
\begin{align*}
\Pr\left(\random y\leqslant y\right)
& = \Pr\left(\random x^2\leqslant y\right)=
\Pr\left(-\sqrt{y}\leqslant\random x\leqslant\sqrt{y}\,\right) \\
& = \Pr\left(\random x\leqslant\sqrt{y}\,)-\Pr(\random x < -\sqrt{x}\,\right)
\end{align*}
so that (because $\Pr(\random x=-\sqrt{y})=0$)
$F_{\random y}(y) = F_{\random x}\left(\sqrt{y}\,\right) - F_{\random x}\left(-\sqrt{y}\,\right)$
and on differentiation
$p_{\random y}(y)=\half y^{-\frac{1}{2}}p_{\random x}\left(\sqrt{y}\,\right)+ \half y^{-\frac{1}{2}}p_{\random x}\left(-\sqrt{y}\,\right).$
Alternatively, you could argue that
$\Pr(y < \random y \leqslant y + \dy) = \Pr(x < \random x \leqslant x + \dx) + \Pr(-x -\dx \leqslant \random x < -x)$
implying
$p_{\random y}(y)\dy = p_{\random x}(x)\dx + p_{\random x}(-x)\dx$
which as $\dy/\dx = 2x = 2\sqrt{y}$ leads to the same conclusion.

In the case where $x\sim\N(0,1)$ this gives
$p_{\random y}(y)=(2\pi y)^{-\frac{1}{2}}\exp\left(-\half y\right)$
which is the density of $\chi_1^2$.

\nextq By independence, since $\random M \leqslant M$ iff every one of the
indvidual $X_i$ are less than or equal to $M$
$\begin{array}{ll} F_M(M)=\Pr(X_i\leqslant M \quad\forall i) = (F(M))^n; & F_m(m)=1-(1-F(m))^n; \\ p_M(M)=nf(M)(F(M))^{n-1}; & p_m(m)=nf(m)(1-F(m))^{n-1} \end{array}$

\nextq Let $m$ be the minimum of $u$ and $v$ and $c$ be the length of
the centre section.  Then
$p(m, c) =p_{\random u,\,\random v}(m,c)+p_{\random u,\,\random v}(c,m) =2$ if $m+c\leqslant1$ and 0 otherwise.

\nextq If $F(x,y)=F(x)F(y)$ then by differentiation with respect to $x$
and with respect to $y$ we have $p(x,y)=p(x)p(y)$.  The converse follows
by integration.

In discrete cases if $F(x,y)=F(x)F(y)$ then
$p(x,y)=F(x,y)-F(x-1,y)-F(x,y-1)+F(x-1,y-1)= (F_{\random x}(x)-F_{\random x}(x-1))(F_{\random y}(y)-F_{\random y}(y-1))= p(x)p(y)$.  Coversely if $p(x,y)=p(x)p(y)$ then
$F(x,y)=\sum_{\xi\leqslant x}\sum_{\eta\leqslant y} p_{\random x,\random y}(\xi,\eta)$ and so
$F(x,y)=\sum_{\xi\leqslant x}p_{\random x}(\xi) \sum_{\eta\leqslant y}p_{\random y}(\eta)=F(x)F(y)$.

\nextq $\E X=n(1-\pi)/\pi;\ \ \Var X=n(1-\pi)\pi^2.$

\nextq $\E X=\sum\E Z_i^2=\sum 1=\nu$, while
$\E X^2=\sum\E Z_i^4+\sum_{i\ne j}\E X_i^2X_j^2=3\nu+\nu(\nu-1)$ so that
$\Var Z=\E X^2-(\E X)^2=2\nu$.  Similarly by integration.

\nextq $\E X=n\pi$, $\E X(X-1)=n(n-1)\pi^2$ and
$\E X(X-1)(X-2)=n(n-1)(n-2)\pi^3$, so $\E X=n\pi$,
$\E X^2=\E X(X-1)+\E X=n(n-1)\pi+n\pi$ and
\begin{align*}
\E(X-\E X)^3 & = \E X^3-3(\E X^2)(\E X)+2(\E X)^3 \\
& = \E X(X-1)(X-2)-3(\E X^2)(\E X)+3\E X^2+2(\E X)^3-2\E X \\
& = n\pi(1-\pi)(1-2\pi),
\end{align*}
and thus
$\gamma_1=(1-2\pi)/\sqrt{[n\pi(1-\pi)]}$.

\nextq $\phi\geqslant \int_{\{x;\,|x-\mu|\geqslant c\}}c^2p(x)\dx =c^2\Pr(|x-\mu|\geqslant c)$.

\nextq By symmetry $\E x=\E y=\E xy=0$ so $\Cov(x,y)=\E xy-\E x\E y=0$.
However $0=\Pr(x=0,\,y=0)\ne\Pr(x=0)\,\Pr(y=0)=\half\times\half=\quarter$.

\nextq $p(y|x)=\{2\pi(1-\rho^2)\}^{-\frac{1}{2}} \exp\{-\half(y-\rho x)^2/(1-\rho^2)\}$ so conditional on $\random x=x$
we have $y\sim\N(\rho x,1-\rho^2)$, so $\E (y|x)=\rho x$.
$\E (y^2|x)=\Var (y|x)+\{\E(y|x)\}^2=1-\rho^2+\rho^2x^2$, hence
$\E (xy|x)=\rho x^2$, $\E(x^2y^2|x)=x^2-\rho^2x^2+\rho^2x^4$.  Therefore
$\E xy=\rho$ and (as $\E x^4=3$) $\E x^2y^2=1+2\rho^2$, so $\E xy-(\E x)(\E y)=\rho$ and $\E x^2y^2-(\E x^2)(\E y^2)=2\rho^2$.  As $\Var x=1$
and $\Var x^2=\E x^4-(\E x^2)^2=3-1=2=\Var y$, the result follows.

\nextq \!\!(a)\quad $p(x,y)= (\lambda^x\text{e}^{-x}/x!)\binom{x}{y}\pi^y(1-\pi)^{x-y}$
so adding over $x=y, y+1, \dots$ and using
$\sum\lambda^{x-y}(1-\pi)^{x-y}=\text{e}^{\lambda(1-\pi)}$ we get
$p(y)=(\lambda\pi)^y\text{e}^{-\lambda\pi}/y!$ so that
$\random y\sim\P(\lambda\pi)$.  Now note that
$\E_{\random y|\random x}(\random y|\random x)=x\pi$ and this has
expectation $\lambda\pi$.

$\Var_{\random y|\random x}(\random y|\random x)=x\pi(1-\pi)$ which has
expectation $\lambda\pi(1-\pi)$ and that
$\E_{\random y|\random x}(\random y|\random x)=x\pi$ which has
variance $\lambda\pi^2$ so that the right hand side adds to $\lambda\pi$,
the variance of $\random y$.

\nextq We note that
$I=\int_{0}^{\infty}\exp(-\half z^2)\,dz= \int_{0}^{\infty}\exp(-\half(xy)^2)\,y\,dx$
for any $y$ (on setting $z=xy$).  Putting $z$ in place of $y$, it
follows that for any $z$
$I = \int_{0}^{\infty}\exp(-\half(zx)^2)\,z\,dx$
so that
$I^2=\left(\int_{0}^{\infty}\exp(-\half z^2)\,dz\right) \left(\int_{0}^{\infty}\exp(-\half(zx)^2)\,dx\right) =\int_{0}^{\infty}\int_{0}^{\infty} \exp\{-\half(x^2+1)z^2\}\,z\,dz\,dx.$
Now set $(1+x^2)z^2=2t$ so that $z\,dz=dt/(1+x^2)$ to get
\begin{align*}
I^2&=\int_{0}^{\infty}\int_{0}^{\infty}
\exp(-t)\,\frac{dt}{(1+x^2)}\,dx
=\left(\int_{0}^{\infty}\exp(-t)\,dt\right)
\left(\int_{0}^{\infty}\frac{dx}{(1+x^2)}\right) \\
&=\left[-\exp(-t)\right]_0^{\infty}\,
\left[\tan^{-1}x\right]_0^{\infty}
=\bigl[1\bigr]\bigl[\half\pi\bigr] \\
&=\frac{\pi}{2}
\end{align*}
and hence $I=\sqrt{\pi/2}$ so that the integral of $\phi$ from $-\infty$
to $\infty$ is 1, and hence $\phi$ \textit{is} a probability density
function.  This method is apparently due to Laplace (1812, Section 24,
pages 94--95 in the first edition).

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq $p(\pi)=\bigl(\Betafn(k+1,n-k+1)\bigr)^{-1}\pi^k(1-\pi)^{n-k}$ or
$\pi\sim\Be(k+1,n-k+1)$.

\nextq $\mean x=16.35525$, so assuming uniform prior, posterior is
$\N(16.35525,\,1/12)$.  A 90\% HDR is $16.35525\pm1.6449/\sqrt{12}$ or
$16.35525\pm0.47484$, that is, the interval $(15.88041, 16.83009)$.

\nextq $x-\theta\sim\N(0,1)$ and $\theta\sim\N(16.35525,\,1/12)$, so
$x\sim\N(16.35525,\,13/12)$.

\nextq  Assumming uniform prior, posterior is $\N(\mean x,\phi/n)$, so
take $n=k$.  If prior variance is $\phi_0$, posterior is
$\{1/\phi_0+n/\phi\}^{-1}$, so take $n$ the least integer such that
$n\geqslant(k-1)\phi/\phi_0$.

\nextq Posterior is
$k(2\pi/25)^{-\frac{1}{2}}\exp\{-\half(\theta-0.33)^2\times25\}$ for
$\theta > 0$ and 0 for $\theta < 0$.  Integrating we find $1=k\Pr(X > 0)$ where
$X\sim\N(0.33,\,1/25)$, so $k=\{1-\Phi(-1.65)\}^{-1}=1.052$.  We now seek
$\theta_1$ such that $p(\theta\leqslant\theta_1|\vect x)=0.95$ or
equivalently $k\Pr(0 < X\leqslant\theta_1)=1$ with $k$ and $X$ as before.
This results in $0.95=1.052\{\Phi(5\theta_1-1.65)-\Phi(-1.65)\}$, so
$\Phi(5\theta_1-1.65)=0.95/1.052+0.0495=0.9525$ leading to
$5\theta_1-1.65=1.67$ and so to $\theta_1=0.664$.  The required interval
is thus $[0, 0.664)$.n

\nextq From the point of view of someone starting from prior ignorance,
my beliefs are equivalent to an observation of mean $\lambda$ and
variance $\phi$ and yours to one of mean $\mu$ and variance $\psi$, so
after taking into account my beliefs such a person is able to use
Section 2.2 and conclude $\theta\sim\N(\lambda_1,\phi_1)$ where
$1/\phi_1=1/\phi+1/\psi$ and $\lambda_1/\phi_1=\lambda/\phi+\mu/\psi$.

\nextq The likelihood (after inserting a convenient constant) is
$l(\vect x|\theta)=(2\pi\phi/n)^{-\frac{1}{2}} \exp\{-\half (\mean x-\theta)^2/(\phi/n)\}.$
Hence by Bayes' Theorem, within $I_{\alpha}$
\begin{gather*}
Ac(1-\varepsilon)(2\pi\phi/n)^{-\frac{1}{2}}
\exp\{-\half(\mean x-\theta)^2/(\phi/n)\}
\leqslant p(\theta|\vect x) \\
\leqslant
Ac(1+\varepsilon)(2\pi\phi/n)^{-\frac{1}{2}}
\exp\{-\half(\mean x-\theta)^2/(\phi/n)\}
\end{gather*}
and outside $I_{\alpha}$
$0\leqslant p(\theta|\vect x)\leqslant AMc(2\pi\phi/n)^{-\frac{1}{2}} \exp\{-\half(\mean x-\theta)^2/(\phi/n)\},$
where $A$ is a constant equal to $p(\vect x)^{-1}$.  Using the right
hand inequality for the region inside $I_{\alpha}$ we get
\begin{align*}
\int_{I_{\alpha}} p(\theta|\vect x)\dtheta
& \leqslant Ac(1+\varepsilon)
\int_{I_{\alpha}} (2\pi\phi/n)^{-\frac{1}{2}}
\exp\{-\half(\mean x-\theta)^2/(\phi/n)\}\dtheta \\
& = Ac(1+\varepsilon)\int_{-\lambda_{\alpha}}^{\lambda_{\alpha}}
(2\pi)^{-\frac{1}{2}}\exp(-\half t^2)\dt,\text{\ where\ }
t=(\mean x-\theta)/\sqrt{\phi/n} \\
& = Ac(1+\varepsilon)[\Phi(\lambda_{\alpha})-\Phi(-\lambda_{\alpha})]
=Ac(1+\varepsilon)(1-\alpha)
\end{align*}
since $\Phi(-\lambda_{\alpha})=1-\Phi(\lambda_{\alpha})$.  Similarly,
the same integral exceeds $AQc(1-\varepsilon)(1-\alpha)$, and, if
$J_{\alpha}$ is the outside of $I_{\alpha}$,
$0\leqslant \int_{J_{\alpha}} p(\theta|\vect x)\dtheta \leqslant AMc\alpha.$
Combining these results we have, since
$\int_{I_{\alpha}\cup J_{\alpha}} p(\theta|\vect x)\dtheta=1$,
$Ac(1-\varepsilon)(1-\alpha)\leqslant1\leqslant Ac[(1+\varepsilon)(1-\alpha)+M\alpha],$
and hence
$\frac{1}{(1+\varepsilon)(1-\alpha)+M\alpha}\leqslant Ac\leqslant \frac{1}{(1-\varepsilon)(1-\alpha)}.$
The result now follows on remarking that the maximum value of the
exponential in $J_{\alpha}$ occurs at the end-points
$\theta\pm \lambda_{\alpha}\sqrt{\phi/n}$, where it has the value
$\exp(-\half\lambda_{\alpha}^2)$.

\nextq The likelihood is
\begin{align*}
l(\theta|\vect x) & \propto h(\theta)^n
\exp\left\{\sum t(x_i)\psi(\theta)\right\} \\
& \propto h(\theta)^n
\exp\left\{\exp\left[\log\psi(\theta)-
\log\left(1\left/\,\sum t(x_i)\right.\right)\right]\right\}.
\end{align*}
This is of the data-translated form $g(\psi(\theta)-t(\vect x))$
\textit{if} the function $h(x)\equiv1$.

\nextq Take prior $S_0\chi_{\nu}^{-2}$ where $\nu=8$ and
$S_0/(\nu-2)=100$ so $S_0=600$ and take $n=30$ and $S=13.2^2(n-1)=5052.96$.
Then (cf.\ Section 2.7) posterior is $(S_0+S)\chi_{\nu+n}^{-2}$ where
$\nu+n=38$ and $S_0+S=5652.96$.  Values of $\chi^2$ corresponding to a
90\% HDR for log $\chi_{38}^2$ are (interpolating between 35 and 40
degrees of freedom) 25.365 and 54.269 so 90\% interval is $(104,223)$.

\nextq $n=10$, $\mean x=5.032$, $S=3.05996$, $s/\sqrt{n}=0.1844$.
Posterior for mean $\theta$ is such that
$(\theta-5.032)/0.1844\sim\t_{\nu}$, so as $\t_{9,0.95}=1.833$ a 90\% HDR is
$(4.69,5.37)$.  Posterior for variance $\phi$ is $S\chi_{\nu}^{-2}$, so
as values of $\chi^2$ corresponding to a 90\% HDR for $\log\chi_9^2$ are
3.628 and 18.087 required interval is $(0.17,0.84)$.

\nextq Sufficient statistic is $\sum_{i=1}^n x_i$, or equivalently
$\mean x$.

\nextq Sufficient statistic is
$\left(\sum_{i=1}^n x_i, \prod_{i=1}^n x_i\right)$ or equivalently
$(\mean x,\check x)$ where $\check x$ is the geometric mean.

\nextq $p(\beta)\propto\beta^{-\alpha_0}\exp(-\xi/\beta)$.

\nextq If $\theta\sim\U(-\pi/2,\pi/2)$ and $x=\tan\theta$, then by the usual
change-of-variable rule
$p(x)=\pi^{-1}|\!\dtheta/\dx|=\pi^{-1}\left(1+x^2\right)^{-1}$,
and similarly if $\theta\sim\U(\pi/2,3\pi/2)$.  The result follows.

\nextq Straightforwardly
\begin{align*}
p(\vect x|\vect \pi) & = \frac{n!}{x!\,y!\,z!}\pi^x\rho^y\sigma^z     \\
& = \left(\frac{n!}{x!\,y!\,(n-x-y)!}\right)
\times\exp\{n\log(1-\pi-\rho)\}        \\
& = g(x,y)\times h(\pi,\rho)\times
\exp[t(x,y)\psi(\pi,\rho)+u(x,y)\chi(\pi,\rho)]
\end{align*}

\nextq $\nu_1=\nu_0+n=104$; $n_1=n_0+n=101$;
$\theta_1=(n_0\theta_0+n\mean x)/n_1=88.96$; $S=(n-1)s^2=2970$;
$S_1=S_0+S+(n_0^{-1}+n^{-1})^{-1}(\theta_0-\mean x)^2=3336$;
$s_1/\sqrt{n_1}=\sqrt{3336/(101\times104)}=0.56$.  It follows that
\textit{a posteriori}
$\frac{\mu-\theta_1}{s_1/\sqrt{n_1}}\sim\t_{\nu_1},\qquad\ \phi\sim S_1\chi_{\nu_1}^{-2}.$
For prior, find 75\% point of $\t_4$ from, e.g., Neave's Table 3.1 as 0.741.
For posterior, as degrees of freedom are large, can approximate $\t$ by
normal, noting that the 75\% point of the standard normal distribution is
0.6745.  Hence a 50\% prior HDR for $\mu$ is
$85\pm0.741\sqrt{(350/4\times1)}$, that is
$(75.6,94.4)$, while a 50\% posterior HDR for $\mu$ is
$88.96\pm0.6745\times0.56$, that is, $(88.58, 89.34)$.

\nextq With $p_1(\theta)$ being $\N(0,1)$ we see that $p_1(\theta|x)$
is $\N(2,2)$, and with $p_2(\theta)$ being $\N(1,1)$ we see that
$p_2(\theta|x)$ is $\N(3,1)$.  As
\begin{align*}
\int p(x|\theta)p_1(\theta)\dtheta
& = \int \frac{1}{2\pi}\exp\{-\half(x-\theta)^2-\half\theta^2\}\dtheta \\
& = \frac{\sqrt{\half}}{\sqrt{2\pi}}\exp\{-\quarter x^2\} \\
\exp\{-\half(\theta-\half x)^2/\half\}\dtheta \\
& = \frac{\sqrt{\half}}{\sqrt{2\pi}}\exp\{-\quarter x^2\}
=\frac{1}{2\sqrt{\pi}}\exp\{-2\}
\end{align*}
and similarly
\begin{align*}
\int p(x|\theta)p_1(\theta)\dtheta
& =\int \frac{1}{2\pi}\exp\{-\half(x-\theta)^2-\half(\theta-1)^2\}\dtheta \\
& =\frac{\sqrt{\half}}{\sqrt{2\pi}}\exp\{-\quarter (x-1)^2\} \\
\exp\{-\half(\theta-\half (x+1))^2/\half\}\dtheta \\
& =\frac{\sqrt{\half}}{\sqrt{2\pi}}\exp\{-\quarter (x-1)^2\}
=\frac{1}{2\sqrt{\pi}}\exp\{-\quarter\}
\end{align*}
so that just as in Section 2.13 we see that the posterior is an
$\alpha'$ to $\beta'$ mixture of $\N(2,1)$ and $\N(3,1)$ where
$\alpha'\propto\twothirds\text{e}^{-2}=0.09$ and
$\beta'\propto\third\text{e}^{-1/4}=0.26$, so that $\alpha'=0.26$
and $\beta'=0.74$.  It follows that
$\Pr(\theta > 1)=0.26\times0.1587+0.74\times0.0228=0.058.$

\nextq Elementary manipulation gives
\begin{align*}
& n\mean X^2+n_0\theta_0^2 -
(n+n_0)\left(\frac{n\mean X+n_0\theta_0}{n+n_0}\right)^2          \\
& = \frac{1}{n+n_0}[\{n(n+n_0)-n^2\}\mean X^2+
\{n_0(n+n_0)-n_o^2\}\theta_0^2-2(nn_0)\mean X\theta] \\
& =\frac{nn_0}{n+n_0}[\mean X^2+\theta_0^2-2\mean X\theta_0]
= (n_0^{-1}+n^{-1})^{-1}(\mean X-\theta)^2.
\end{align*}

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq Using Bayes postulate $p(\pi)=1$ for $0\leqslant\pi\leqslant1$ we
get a posterior $(n+1)\pi^n$ which has mean $(n+1)/(n+2)$.

\nextq From Table B.5, $\underline F_{\,40,24}=0.55$ and
$\overline F_{40,24}=1.87$, so for $\Be(20,12)$ take lower limit
$20\times0.55/(12+20\times0.55)=0.48$ and upper limit
$20\times1.87/(12+20\times1.87)=0.76$ so 90\% HDR $(0.48.0.76)$.
Similarly by interpolation $\underline F_{\,41,25}=0.56$ and
$\overline F_{41,25}=1.85$, so for $\Be(20.5,12.5)$ quote $(0.48.0.75)$.
Finally by interpolation $\underline F_{\,42,26}=0.56$ and
$\overline F_{42,26}=1.83$, so for $\Be(21,13)$ quote $(0.47.0.75)$.
It does not seem to make much difference whether we use a $\Be(0,0)$, a
$\Be(\half,\half)$ or a $\Be(1,1)$ prior.

\nextq Take $\alpha/(\alpha+\beta)=1/3$ so $\beta=2\alpha$ and
$\Var\pi=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} =\frac{2}{3^2(3\alpha+1)}$
so $\alpha=55/27\cong2$ and $\beta=4$.  Posterior is then
$\Be(2+8,4+12)$, that is $\Be(10,16)$.  The 95\% values for $F_{32,20}$
are 0.45 and 2.30 by interpolation, so those for $F_{20,32}$ are 0.43
and 2.22.  An appropriate interval for $\pi$ is from
$10\times0.43/(16+10\times0.43$ to $10\times2.22/(16+10\times2.22)$,
that is $(0.21,0.58)$.

\nextq Take $\alpha/(\alpha+\beta)=0.4$ and $\alpha+\beta=12$, so
approximately $\alpha=5$, $\beta=7$.  Posterior is then
$\Be(5+12,7+13)$, that is, $\Be(17,20)$.

\nextq Take beta prior with $\alpha/(\alpha+\beta)=\third$ and
$\alpha+\beta=\quarter11=2.75$.  It is convenient to take integer
$\alpha$ and $\beta$, so take $\alpha=1$ and $\beta=2$, giving
$\alpha/(\alpha+\beta)=\third$ and $(\alpha+\beta)/11=0.273$.  Variance
of prior is $2/36=0.0555$ so standard deviation is 0.236.  Data is such
that $n=11$ and $X=3$, so posterior is $\Be(4,10)$.  From tables, values
of $\F$ corresponding to a 50\% HDR for $\log\F_{20,8}$ are $\underline F=0.67$ and $\overline F=1.52$.  It follows that the appropriate
interval of values of $\F_{8,20}$ is $(1/\overline F, 1/\underline F)$,
that is $(0.66, 1.49)$.  Consequently and appropriate interval for the
proportion $\pi$ required is
$4\times0.66/(10+4\times0.66)\leqslant\pi\leqslant4\times1.49/(10+4\times1.49)$,
that is $(0.21, 0.37)$.  The posterior mean is $4/(4+10)=0.29$, the
posterior mode is $(4-1)/(4+10-2)=0.25$ and using the relationship
$\text{median}\cong(2\,\text{mean}+\text{mode})/3$ the posterior mode is
approximately 0.28.  The actual overall proportion $86/433=0.20$ is
consistent with the above.  [Data from the York A.U.T.\ Membership List
for 1990/91.]

\nextq By Appendix A.13, $\E x=n(1-\pi)/\pi$ and $\Var x=n(1-\pi)/\pi^2$, so $g'(\E x)=\half n^{-1}[(1-\pi)/\pi^2]^{-\frac{1}{2}}$,
so $\Var x=1/4n$ (cf.\ Section 1.5).  By analogy with the argument that an
arc-sine prior for the binomial distribution is approximately data-translated,
this suggests a prior uniform in $\sinh^{-1}\sqrt{\pi}$ so with density
$\half\pi^{-\frac{1}{2}}(1+\pi)^{-\frac{1}{2}}$.  But note Jeffreys' Rule
suggests $\Be(0,\half)$ as remarked in Section 7.4.

\nextq As this is a rare event, we use the Poisson distribution.  We
have $n=280$, $T=\sum x_i=196$.  Posterior is $S_1^{-1}\chi_{\nu^{\,\prime}}^2$
where $S_1=S_0+2n$, $\nu^{\,\prime}=\nu+2T$.  For reference prior take $\nu=1$,
$S_0=0$ we get a posterior which is $560^{-1}\chi_{393}^2$.  Using the
approximation in Appendix A.3, we find that a 95\% interval is
$\half(\sqrt{785} \pm 1.96)^2/560$, that is, $(0.61,0.80)$.

\nextq Prior is such that $\nu/S_0=0.66$, $2\nu/S_0^2=0.115^2$, so
$\nu=66$, $S_0=100$, so posterior has $S_1=660$, $\nu^{\,\prime}=458$.
This gives a posterior 95\% interval $\half(\sqrt{915}\pm1.96)^2/660$,
that is, $(0.61,0.79)$.

\nextq $\partial p(x|\alpha)/\partial\alpha=3/2\alpha-\half x^2$ and
$\partial^2 p(x|\alpha)/\partial\alpha^2=-3/2\alpha^2$, so
$I(\alpha|x)=3/2\alpha^2$ and we take $p(\alpha)\propto1/\alpha$ or
equivalently a uniform prior in $\psi=\log\alpha$.

\nextq We have $\partial^2L/\partial\pi^2=-x/\pi^2-z/(1-\pi-\rho)^2$,
etc., so
$I(\pi,\rho\,|\,x,y,z)=\left(\begin{array}{cc} n/\pi+n/(1-\pi-\rho) & n/(1-\pi-\rho) \\ n/(1-\pi-\rho) & n/\rho+n/(1-\pi-\rho) \end{array}\right)$
and so after a little manipulation
$\det I(\pi,\rho\,|\,x,y,z)=n\{\pi\rho(1-\pi-\rho)\}^{-1}$, suggesting a prior
$p(\pi,\rho)\propto\pi^{-\frac{1}{2}}\rho^{-\frac{1}{2}} (1-\pi-\rho)^{-\frac{1}{2}}$
which is evidently related to the arc-sine distribution.

\nextq $\partial p(x|\gamma)/\partial\gamma=1/\gamma+\log\xi-\log x$ and
$\partial^2 p(x|\gamma)/\partial\gamma^2=-1/\gamma^2$, so
$I(\gamma|x)=1/\gamma^2$ and we take $p(\gamma)\propto1/\gamma$ or
equivalently a uniform prior in $\psi=\log\gamma$.

\nextq Using Appendix A.16, coefficient of variation is
$\sqrt{}\{2/(\gamma+1)(\gamma-2)\}$.  This is less than 0.01 if
$\half(\gamma+1)(\gamma-2)>1/0.01^2$ or $\gamma^2-\gamma-20,002 > 0$,
so if $\gamma > \half(1+\sqrt{80,009})=141.9$.  Taking the reference
prior $p(\alpha,\beta)\propto(\beta-\alpha)^{-2}$, that is,
$\Pabb(-\infty,\infty,-1)$ (cf.\ Section 3.6), we need $\gamma'=n-1>141.9$,
that is, $n$ at least 143.

\nextq Take prior $p(\theta)=(d-1)\theta^{-d}$ for
$\theta_0 < \theta < \infty$.
Then posterior is $p(\theta|\vect x)=(d'-1)\theta^{-d'}$ for
$\theta_1 < \theta < \infty$ where $d'=d+n(c+1)$ and
$\theta_1=\max\{\theta_0,M\}$ where $M=\max\{x_i\}$.

\nextq Prior $p(\nu)\propto1/\nu$ as before, but likelihood is now
$p(71,100\,|\,\nu)=1/\nu^2$ for $\nu\geqslant100$, so posterior is
approximately
$p(\nu\,|\,71,100)\propto \nu^{-3}/\left(\sum_{\mu\geqslant100}\mu^{-3}\right).$
Approximating sums by integrals, the posterior probability
$\Pr(\nu\geqslant \nu_0\,|\,71,100)=100^2/\nu_0^2$,
and in particular the posterior median is $100\sqrt{2}$ or about 140.

\nextq We have $p(\theta)=1/\theta$ and setting $\psi=\log\theta$ we get
$p(\psi)=p(\theta)\,|\!\dtheta/\!\dpsi|=1$.  Thus we expect all pages to
be more or less equally dirty.

\nextq The group operation is $(x,y)\mapsto x+y\mod2\pi$ and the
appropriate Haar measure is Lebesgue measure on the circle, that is, arc
length around the circle.

\nextq Table of $c^2\{c^2+(x_1-\mu)^2\}^{-1}\{c^2+(x_2-\mu)^2\}^{-1}$:
$\begin{array}{cccccc} \mu\backslash c \ & 1 & 2 & 3 & 4 & 4 \\ 0 & 0.00 & 0.01 & 0.01 & 0.01 & 0.01 \\ 2 & 0.06 & 0.05 & 0.04 & 0.03 & 0.02 \\ 4 & 0.04 & 0.06 & 0.05 & 0.04 & 0.03 \\ 6 & 0.06 & 0.05 & 0.04 & 0.03 & 0.02 \\ 8 & 0.00 & 0.01 & 0.01 & 0.01 & 0.01 \end{array}$
Integrating over $c$ using Simpson's Rule (and ignoring the constant)
for the above values of $\mu$ we get 0.11, 0.48, 0.57, 0.48 and 0.11
respectively.  Integrating again we get for intervals as shown:
$\begin{array}{cccccc} (-1,1) & (1,3) & (3,5) & (5,7) & (7,9) & \text{Total} \\ 0.92 & 2.60 & 3.24 & 2.60 & 0.92 & 10.28 \end{array}$
so the required posterior probability is $3.24/10.28=0.31$.

\nextq First part follows as sum of concave functions is concave and a
concave function has a unique maximum.  For the example, note that
\begin{align*}
p(x|\theta) & =\exp(\theta-x)/\{1+\exp(\theta-x)\}^2
\ \ (-\infty < x < \infty) \\
& =\quarter\sech^2\half(\theta-x)\ \ (-\infty < x < \infty)
\end{align*}
(which is symmetrical about $x=\theta$), so that the log-likelihood is
$L(\theta|x)=\theta-x-2\log\{1+\exp(\theta-x)\}.$
Hence
\begin{align*}
L'(\theta|x)  & = 1 - 2\exp(\theta-x)/\{1+\exp(\theta-x)\}  \\
& = \{1-\exp(\theta-x)\}/\{1+\exp(\theta-x)\} \\
& = 1 - 2/\{1+\exp(x-\theta)\}                \\
L''(\theta|x) & =-2\exp(x-\theta)/\{1+\exp(x-\theta)\}^2    \\
& =-2\exp(\theta-x)/\{1+\exp(\theta-x)\}^2.
\end{align*}
As this is clearly always negative, the log-likelihood is concave.  Also
\begin{align*}
L'(\theta|x)/L''(\theta|x) & =\half\{\exp(\theta-x)-\exp(x-\theta)\}  \\
I(\theta|x) & = 2\int_{-\infty}^{\infty}
\exp(2(\theta-x))/\{1+\exp(\theta-x)\}^4\dx                           \\
& = (1/8) \int_{-\infty}^{\infty} \sech^4(\theta-x) \dx               \\
& = (1/8) \int_{-\infty}^{\infty} \sech^4 y \dy                       \\
& = (1/24) [\sinh y \sech^3 y + 2 \tanh y]_{-\infty}^{\infty}=1/6.
\end{align*}
(The integration can be checked by differentiation of the result).  Now
proceed as in Section 3.10.

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq $1-(1-p_0)=p_0=\left[1+(1-\pi_0)\pi_0^{-1}B^{-1}\right]^{-1} =1-(1-\pi_0)\pi_0^{-1}B^{-1}+(1-\pi_0)^2\pi_0^{-2}B^{-2}$.
Result then follows on noting $\pi_0^{-1}=\left\{1-(1-\pi_0)\right\}^{-1} \cong1+(1-\pi_0)$.

\nextq Substituting in the formulae in the text we get
\begin{align*}
\phi_1 & = (0.9^{-2}+1.8^{-2})^{-1}=0.648=0.80^2; \\
\theta_1 & = 0.648(93.3/0.9^2+93.0/1.8^2)=93.2; \\
\pi_o & = \Phi((93.0-93.3)/0.9)=\Phi(-0.333)=0.3696;\ \
\pi_0/(1-\pi_0)=0.59; \\
p_0 & = \Phi((93.0-93.2)/0.8)=\Phi(-0.25)=0.4013;\ \
p_0/(1-p_0)=0.67; \\
B & = 0.67/0.59=1.14.
\end{align*}

\nextq $n=12$, $\mean x=118.58$, $S=12969$, $s/\sqrt{n}=9.91$,
$(\mean x-100)/(s/\sqrt{n})=1.875$.  Taking a normal approximation this
gives a $P$-value of $1-\Phi(1.875)=0.0303$.

\nextq $n=300$, $n\pi=900/16=56.25$, $n\pi(1-\pi)=45.70$,
$(56-56.25)/\sqrt{45.70}=-0.037$, so certainly not significant.

\nextq Posterior with reference prior is $S_1^{-1}\chi_{\nu^{\,\prime}}^2$
with $S_1=12$ and $\nu^{\,\prime}=31$.  Values of $\chi^2$ corresponding to
a 90\% HDR for $\log\chi_{31}^2$ are (by interpolation) 19.741 and 45.898,
so a 90\% posterior HDR is from 1.65 to 3.82 which includes 3.  So not
appropriate to reject null hypothesis.

\nextq If $k=0.048$ then $\exp(2k)=1.101=1/0.908$, so take $\theta$ within
$\pm\varepsilon=k\sqrt{(\phi/n)}\z=0.048\sqrt{(\phi/10)}/2.5=0.006\sqrt{\phi}$.

\nextq Using standard power series expansions
\begin{align*}
B & = (1+1/\lambda)^{\frac{1}{2}}\exp[-\half(1+\lambda)^{-1}] \\
& = \lambda^{-\frac{1}{2}}(1+\lambda)^{\frac{1}{2}}\exp(-\half z^2)
\exp[\half\lambda z^2(1+\lambda)^{-1}] \\
& = \lambda^{\frac{1}{2}}(1+\half\lambda+\dots)\exp(-\half z^2)
[1+\half z^2(1+\lambda)^{-1}+\dots] \\
& = \lambda^{\frac{1}{2}}\exp(-\half z^2)(1+\half\lambda(z^2+1)+\dots).
\end{align*}

\nextq Likelihood ratio is
\begin{align*}
\frac
{\{2\pi(\phi+\varepsilon)\}^{-n/2}\exp[-\half\sum x_i^2/(\phi+\varepsilon)]}
{\{2\pi(\phi-\varepsilon)\}^{-n/2}\exp[-\half\sum x_i^2/(\phi-\varepsilon)]}
& \cong\left(1+\frac{\varepsilon}{\phi}\right)^{-n}
\exp\left[\varepsilon\sum x_i^2/\phi^2\right] \\
& \cong\exp\left[\varepsilon\left(\sum x_i^2-n\phi\right)/\phi^2\right] \\
& \cong\exp\left[\frac{n\varepsilon}{\phi}
\left(\frac{\sum x_i^2/n}{\phi}-1\right)\right].
\end{align*}

\nextq $\partial p_1(\mean x)/\partial\psi=\{-\half(\psi+\phi/n)^{-1}+ \half(\mean x-\theta)^2/(\psi+\phi/n)^2\}p_1(\mean x)$ which vanishes
if $\psi+\phi/n=(\mean x-\theta)^2=z^2\phi/n$ so if
$\psi=(z^2-1)\phi/n$.  Hence
$p_1(\mean x)\leqslant(2\pi\phi/n)^{-\frac{1}{2}}z^{-1}\exp(-\half)$.
Consequently $B=(2\pi\phi/n)^{-\frac{1}{2}}\exp(-\half z^2)/p_1(\mean x) \geqslant\sqrt{\text{e}}\,z\exp(-\half z^2)$.  For last part use
$p_0=\left[1+(\pi_1/\pi_0)B^{-1}\right]$.

\nextq Posterior probability is a minimum when $B$ is a minimum, hence
when $2\log B$ is a minimum, and
$\d\,(2\log B)/\d n= \{1+n\}^{-1}+z^2\{1+1/n\}^{-2}(-1/n^2)$
which vanishes when $n=z^2-1$.  Since $z^2-1$ is not necessarily an
integer, this answer is only approximate.

\nextq Test against $\B(7324,0.75)$ with mean 5493 and variance 1373, so
$z=(5474-5493)/\sqrt{1373}=-0.51$.  Not significant so theory confirmed.

\nextq Likelihood ratio is
\begin{align*}
& \frac{(2\pi2\phi)^{-\frac{1}{2}}\exp[-\half u^2/(2\phi)]
(2\pi\psi)^{-\frac{1}{2}}\exp[-\half(z-\mu)/\psi]}
{(2\pi2\psi)^{-\frac{1}{2}}\exp[-\half u^2/(2\psi)]
(2\pi\psi/2)^{-\frac{1}{2}}\exp[-\half(z-\mu)/(\psi/2)]} \\
& = (\psi/2\phi)^{\frac{1}{2}}
\exp[-\half u^2(1//2\phi-1/2\psi)+\half(z-\mu)^2/\psi]  \\
& \cong (\psi/2\phi)^{\frac{1}{2}}
\exp[-\half u^2/(2\phi)+\half(z-\mu)^2/\psi]
\end{align*}
With $\sqrt{(\psi/\phi)}=100$, $u=2\times\sqrt{(2\phi)}$ and $z=\mu$, we
get $B=\left(100/\sqrt{2}\right)\exp(-2)=9.57$, although 2 standard deviation
is beyond 1.96, the 5\% level.

\nextq $p_1(\mean x)=\int \rho_1(\theta) p(\mean x|\theta)\dtheta$ which is
$1/\tau$ times the integral between $\mu+\tau/2$ and $\mu-\tau/2$ of an
$\N(\theta,\phi/n)$ density and so as $\tau\gg\phi/n$ is nearly the
whole integral.  Hence $p_1(\mean x)\cong1/\tau$, from which it easily follows
that $B=(2\pi\phi/n\tau^2)^{-\frac{1}{2}}\exp(-\half z^2)$.  In Section 4.5 we
found $B=(1+n\psi/\phi)^{\frac{1}{2}}\exp[-\half z^2(1+\phi/n\psi)^{-1}]$,
which for large $n$ is about
$B=(\phi/n\psi)^{-\frac{1}{2}}\exp(-\half z^2)$.  This agrees with the
first form found here if $\tau^2=2\pi\phi$.  As the variance of a
uniform distribution on $(\mu-\tau/2,\mu+\tau/2)$ is $\tau^2/12$, this
may be better expressed as $\tau^2/12=(\pi/6)\phi=0.52\phi$.

\nextq Jeffreys considers the case where both the mean $\theta$ and the
variance $\phi$ are unknown, and wishes to test $\Hyp_0:\theta=0$ versus
$\Hyp_1: \theta\neq0$ using the conventional choice of prior odds
$\pi_0/\pi_1=1$, so that $B=p_0/p_1$.  Under both hypotheses he uses the
standard conventional prior for $\phi$, namely $p(\phi)\propto1/\phi$.
He assumes that the prior for $\theta$ is dependent on
$\sigma=\sqrt{\phi}$ as a scale factor.  Thus if $\gamma=\theta/\sigma$
he assumes that $\pi(\gamma,\theta)=p(\gamma)p(\theta)$ so that $\gamma$
and $\theta$ are independent.  He then assumes that

\blankline(i) if the sample size $n=1$ then $B=1$, and

\blankline(ii) if the sample size $n\geqslant2$ and
$S=\sum(X_i-\mean X)^2=0$, then $B=0$, that is, $p_1=1$.

\blankline
From (i) he deduces that $p(\gamma)$ must be an even function with
integral 1, while he shows that (ii) is equivalent to the condition
$\int_0^{\infty} p(\gamma)\gamma^{n-2}\dgamma=\infty.$
He then shows that the simplest function satisfying these conditions is
$p(\gamma)=\pi^{-1}(1+\gamma^2)^{-1}$ from which it follows that
$p(\theta)=p(\gamma)|\!\dgamma/\!\dtheta|= \pi^{-1}\sigma(\sigma^2+\theta^2)^{-1}$.
Putting $\sigma=\sqrt{\phi}$ and generalizing to the case where $\Hyp_0$
is that $\theta=\theta_0$ we get the distribution in the question.

There is some arbitrariness in Jeffreys' simplest function''; if
instead he had taken $p(\gamma)=\pi^{-1}\kappa(\kappa^2+\gamma^2)^{-1}$
he would have ended up with
$p(\theta)=\pi^{-1}\tau(\tau^2+(\theta-\theta_0)^2)^{-1}$ where
$\tau=\kappa\sigma$.  However, even after this generalization, the
argument is not overwhelmingly convincing.

\nextq \!\!(a)\quad Maximum likelihood estimator $\est\theta$ of $\theta$
is $x/n$, so
$B\geqslant\left(\frac{\theta_0}{\est\theta}\right)^x \left(\frac{1-\theta_0}{1-\est\theta}\right)^{n-x};\qquad\ p_0\geqslant\left[1+\frac{1-\pi_0}{\pi_0} \left(\frac{\est\theta}{\theta_0}\right)^x \left(\frac{1-\est\theta}{1-\theta_0}\right)^{n-x}\right]^{-1}.$

\blankline (b)\quad From tables (e.g.\ D.~V.~Lindley and W.~F.~Scott,
\textit{New Cambridge Elementary Statistical Tables}, Cambridge:
University Press 1995 [1st edn (1984)], Table 1, or H.~R.~Neave,
\textit{Statistics Tables for mathematicians, engineers, economists
and the behavioural and management sciences}, London: George Allen
\& Unwin (1978), Table 1.1) the probability of a binomial observation
$\leqslant14$ from $\B(20,0.5)$ is 0.9793, so the appropriate (two-tailed)
$P$-value is $2(1-0.9793)=0.0414\cong1/24$. The maximum likelihood
estimate is $\est\theta=15/20=0.75$, so the lower bound on $B$ is
$(0.5/0.75)^{15}(0.5/0.25)^5=2^{20}/3^{15}=0.0731$, implying a lower
bound on $p_0$ of 0.0681 or just over 1/15.

\nextq \!\!(a)\quad $n=12$, $\nu=11$, $t=\mean x/(s/\sqrt{n})=1.2/\sqrt{(1.1/12)}=3.96$ and if we take $k=1$ the
Bayes factor $B$ is
$\frac{(1+t^2/\nu)^{-(\nu+1)/2}} {(1+nk)^{-\frac{1}{2}}(1+t^2(1+nk)^{-1}/\nu)^{-(\nu+1)/2}} =\frac{0.004910}{(0.2774)(0.5356)} =0.033.$

\blankline (b)\quad $z=\mean x/(s/\sqrt{n})=1.2/\sqrt{12}=4.16$ and
(taking $\psi=\phi$ as usual)
\begin{align*}
B & = (1+n)^{\frac{1}{2}}\exp\left[-\half z^2(1+1/n)^{-1}\right] \\
& = (1+12)^{\frac{1}{2}}\exp\left[-\half(4.16)^2(1+1/12)^{-1}\right]=0.001.
\end{align*}

\nextq Two-tailed $P$-value is 0.0432 (cf.\ D.~V.~Lindley and W.~F.~Scott,
\textit{New Cambridge Elementary Statistical Tables}, Cambridge:
University Press 1995 [1st edn (1984)], Table 9), while Bayes factor is
$\frac{(1+t^2/\nu)^{-(\nu+1)/2}} {(1+nk)^{-\frac{1}{2}}(1+t^2(1+nk)^{-1}/\nu)^{-(\nu+1)/2}} =\frac{0.08712}{(0.3162)(0.7313)} =0.377$
so $F=1/B=2.65$.  Range $(1/30P,3/10P)$ is $(0.77, 6.94)$, so $F$ \textit{is}
inside it and we do not need to think again''.

\nextq For $P$-value 0.1 think again if $F$ not in $(\third,3)$.  As
$p_0=[1+F]^{-1}$ this means if $p_0$ not in $(0.250,0.750)$, so if $n=1000$.
Similarly if $P$-value 0.05, if $p_0$ not in $(0.143,0.600)$, so if
$n=1000$ (and the case $n=100$ is marginal); if $P$-value 0.01, if $p_0$
not in $(0.032,0.231)$, so if $n=100$ or $n=1000$; if $P$-value 0.001,
if $p_0$ not in $(0.003,0.029)$, so if $n=50$, $n=100$ or $n=1000$.

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq Mean difference $\mean w=0.05\dot{3}$, $S=0.0498$, $s=0.0789$.
Assuming a standard reference prior for $\theta$ and a variance known
equal to $0.0789^2$, the posterior distribution of the effect $\theta$
of Analyst $A$ over Analyst $B$ is $\N(0.05\dot{3},0.0789^2/9)$ leading
to a 90\% interval $0.05\dot{3}\pm1.6449\times0.0789/3$, that is,
$(0.010,0.097)$.  If the variance is not assumed known then the normal
distribution should be replaced by $\t_8$.

\nextq If variance assumed known, $z=0.05\dot{3}/(0.0789/3)=2.028$ so
with $\psi=\phi$
\begin{align*}
B & =(1+n)^{\frac{1}{2}}\exp[-\half z^2(1+1/n)^{-1}] \\
& =(1+9)^{\frac{1}{2}}\exp[-\half(2.028)^2(1+\ninth)^{-1}]
=0.497
\end{align*}
and $p_0=\left[1+0.497^{-1}\right]^{-1}=0.33$ with the usual assumptions.
If the variance is not assumed known,
$B=\frac{\{1+2.028^2/8\}^{-9/2}} {10^{-\frac{1}{2}}\{1+2.028^2{10}^{-1}/8\}^{-9/2}} =\frac{0.1546}{(0.3162)(0.7980)}=0.613$
and $p_0=\left[1+0.613^{-1}\right]^{-1}=0.38$.

\nextq We know $\sum(x_i+y_i)\sim\N(2\theta,2\sum\phi_i)$ and
$\sum(x_i-y_i)$ is independently $\N(0,2\sum\phi_i)$, so
$\sum(x_i-y_i)^2/2\sum\phi_i\sim\chi_n^2$
and hence if $\theta=0$ then
$\sum(x_i+y_i)/\sqrt{\sum(x_i-y_i)^2}\sim\t_n.$
Hence test whether $\theta=0$.  If $\theta\ne0$, it can be estimated
as $\half\sum(x_i+y_i)$.

\nextq In that case
\begin{align*}
B & = (1+440.5/99)^{\frac{1}{2}}\exp[-\half(1.91)^2\{1+99/440.5\}^{-1}] \\
& = 5.4495^{\frac{1}{2}}\exp(-1.4893)=0.53.
\end{align*}
If the prior probability of the null hypothesis is taken as
$\pi_0=\half$, then this gives a posterior probability of
$p_0=(1+0.53^{-1})^{-1}=0.35$.

\nextq $m=9$, $n=12$, $\mean x=12.42$, $\mean y=12.27$, $s_x=0.1054$,
$s_y=0.0989$.

\blankline (a)\quad Posterior of $\delta$ is $\N(12.42-12.27,0.1^2(1/9+1/12))$
so 90\% interval is $0.15\pm1.6449\times\sqrt{0.00194}$, that is,
$(0.077,0.223)$.

\blankline (b)\quad $S=8\times0.1054^2+11\times0.0989^2=0.1965$,
$s=\sqrt{0.1965/19}=0.102$, so $s\sqrt{\phantom{|}}(m^{-1}+n^{-1})=0.045$,
so from tables of $\t_{19}$ a 90\% interval is $0.15\pm1.729\times0.045$,
that is $(0.072,0.228)$.

\nextq With independent priors uniform in $\lambda$, $\mu$ and
$\log\phi$, that is, $p(\lambda,\mu,\phi)\propto1/\phi$,
\begin{align*}
p(\lambda,\mu,\phi\, & |\,\vect x,\vect y) \propto p(\lambda,\mu,\phi)\,
p(\vect x|\lambda,\phi)\,p(\vect y|\mu,\phi)                       \\
& \propto (1/\phi)(2\pi\phi)^{-(m+n)/2}
\exp\left[-\half
\left\{\sum(x_i-\lambda)^2+\half\sum(y_i-\mu)^2\right\}/\phi\right]  \\
& \propto \phi^{-(m+n)/2-1}
\exp[-\half\{S_x+m(\mean x-\lambda)^2+
\half S_y+\half n(\mean y-\mu)^2\}/\phi]
\end{align*}
Writing $S=S_x+\half S_y$ and $\nu=m+n-2$ we get
\begin{align*}
p(\lambda,\mu,\phi|\vect x,\vect y) & \propto \phi^{-\nu/2-1}
\exp[-\half S/\phi]\,\,
(2\pi\phi/m)^{-\frac{1}{2}}\exp[-\half m(\lambda-\mean x)^2)/2\phi]      \\
\exp[-\half m(\mu-\mean y)^2)/2\phi] \\
& \propto p(\phi|S)\, p(\lambda|\phi,\mean x)\, p(\mu|\phi,\mean y)
\end{align*}
where
$\begin{array}{ll} p(\phi\,|\,S) & \text{is an S\chi_{\nu}^{-2} density,} \\ p(\lambda\,|\,\phi,\mean x) & \text{is an \N(\mean x,\phi/m) density,} \\ p(\mu\,|\,\phi,\mean x) & \text{is an \N(\mean y,2\phi/m) density,} \end{array}$
It follows that, for given $\phi$, the parameters $\lambda$ and $\mu$
have independent normal distributions, and so that the joint density of
$\delta=\lambda-\mu$ and $\phi$ is
$p(\delta,\phi|\vect x,\vect y) = p(\phi|S)\, p(\delta|\mean x-\mean y, \phi)$
where $p(\delta\,|\,\mean x-\mean y, \phi)$ is an $\N(\mean x-\mean y, \ \phi(m^{-1}+2n^{-1})$ density.  This variance can now be integrated
out just as in Sections 2.12 and 5.2, giving a very similar conclusion,
that is, that if
$t=\frac{\delta-(\mean x-\mean y)}{s\{m^{-1}+2n^{-1}\}^{\frac{1}{2}}}$
where $s^2=S/\nu$, then $t\sim\t_{\nu}$.

\nextq We find that
$S_1=S_0+S_x+S_y+m_0\lambda_0^2+m\mean x^2+ n_0\mu_0^2+n\mean y^2-m_1\lambda_1^2-n_1\mu_1^2$
and then proceed as in Exercise 18 on Chapter 2 to show that
$S_1=S_0+S_x+S_y+(m_0^{-1}+m^{-1})^{-1}(\mean x-\lambda_0)^2+ (n_0^{-1}+n^{-1})^{-1}(\mean y-\mu_0)^2.$

\nextq $m=10$, $n=9$, $\mean x=22.2$, $\mean y=23.1$, $s_x=1.253$,
$s_y=0.650$.  Consequently $\sqrt{\phantom{|}}(s_x^2/m+s_y^2/n)=0.452$ and
$\tan\theta==(s_y/\sqrt{n})/(s_x/\sqrt{m})=0.547$, so
$\theta\cong30^{\circ}$.  Interpolating in Table
B.1 with $\nu_1=8$ and $\nu_2=9$ the 90\% point of the Behrens'
distribution is 1.42, so a 90\% HDR is $22.2-23.1\pm1.42\times0.452$,
that is, $(-1.54,-0.26)$.

\nextq Evidently $f_1=(m-1)/(m-3)$ and
$f_2=\frac{(m-1)^2}{(m-3)^2(m-5)}(\sin^4\theta+\cos^4\theta).$
Also $1=(\sin^2\theta+\cos^2\theta)^2=\sin^4\theta+\cos^4\theta+ 2\sin^2\theta\cos^2\theta$, so that $\sin^4\theta+\cos^4\theta= 1-\sin^2 2\theta=\cos^2 2\theta$.  The result follows.

\nextq As $T_x\sim\t_{\nu(x)}$ and $T_y\sim\t_{\nu(y)}$ we have
$p(T_x,T_y|\vect x,\vect y)\propto (1+T_x^2/\nu(x))^{-(\nu(x)+1)/2} (1+T_y^2/\nu(y))^{-(\nu(y)+1)/2}$
Jacobian is trivial, and the result follows (cf.\ Appendix A.18).

\nextq Set $y=\nu_1x/(\nu_2+\nu_1x)$ so that $1-y=1/(\nu_2+\nu_1x)$ and
$\!\dy/\!\dx=-\nu_1/(\nu_2+\nu_1x)^2$.  Then using Appendix A.19 for the
density of $x$
$p(y)=p(x)\left|\frac{\!\dx}{\!\dy}\right| \propto \frac{x^{\nu_1/2-1}}{(\nu_2+\nu_1x)^{(\nu_1+\nu_2)/2-2}} \propto y^{\nu_1/2-1}(1-y)^{\nu_2/2-1}$
from which the result follows.

\nextq $k=s_1^2/s_2^2=6.77$, so
$\eta^{-1}=k/\kappa=6.77/\kappa\sim \F_{24,14}$.  A 95\% interval for
$\F_{24,14}$ from Table B.5 is $(0.40,2.73)$ so one for $\kappa$ is
$(2.5,16.9)$.

\nextq $k=3$ with $\nu_x=\nu_y=9$.  A an interval corresponding to a
99\% HDR for $\log\F$ for $\F_{9,9}$ is $(0.15,6.54)$, so a 99\%
interval for $\sqrt{\kappa}$ is from $\sqrt{3\times0.15}$ to
$\sqrt{3\times6.54}$, that is, $(0.67,4.43)$.

\nextq Take $\alpha_0+\beta_0=6$, $\alpha_0/(\alpha_0+\beta_0)=\half$,
$\gamma_0+\delta_0=6$, $\gamma_0/(\gamma_0+\delta_0)=\twothirds$, so
$\alpha_0=3$, $\beta_0=3$, $\gamma_0=4$, $\delta_0=2$ and hence
$\alpha=3+a=11$, $\beta=3+b=15$, $\gamma=4+c=52$, $\delta=2+d=64$ so that
$\log\{(\alpha-\half)(\delta-\half)/(\beta-\half)(\gamma-\half)\} =\log0.983=-0.113$
while $\alpha^{-1}+\beta^{-1}+\gamma^{-1}+\delta^{-1}=0.192$.  Hence
posterior of log-odds ratio is $\Lambda-\Lambda'\sim\N(-0.113,0.192)$.
The posterior probability that $\pi > \rho$ is
$\Phi(-0.113/\sqrt{0.192})=\Phi(-0.257)=0.3986.$

\nextq Cross-ratio is $(45\times29)/(28\times30)=1.554$ so its logarithm
is 0.441.  More accurately, adjusted value is
$(44.5\times28.5)/(27.5\times29.5)=1.563$ so its logarithm is 0.446.
Value of $a^{-1}+b^{-1}+c^{-1}+d^{-1}$ is 0.126, and so the posterior
distribution of the log-odds ratio is $\Lambda-\Lambda'\sim\N(0.441,0.126)$
or more accurately $\N(0.446, 0.126)$.  The posterior probability that the
log-odds ratio is positive is $\Phi(0.441/\sqrt{0.126})=\Phi(1.242)=0.8929$
or more accurately $\Phi(0.446/\sqrt{0.126})=\Phi(1.256)=0.8955$.  With the
same data $\sin^{-1}\sqrt{(45/73)}=0.903$ radians and
$\sin^{-1}\sqrt{(30/59)}=0.794$ radians, and $1/4m+1/4n=0.00766$, so the
posterior probability that $\pi > \rho$ is
$\Phi((0.903-0.794)/\sqrt{0.00766})=\Phi(1.245)=0.8934$.

\nextq $\Lambda-\Lambda'=\log(\pi/\rho)-\log\{(1-\pi)/(1-\rho)\}$ so if
$\pi-\rho=\alpha$ we get
\begin{align*}
\Lambda-\Lambda'
&=\log\{\pi/(\pi-\alpha)\}-\log\{(1-\pi)/(1-\pi+\alpha)\} \\
&=-\log\{1-\alpha)/\pi\}+\log\{1+\alpha/(1-\pi)\} \\
&\cong\alpha/\pi+\alpha/(1-\pi)=\alpha/\{\pi(1-\pi)\}.
\end{align*}

\nextq With conventional priors, posteriors are near $\pi\sim\Be(56,252)$
and $\rho\sim\Be(34,212)$, so approximating by normals of the same means
and variances $\pi\sim\N(0.1818,0.0004814)$,
$\rho\sim\N(0.1382,0.0004822)$, so $\pi-\rho\sim\N(0.0436,0.000963)$ so
$\Pr(\pi-\rho > 0.01)=\Phi((0.0436-0.01)/\sqrt{0.000963})= \Phi(1.083)=0.8606$ and so the posterior odds are $0.8606/(1-0.8606)=6.174$.

\nextq Using a normal approximation, $x\sim\N(8.5,8.5)$ and
$y\sim\N(11.0,11.0)$, so that $x-y\sim\N(-2.5,19.5)$.

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq Straightforward substitution gives
{\catcode?=\active
\def?{\kern\digitwidth}
$\begin{array}{lllllll} \text{Sample} & ?1 & ?2 & ?3 & ?4 & ?5 &\ \text{Total} \\ n & 12 & 45 & 23 & 19 & 30 &\ 129 \\ r & ?0.631 & ?0.712 & ?0.445 & ?0.696 & ?0.535 \\ \tanh^{-1}z & ?0.743 & ?0.891 & ?0.478 & ?0.860 & ?0.597 \\ n\tanh^{-1}z & ?8.916 & 40.095 & 10.994 & 16.340 & 17.910 &\ ?94.255 \end{array}$
}
Posterior for $\zeta$ is $\N(94.255/129,1/129)$ and 95\% HDR is
$0.7307\pm1.96\times0.0880$, that is, $(0.5582,0.9032)$.  A
corresponding interval for $\rho$ is $(0.507,0.718)$.

\nextq Another straightforward substitution gives
{\catcode?=\active
\def?{\kern\digitwidth}
$\begin{array}{lllllll} n & 45 & 34 & 49 \\ 1/n & ?0.0222 & ?0.0294 & ?0.0204 \\ r & ?0.489 & ?0.545 & ?0.601 \\ \tanh^{-1}z & ?0.535 & ?0.611 & ?0.695 \end{array}$
}
Hence $\zeta_1-\zeta_2\sim\N(0.535-0.611,\ 0.0222+0.0294)$, that is,
$\N(-0.076,0.227^2)$.  Similarely
$\zeta_2-\zeta_3\sim\N(-0.084,0.223^2)$ and
$\zeta_3-\zeta_1\sim\N(0.160,0.206^2)$.  It follows without detailed
examination that there is no evidence of difference.

\nextq
$\zeta\sim\N\left(\sum n_i\tanh^{-1}r_i)/\sum n_i\,,\,1/\sum n_i\right)$ so
required interval is
$\frac{\sum n_i\tanh^{-1}r_i}{\sum n_i}\pm\frac{1.96}{\sqrt{\sum n_i}}.$

\nextq We found in Section 6.1 that
$p(\rho\,|\,\vect x,\vect y)\propto p(\rho)\frac{(1-\rho^2)^{(n-1)/2}} {(1-\rho r)^{n-(3/2)}}$
As $p(\rho)(1-\rho^2)^{-\frac{1}{2}}(1-\rho r)^{3/2}$ does not depend on
$n$, for large $n$
$p(\rho\,|\,\vect x,\vect y)\propto\frac{(1-\rho^2)^{n/2}} {(1-\rho r)^n}$
so that
$\log l(\rho\,|\,\vect x,\vect y)=c+\half n\log(1-\rho^2)-n\log(1-\rho r)$,
and hence
$(\partial/\partial\rho)\log l(\rho\,|\,\vect x,\vect y)= -n\rho(1-\rho^2)^{-1}+nr(1-\rho r)^{-1}.$
implying that the maximum likelihood estimator $\est\rho\cong r$.  Further
$(\partial^2/\partial\rho^2)\log l(\rho\,|\,\vect x,\vect y)= -n(1-\rho^2)^{-1}-2n\rho^2(1-\rho^2)^{-2}+nr^2(1-\rho r)^{-2},$
so that if $\rho=r$ we have
$(\partial^2/\partial\rho^2)\log l(\rho\,|\,\vect x,\vect y)= -n(1-\rho^2)^{-2}$.
This implies that the information should be about
$I(\rho\,|\,\vect x,\vect y)=n(1-\rho^2)^{-2}$,
and so leads to a prior $p(\rho)\propto(1-\rho^2)^{-1}$.

\nextq We have
$p(\rho|\vect x,\vect y)\propto p(\rho)(1-\rho^2)^{(n-1)/2} \int_0^{\infty} (\cosh t-\rho r)^{-(n-1)}\dt$
Now write $-\rho r=\cos\theta$ so that
$p(\rho|\vect x,\vect y)\propto p(\rho)(1-\rho^2)^{(n-1)/2} \int_0^{\infty} (\cosh t+\cos\theta)^{-(n-1)}\dt$
and set
$I_k=\int_0^{\infty} (\cosh t + \cos\theta)^{-k} \dt$
We know that $I_1=\theta/\sin\theta$ (cf.\ J.~A.~Edwards, \textit{A
Treatise on the Integral Calcuous} (2 vols), London: Macmillan (1921)
[reprinted New York: Chelsea (1955)], art.\ 180).  Moreover
$(\partial/\partial\theta)(\cosh t+\cos\theta)^{-k}= k\sin\theta(\cosh t+\cos\theta)^{-(k+1)}$
and by induction
$(\partial/\sin\theta\partial\theta)^k(\cosh t+\cos\theta)^{-1}= k!(\cosh t+\cos\theta)^{-(k+1)}$
Differentiating under the integral sign, we conclude that
$(\partial/\sin\theta\partial\theta)^kI_1=k!I_{k+1}\qquad\ (k\geqslant0).$
Taking $k=n-2$, or $k+1=n-1$, we get
$I_k=\int_0^{\infty} (\cosh t + \cos\theta)^{-k} \dt\propto (\partial/\sin\theta\partial\theta)^k(\theta/\sin\theta).$
(ignoring the factorial).  Consequently
$p(\rho|\vect x,\vect y)\propto p(\rho)(1-\rho^2)^{(n-1)/2} (\partial/\sin\theta\partial\theta)^k(\theta/\sin\theta).$
Since $\d(r\rho)/\!\dtheta=\d(-\cos\theta)/\!\dtheta=\sin\theta$ and so
$\partial/\sin\theta\partial\theta=\d/\d(r\rho)$, we could alternatively write
$p(\rho|\vect x,\vect y)\propto p(\rho)(1-\rho^2)^{(n-1)/2} \frac{\d^{n-2}}{\d(\rho r)^{n-2}} \left(\frac{\text{arc\ }\cos(-\rho r)}{(1-\rho^2r^2)^{\frac{1}{2}}}\right)$

\nextq Supposing that the prior is
$p(\rho) \propto (1-\rho^2)^{k/2}$
and $r=0$ then
$p(\rho|\vect x,\vect y) \propto p(\rho)(1-\rho^2)^{(n-1)/2}$
so with the prior as in the question we get
$p(\rho|\vect x,\vect y) \propto (1-\rho^2)^{(k+n-1)/2}$
If we define
$t=\sqrt{(k+n+1)}\frac{\rho}{\sqrt{(1-\rho^2)}},\qquad \rho=\frac{t}{\sqrt{\{(k+n+1)+t^2\}}}$
so that
\begin{gather*}
2\rho\frac{\drho}{\dt}=\frac{(k+n+1)2t}{\{(k+n+1)+t^2\}^2}     \\
\frac{\drho}{\dt}=\frac{(k+n+1)}{\{(k+n+1)+t^2\}^{3/2}}
\end{gather*}
and hence
\begin{align*}
p(t|\vect X,\vect Y) & \propto p(\rho|\vect X,\vect Y)\drho/\dt \\
& \propto\left[\frac{(k+n+1)}{(k+n-1)+t^2}\right]^{(k+n-1)/2}
\frac{(k+n+1)}{\{(k+n+1)+t^2\}^{3/2}}            \\
& \propto\left[1+\frac{t^2}{k+n+1}\right]^{-(k+n+2)/2}
\end{align*}
This can be recognized as Student's $\t$ distribution on $k+n+1$ degrees
of freedom (see Section 2.12).

\nextq See G.~E.~P.~Box and G.~C.~Tiao, \textit{Bayesian Inference in
8.5.4---the equation given in the question is implicit in their equation
(8.5.49)).

\nextq See G.~E.~P.~Box and G.~C.~Tiao, \textit{Bayesian Inference in
and in particular equation (8.5.43)).

\nextq For a bivariate normal distribution
\begin{align*}
\log l(\alpha,\beta,\gamma|x,y) & =-\log2\pi+\half\log\delta \\
\end{align*}
where $\delta=\alpha\beta-\gamma^2=1/\Delta$.  Then
\begin{align*}
(\partial/\partial\alpha)\log l & =\half\beta/\delta-\half(x-\lambda)^2   \\
(\partial/\partial\beta)\log l  & ==\half\alpha/\delta-\half(y-\mu)^2,    \\
(\partial/\partial\gamma)\log l & =-\gamma/\delta-(x-\lambda)(y-\mu),
\end{align*}
Consequently the information matrix, i.e.\ minus the matrix of second
derivatives (taking expectations is trivial as all elements are
constant) is
\begin{align*}
I & =\left(\begin{array}{ccc}
\smallhalf\beta^2\delta^{-2}
& -\smallhalf\delta^{-1}+\smallhalf\alpha\beta\delta^{-2}
& -\beta\gamma\delta^{-2} \\
-\smallhalf\delta^{-1}+\smallhalf\alpha\beta\delta^{-2}
&  \smallhalf\alpha^2\delta^{-2}
& -\alpha\gamma\delta^{-2} \\
-\beta\gamma\delta^{-2}
& -\alpha\gamma\delta^{-2}
&  \delta^{-1}+2\gamma^2\delta^{-2}
\end{array}\right)                                           \\
& =\frac{1}{2\delta^2}\left(\begin{array}{ccc}
\beta^2      &  \gamma^2      & -2\beta\gamma                  \\
\gamma^2     &  \alpha^2      & -2\alpha\gamma                 \\
-2\beta\gamma & -2\alpha\gamma &  2(\alpha\beta+\gamma^2)
\end{array}\right)
\end{align*}
so that its determinant is
$\det I=\frac{1}{8\delta^6}\left|\begin{array}{ccc} \beta^2 & \gamma^2 & -2\beta\gamma \\ \gamma^2 & \alpha^2 & -2\alpha\gamma \\ -2\beta\gamma & -2\alpha\gamma & -2(\alpha\beta+\gamma^2) \end{array}\right|$
Adding $\half\beta/\gamma$ times the last column to the first and
$\half\gamma/\beta$ times the last column to the second we get
\begin{align*}
\det I & =\frac{1}{8\delta^6}\left(\begin{array}{ccc}
0                      &  0                      & -2\beta\gamma     \\
-\delta                 &  \alpha\beta^{-1}\delta & -2\alpha\gamma    \\
\beta\gamma^{-1}\delta & -\beta^{-1}\gamma\delta &  2(\alpha\beta+\gamma^2)
\end{array}\right)                    \\
& =\frac{1}{8\delta^6}(-2\beta\gamma)
\left(-\frac{\alpha}{\gamma}\delta^2+\frac{\gamma}{\beta}\delta^2\right)
=\frac{1}{4\delta^3}
\end{align*}
We thus conclude that $\det I=1/4\delta^3$ implying a reference prior
$p(\alpha,\beta,\gamma\propto\delta^{-3/2}.$
Rather similarly we get
\begin{align*}
\frac{\partial(\alpha,\beta,\gamma)}{\partial(\phi,\psi,\kappa)}
& =\left|\begin{array}{ccc}
-\psi^2/\Delta^2 & 1/\Delta-\phi\psi/\Delta^2 & 2\psi\kappa/\Delta^2 \\
1/\Delta-\phi\psi/\Delta^2 & -\phi^2/\Delta^2 & 2\phi\kappa/\Delta^2 \\
\psi\kappa/\Delta^2 & \phi\kappa/\Delta^2 & -1/\Delta-2\kappa^2/\Delta^2
\end{array}\right|                            \\
& =\frac{1}{\Delta^6}\left|\begin{array}{ccc}
\psi^2      &  \kappa^2    & -2\psi\kappa                   \\
\kappa^2    &  \phi^2      & -2\phi\kappa                   \\
-\psi\kappa  & -\phi\kappa  &  \phi\psi+\kappa^2
\end{array}\right|
\end{align*}
and evaluate this as $-1/\Delta^3$.  It then follows that
$p(\phi,\psi,\kappa)\propto |-1/\Delta^3|\,p(\alpha,\beta,\gamma) \propto\Delta^{-3/2}.$

\nextq Age is $y$, weight is $x$.  $n=12$, $\mean x=2911$, $\mean y=38.75$,
$S_{xx}=865,127$, $S_{yy}=36.25$, $S_{xy}=4727$.  Hence $a=\mean y=38.75$, $b=S_{xy}/S_{xx}=0.005464$,
$r=S_{xy}/\sqrt{(S_{xx}S_{yy})}=0.8441$,
$S_{ee}=S_{yy}-S_{xy}^2/S_{xx}=S_{yy}(1-r^2)=10.422$,
and $s^2=S_{ee}/(n-2)=1.0422$.  Take $x_0=3000$ and get
$a+b(x_0-\mean x)=39.24$ as mean age for weight 3000.  For a particular
baby, note that the 95\% point of $\t_{10}$ is 1.812 and
$s\sqrt{\phantom{|}}\{1+n^{-1}+(x_0-\mean x)^2/S_{xx}\}=1.067$
so a 90\% interval is $39.24\pm1.812\times1.067$, that is, $(37.31,41.17)$.
For the mean weight of all such babies note
$s\sqrt{\phantom{|}}\{n^{-1}+(x_0-\mean x)^2/S_{xx}\}=0.310$,
so interval is $(38.68, 39.80)$.

\nextq From the formulae in the text
$S_{ee}=S_{yy}-S_{xy}^2/S_{xx}=S_{yy}-2S_{xy}^2/S_{xx}+S_{xy}^2/S_{xx} = S_{yy}-2bS_{xy}+b^2S_{xx}$
where $b=S_{xy}/S_{xx}$.  Hence
\begin{align*}
S_{ee} & =\sum(y_i-\mean y)^2-2b\sum(y_i-\mean y)(x_i-\mean x)
+b^2\sum(x_i-\mean x)^2                                         \\
& =\sum\{(y_i-\mean y)-b(x_i-\mean x)\}^2.
\end{align*}
The result now follows as $y_i=a$.

\nextq Clearly (as stated in the hint) $\sum u_i=\sum v_i=\sum u_iv_i=0$,
hence $\mean u=\mean v=0$ and $S_{uv}=0$.  We now proceed on the lines
of Section 6.3, redefining $S_{ee}$ as
$S_{yy}-S{uy}^2/S_{uu}-S_{vy}^2/S_{vv}$ and noting that
\begin{align*}
& \sum (y_i-\alpha-\beta u_i-\gamma v_i)^2=
\sum\left\{(y_i-\mean y)+(\mean y-\alpha)-
\beta u_i-\gamma v_i\right\}^2                                  \\
& =S_{yy}+n(y_i-\mean y)^2+\beta^2S_{uu}+\gamma^2S_{vv}
-2\beta S_{uy}-2\gamma S_{vy}                                       \\
& =S_{yy}-S_{uy}^2/S_{uu}-S_{vy}^2/S_{vv}+n(y_i-\mean y)^2              \\
& =S_{ee}+n(y_i-\mean y)^2+
S_{uu}(\beta-S_{uy}/S_{uu})^2+S_{vv}(\gamma-S_{vy}/S_{vv})^2
\end{align*}
We consequently get a density $p(\alpha,\beta,\gamma,\phi\,|\,\vect x,\vect y)$
from which we integrate out \textit{both} $\beta$ and $\gamma$ to get
$p(\alpha,\phi\,|\,\vect x,\vect y)\propto\phi^{-n/2} \exp[-\half\{S_{ee}+n(\alpha-a)^2\}/\phi].$
The result now follows.

\nextq $I=4$, $\sum K_i=28$, $G=1779$, $\sum\sum x_{ik}^2=114,569$,
$C=G^2/N=113,030$.  Further, $S_T=1539$, $S_t=(426^2+461^2+450^2+442^2)/7-C=93$
and hence the analysis of variance table is as follows:
{\catcode?=\active
\def?{\kern\digitwidth}
\begin{center}
\begin{tabular}{lllll}
\multicolumn{5}{c}{ANOVA Table}                             \\ \hline
&         & Degrees                              \\
& Sum of  & of       & Mean                      \\
Source     & squares & freedom  & square & Ratio            \\ \hline
Treatments & ??93    & ?6       & ?31    & $( < 1)$         \\
Error      & 1446    & 24       & ?60                       \\ \hline
TOTAL      & 1539    & 27
\end{tabular}
\end{center}
}
\noindent We conclude that the results from the four samples agree.

\nextq $d=\theta_2+\theta_4+\theta_6-\theta_3-\theta_5-\theta_7$ with
$\est d=-1.4$ and $K_d=(6/4)^{-1}$ so that $0.82(d+1.4)/2.74\sim\t_{25}$.

\nextq The analysis of variance table is as follows:
{\catcode?=\active
\def?{\kern\digitwidth}
\begin{center}
\begin{tabular}{lllll}
\multicolumn{5}{c}{ANOVA Table}                           \\ \hline
&         & Degrees                            \\
& Sum of  & of       & Mean                    \\
Source     & squares & freedom  & square & Ratio          \\ \hline
Treatments & ?49,884 & ?2       & 24,942 & 13.3           \\
Blocks     & 149,700 & ?5       & 29,940 & 16.0           \\
Error      & ?18,725 & 10       & ?1,872                  \\ \hline
TOTAL      & 218,309 & 17
\end{tabular}
\end{center}
}
\noindent We conclude that the treatments differ significantly (an
$\F_{2,10}$ variable exceeds 9.43 with probability 0.5\%).

\nextq This is probably seen most clearly by example.  When $r=t=2$ we take
$\vect x=\left(\begin{array}{c}x_{111}\\x_{112}\\x_{121}\\x_{122}\\ x_{211}\\x_{212}\\x_{221}\\x_{222} \end{array}\right),\quad \matr A=\left(\begin{array}{cccccc} 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ \end{array} \right),\quad \vect\beta=\left(\begin{array}{c}\mu\\ \alpha_1 \\ \alpha_2 \\ \beta_1 \\ \beta_2 \\ \kappa_{12} \end{array}\right).$

\nextq Evidently $\matr A^{+}\matr A=\matr I$ from which (a) and (b) and
(d) are immediate.  For (c), note
$\matr A\matr A^{+}=\matr A(\matr A\transpose\matr A)^{-1}\matr A\transpose$
which is clearly symmetric.

\nextq We take
$\matr A=\left(\begin{array}{cc}1 & x_1\\1 & x_2\\ \multicolumn{2}{c}{\ddots}\\ 1 & x_n\end{array}\right),\qquad \matr A\transpose\matr A=\left(\begin{array}{cc} n & \sum x_i\\ \sum x_i & \sum x_i^2\end{array}\right)$
so that writing $S=\sum(x_i-\mean x)^2$ we see that
$\det(\matr A\transpose\matr A)=nS$ and hence
$(\matr A\transpose\matr A)^{-1}=\frac{1}{nS}\left(\begin{array}{cc} \sum x_i^2 & -\sum x_i\\ -\sum x_i & n\end{array}\right),\qquad \matr A\transpose\vect y=\left(\begin{array}{c} \sum y_i\\ \sum x_iy_y\end{array}\right)$
from which
$\est\beeta=(\matr A\transpose\matr A)^{-1}\matr A\transpose\vect y$
is easily found.  In particular
$\est\eta_1=\frac{1}{nS}\left(-\sum x_i\sum y_i+n\sum x_iy_i\right)= \frac{\sum(y_i-\mean y)(x_i-\mean x)}{\sum(x_i-\mean x)^2}.$

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq Recall that $t$ is sufficient for $\theta$ given $x$ if
$p(x|\theta)=f(t,\theta)g(x)$ (see Section 2.9).  Now
$p(t|\theta)=\left\{\begin{array}{ll} p(x|\theta)+p(y|\theta)=p(z|\theta) & \text{if t=t(z)} \\ p(x|\theta) & \text{if t=t(x) for x\neq y} \end{array}\right.$
so that
$p(t|\theta)=\left\{\begin{array}{ll} 0 & \text{if x=y} \\ p(x|\theta) & \text{if x=t} \end{array}\right.$
Setting $f(t,\theta)=p(t|\theta)$ and $g(x)=1$ ($x\neq y$), $g(y)=0$, it
follows that $t$ is sufficient for $x$ given $\theta$.  It then follows
from a na\"\i ve appliaction of the weak sufficiency principle that
$\Ev(E,y,\theta)=\Ev(E,z,\theta)$.  However if $\random x$ is a continuous
random variable, then $p_{\random x}(y|\theta)=0$ for all $y$, so we may
take $y$ and $z$ as \textit{any} two possible values of $\random x$.

\nextq $l(\theta\,|\,g(x))=l(\theta\,|\,h(x))$ from which the result follows.

\nextq The likelihood function is easily seen to be
$l(\theta|x)=\frac{1}{3} \text{\qquad for\qquad} \theta= \left\{\begin{array}{ll} x/2,\ 2x,\ 2x+1 & \text{when x is even} \\ (x-1)/2,\ 2x,\ 2x+1 & \text{when x\neq1 is odd} \\ 1,\ 2,\ 3 & \text{when x=1.} \end{array}\right.$
The estimators $d_1$, $d_2$ and $d_3$ corresponding to the smallest,
middle and largest possible $\theta$ are
$d_1(x)= \left\{\begin{array}{ll} x/2 & \text{when x is even} \\ (x-1)/2 & \text{when x\neq1 is odd} \\ 1 & \text{when x=1,} \end{array}\right.$
and $d_2(x)=2x$, $d_3(x)=2x+1$.  The values of $p(d_1=\theta)$ given in
the question now follow.  However, a Bayesian analyis leads to a posterior
$p(\theta|x)=\frac{l(\theta|x)\pi(\theta)}{p(x)}= \frac{\pi(\theta)I_A(\theta)}{\pi(d_1(x))+\pi(d_2(x))+\pi(d_3(x))}$
where $\pi(\theta)$ is the prior, $A=\{d_1(x), d_2(x), d_3(x)\}$ and
$I_A(\theta)$ is the indicator function of $A$.  Thus, indeed, the data
conveys nothing to the Bayesian version except that $\theta$ is
$d_1(x)$, $d_2(x)$, or $d_3(x)$.  However, Bayesians are indifferent
between $d_1(x)$, $d_2(x)$, or $d_3(x)$ only if they have equal prior
probabilities, which cannot hold for all $x$ if the prior is proper.
For a discussion, see J.~O.~Berger and R.~L.~Wolpert, \textit{The
Likelihood Principle}, Hayward, CA: Institute of Mathematical Statistics
(1984 and 1988, Example 34).

\nextq Computation of the posterior is straightforward.  For a
discussion, see J.~O.~Berger and R.~L.~Wolpert \textit{The Likelihood
Principle}, Hayward, CA: Institute of Mathematical Statistics
(1984 and 1988, Example 35).

\nextq Rules (a) and (b) \textit{are} stopping times and rule (c) is
\textit{not}.

\nextq $n=4$, $\mean x=1.25$, $z=1.25\sqrt{4}=2.50$.

\blankline (a)\quad Reject at the 5\% level (and indeed possibly do so on the
basis of the first observation alone).

$B=(1+n\psi/\phi)^{\frac{1}{2}}\exp[-\half z^2(1+\phi/n\psi)^{-1}]=1.07$
with $\psi=\phi$, so with $\pi_0=\half$ we get $p_0=(1+B^{-1})^{-1}=0.52$.
Null hypothesis still more probable than not.

\nextq As we do \textit{not} stop the first four times but \textit{do}
the fifth time
\begin{align*}
p(\vect x|\lambda) & =\frac{\lambda^{3+1+2+5+7}}{3!\,1!\,5!\,2!\,7!}
\exp(-5\lambda)\frac{3}{3}\frac{1}{4}\frac{2}{6}\frac{5}{11}\frac{11}{18} \\
l(\lambda|\vect x) & \propto \lambda^{10}\exp(-5\lambda).
\end{align*}

\nextq $\E S=\E (S+1)-1$ and after re-arrangement $\E(S+1)$ is
$(s+1)(R''-2)/(r''-2)$ times a sum of probabilities for the beta-Pascal
distribution with $S$ replaced by $(S+1)$, with $s$ replaced by $(s+1)$,
and with $r''$ replaced by $(r''-1)$.  As probabilities sum to unity,
the result follows.

\nextq Up to a constant
$L(\pi|x,y)=\log l(\pi|x,y)=(x+n)\log\pi+(n-x+y)\log(1-\pi)$
so $-(\partial^2L(\pi)|x,y)/\partial\pi^2)=(x+n)/\pi^2+(n-x+y)/(1-\pi)^2$.
Because the expectations of $x$ and $y$ are $n\pi$ and $n(1-\pi)/\pi$
respectively, we get $I(\pi|x,y)=n(1+\pi)/\pi^2(1-\pi)$, so that
$p(\pi|x,y)\propto (1+\pi)^{\frac{1}{2}}\pi^{-1}(1-\pi)^{-\frac{1}{2}}.$

\nextq $\E u(x)=\sum u(x)p(x)=\sum u(x)2^{-x}$ suggesting you would
enter provided $e < \E u(x)$.  If $u(x)\propto x$ then
$\E u(x)\propto\sum2^x2^{-x}=\infty$ resulting in the implausible proposition
that you would pay an arbitrarily large entry fee $\pounds e$.

\nextq By differentiation of the log-likelihood
$L(\pi|x)=x\log\theta+(n-x)\log(1-\theta)$ with respect to $\theta$ we
see that $x/n$ is the maximum likelihood estimator.

Because prior for $\theta$ is uniform, that is, $\Be(1,1)$, posterior
is $\Be(x+1,n-x+1)$.  The question deals with a particular case of weighted
quadratic loss, so we find $d(x)$ as
$\E^w(\theta|x)= \frac{\E((1-\theta)^{-1}|x)}{\E (\theta^{-1}(1-\theta)^{-1}\,|\,x)}= \frac{\Betafn(x+1,\,n-x)}{\Betafn(x+1,\,n-x+1)} \frac{\Betafn(x+1,\,n-x+1)}{\Betafn(x,\,n-x)}=\frac{x}{n}.$
If $x=0$ or $x=n$ then the posterior loss $\rho(a,x)$ is infinite for
all $a$ because the integral diverges at $x=0$ or at $x=n$ so the
minimum is not well-defined.

\nextq The minimum of $\E (\pi-\rho)^2-2a(\E (\pi-\rho)+a^2$ clearly
occurs when $a=\E (\pi-\rho)$.  But since the prior for $\pi$ is
uniform, that is, $\Be(1,1)$, its posterior is $\Be(x+1,n-x+1)$ and so
its posterior mean is $(x+1)/(n+2)$; similarly for $y$.  We conclude
that the Bayes rule is
$d(x,y)=(x-y)/(n+2).$

\nextq Posterior mean (resulting from quadratic loss) is a weighted mean
of the component means, so with the data in Section 2.10 is
$\alpha'\frac{10}{10+20}+\beta'\frac{20}{10+20}= \frac{115}{129}\frac{10}{10+20}+\frac{14}{129}\frac{20}{10+20}=0.370.$
Posterior median (resulting from absolute error loss) can be found by
computing the weighted mean of the distribution functions of the two
beta distributions for various values and honing in.  Result is 0.343.
Posterior mode (resulting from zero-one loss) is not very meaningful in
a bimodal case, and even in a case where the posterior is not actually
bimodal it is not very useful.

\nextq If we take as loss function
$L(\theta, a) = \left\{\begin{array}{ll} u(a-\theta) & \text{if a\leqslant\theta} \\ v(\theta-a) & \text{if a\geqslant\theta} \end{array}\right.$
Suppose $m(x)$ is a $v/(u+v)$ fractile, so that
$\Pr\bigl(x\leqslant m(x)\bigr)\geqslant v/(u+v),\qquad \Pr\bigl(x\geqslant m(x)\bigr)\geqslant u/(u+v).$
Suuppose further that $d(x)$ is any other rule and, for definiteness,
that $d(x) > m(x)$ for some particular $x$ (the proof is similar if
$d(x) < m(x)$).  Then
$L(\theta, m(x)) - L(\theta, d(x)) = \left\{\begin{array}{ll} u[m(x)-d(x)] & \text{if \theta\leqslant m(x)} \\ (u+v)\theta - [um(x) + vd(x)] & \text{if m(x) < \theta < d(x)} \\ v[d(x)-m(x)] & \text{if \theta\geqslant d(x)} \end{array}\right.$
while for $m(x) < \theta < d(x)$
$(u+v)\theta-[um(x)+vd(x)] < u[\theta-m(x)] < u[d(x)-m(x)]$
so that
$L(\theta,m(x))-L(\theta,d(x))\leqslant\left\{\begin{array}{ll} u[m(x)-d(x)] & \text{if \theta\leqslant m(x)} \\ v[d(x)-m(x)] & \text{if \theta > m(x)} \end{array}\right.$
and hence on taking expectations over $\theta$
\begin{align*}
\rho(m(x),x)-\rho(d(x),x) & \leqslant
\bigl\{u[m(x)-d(x)]\bigr\}\Pr(\theta\leqslant m(x)\,|\,x)              \\
& \qquad+ \bigl\{v[d(x)-m(x)]\bigr\} \Pr(\theta > m(x)\,|\,x)             \\
& = \bigl\{d(x)-m(x)\bigr\} \bigl\{-u\Pr(\theta\leqslant m(x)\,|\,x)      \\
& \leqslant\bigl\{d(x)-m(x)\bigr\}\bigl\{-uv/(u+v)+uv/(u+v)\bigr\} = 0
\end{align*}
from which it follows that taking a $v/(u+v)$ fractile of the posterior
distribution does indeed result in the appropriate Bayes rule for this
loss function.

\nextq By integration by parts, if $\theta\sim\Be(2,k)$ then
\begin{align*}
\Pr(\theta<\alpha)
& = \int_0^{\alpha} k(k+1)\theta(1-\theta)^k\dtheta                \\
& = \left[-(k+1)(1-\theta)^k\theta\right]_0^{\alpha}+
\int_0^{\alpha}(k+1)(1-\theta)^k\dtheta                        \\
& = [-(k+1)(1-\theta)^k\theta-(1-\theta)^{k+1}
= 1-(1-\theta)^k(1+k\alpha).
\end{align*}
In this case, the prior for $\theta$ is $\Be(2,12)$ so that the prior
probability of $\Hyp_0$, that is, that $\theta < 0.1$, is 0.379, while
the posterior is $\Be(2,18)$ so that the posterior probability that
$\theta < 0.1$ is 0.580.

\blankline (a)\quad With 0--1'' loss we accept the hypothesis with the
greater posterior probability, in this case $\Hyp_1$.

\blankline (b)\quad The second suggested loss function is of the 0--$K_i$''
form and the decision depends on the relative sizes of $p_1$ and
$2p_0$.  Again this results in a decision in facour of $\Hyp_1$.

\nextq Posterior expected losses are
$\rho(a_0,x)=10p_1,\qquad\rho(a_1,x)=10p_0,\qquad\rho(a_2,x)=3p_0+3p_1.$
Choose $a_0$ if $0\leqslant p_0\leqslant0.3$, choose $a_1$ if $0.3\leqslant p_0\leqslant0.7$ and choose $a_2$ if $0.7\leqslant p_0\leqslant1$.

\nextq Posterior variance is $(225^{-1}+100^{-1})^{-1}=69.23=8.32^2$ and
the posterior mean is $69.23(100/225+115/100)=110.38$.  Posterior
expected losses are
$p(a_1,x)=\int_{90}^{110} (\theta-90)\pi(\theta|x)\dtheta+ \int_{110}^{\infty} 2(\theta-90)\pi(\theta|x)\dtheta$
Now note that (with $\phi(x)$ the $\N(0,1)$ density function
\begin{align*}
\int_{90}^{110} (\theta-90)\pi(\theta|x)\dtheta & =
\sqrt{69.23}\int_{-2.450}^{-0.046} z\phi(z)\dtheta+
20.38\int_{-2.450}^{-0.046} \phi(z)\dtheta                           \\
& = -\sqrt{(69.23/2\pi)}\{\exp(-\half0.046^2)-\exp(-\half2.450^2)\}  \\
& = -3.15+9.66=6.51.
\end{align*}
By similar manipulations we find that $\rho(a_1,x)=34.3$,
$\rho(a_2,x)=3.6$ and $\rho(a_3,x)=3.3$, and thus conclude that $a_3$ is
the Bayes decision.

\nextq In the negative binomial case
\begin{align*}
p(\pi|x) & = p(\pi) p(x|\pi)/p_{\random x}(x)  \\
& = \binom{n+x-1}{x}(1-\pi)^n\pi^x/p_{\random x}(x)
\end{align*}
It follows that the posterior mean is
\begin{align*}
\E(\pi|x) & = \int \pi\binom{n+x-1}{x}(1-\pi)^n\pi^x/p_{\random x}(x) \\
& = \frac{x+1}{n+x}\binom{n+x}{x+1}(1-\pi)^n\pi^x/p_{\random x}(x) \\
& =\frac{(x+1)}{(n+x)}\frac{p_{\random x}(x+1)}{p_{\random x}(x)}
\end{align*}
This leads in the same way as in Section 7.5 to the estimate
$\delta_n=\frac{(\xi+1)f_n(\xi+1)}{(n+\xi)(f_n(\xi)+1)}.$

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq Write $\gamma=\alpha/(\alpha+\beta)$ and
$\delta=(\alpha+\beta)^{-1/2}$.  The Jacobian is
$\left|\begin{array}{cc} \partial\gamma/\partial\alpha & \partial\gamma/\partial\beta \\ \partial\delta/\partial\alpha & \delta\gamma/\partial\beta \end{array}\right| =\left|\begin{array}{cc} \beta/(\alpha+\beta)^2 & -\alpha/(\alpha+\beta)^2 \\ -\half(\alpha+\beta)^{-3/2} & -\half(\alpha+\beta)^{-3/2} \end{array}\right| =-(\alpha+\beta)^{-5/2}$
from which the result follows.

\nextq $p(\phi,\btheta\,|\,\vect x)\propto p(\phi)\,p(\btheta|\phi)\,p(\vect x|\btheta)$.
So
$p(\phi,\btheta\,|\,\vect x)\propto \phi^n\prod\theta_i^{x_i}\exp(-(\phi+1)\theta_i)$
and by the usual change--of-variable rule (using $|\!\dz/\dphi|=1/(1+\phi)^2$)
$p(z,\btheta\,|\,\vect x)\propto z^{-n-2}(1-z)^n\prod\theta_i^{x_i}\exp(-\theta_i/z).$
Integrating with respect to all the $\theta_i$ using
$\int\theta^x\exp(-\theta/z)\propto z^{x+1}$ we get
$p(z|\vect x)\propto z^{n\mean x-2}(1-z)^n$
or $z\sim\Be(n\mean x-1,n+1)$, from which it follows that
$\E z=(n\mean x-1)/(n\mean x+n)$.  Now note that
$p(\theta_i\,|\,x_i,\phi)\propto p(x_i|\theta_i)\,p(\theta_i|\phi) \propto\theta_i^{x_i}\exp(-\theta_i)\,\phi\exp(-\phi\theta_i) \propto \theta_i^{x_i}\exp(-\theta_i/z)$
which is a $\G(x_i,z)$ or $\half z\chi_{x_i}^2$ distribution, from which
it follows that $\E(\theta_i\,|\,x_i,\phi)=x_iz$ and so
$\E(\theta_i|\vect x)=x_i\E(z|\vect x)=x_i(n\mean x-1)/(n\mean x+n).$

\nextq \!\!(a) Use same estimator $\est\btheta=\bhthetab$.  Expression
for $\rho(\bhthetao,\vect X)$ has $S_1$ replaced by a weighted sum of
squares about $\mu$.

\blankline (b) Use estimator $\est\btheta$ with $\est\theta_i$ the
posterior median of the $i$th component.

\nextq We find the mean square error of the transformed data is 3.14
times smaller and of the equivalent probabilities is 3.09 (as opposed to
corresponding figures of 3.50 and 3.49 for the Efron-Morris estimator)
For a complete analysis, see the program

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/rprogs/baseball.txt}

\bigskip

\noindent
or

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/cprogs/baseball.cpp}

\bigskip

\nextq Evidently $\matr A\transpose\matr A=\matr I$ or equivalently
$\vect\alpha_j\transpose\vect\alpha_k=1$ if $j=k$ and 0 if $j\ne k$.  It
follows that $W_j\sim\N(0,1)$ and that $\Cov(W_j,W_k)=0$ if $j\ne k$ which,
since the $W_j$ have a multivariate normal distribution, implies that $W_j$
and $W_k$ are independent.  Further, as $\matr A\transpose\matr A=\matr I$
we also have $\matr A\matr A\transpose=\matr I$ and so
$\matr W\transpose\matr W= (\vect X-\bmu)\transpose\matr A\matr A\transpose(\vect X-\bmu)= (\vect X-\bmu)\transpose(\vect X-\bmu).$
For the last part, normalize the $\alpha_{ij}$ for fixed $j$ to produce
a unit vector and the rest follows as above.

\nextq We know that
$R(\theta,\bhthetajs)-R(\theta,\bhthetajsplus)= \frac{1}{r}\sum\left[\E\left(\est\theta^{JS^2}-\est\theta^{JS^{{}_+2}}\right) -2\theta_i\E\left(\est\theta_i^{JS}-\est\theta_i^{JS^{{}_+}}\right)\right].$
To show that the experession in brackets is always $> 0$, calculate the
expectation by first conditioning on $S_1$.  For any value $S_1\leqslant r-2$, we have $\est\theta_i^{JS}=\est\theta_i^{JS^{{}_+}}$ so that it is
enough to show that the right hand side is positive when conditional on
any value $S_1=s_1 > r-2$.  Since in that case $\est\theta_i^{JS^{{}_+}}=0$,
it is enough to show that for any $s_1 > r-2$
$\theta_i\E\left[\left.\est\theta_i^{JS}\,\right|\,S_1=s_1\right]= \theta_i\left(1-\frac{r-2}{s_1}\right)\E(X_i-\mu_i\,|\,S_1=s_1)\leqslant0$
and hence it is enough to show that
$\theta_i\E(X_i-\mu_i\,|\,S_1=s_1)\geqslant0$.  Now $S_1=s_1$
is equivalent to
$(X_1-\mu)^2=s_1-\left\{(X_2-\mu)^2+\dots+(X_n-\mu)^2\right\}$.
Conditioning further on $X_2$, \dots, $X_n$ and noting that
\begin{align*}
\Pr(X_1=y\,|\,S_1=s_1,X_2=x_2,\dots,X_n=x_n)
& \propto \exp\left(-\half(y-\theta_1)^2\right) \\
\Pr(X_1=-y\,|\,S_1=s_1,X_2=x_2,\dots,X_n=x_n)
& \propto \exp\left(-\half(y+\theta_1)^2\right)
\end{align*}
we find that
$\E\left[\theta_1(X_1-\mu_1)\,|\,S_1=s_1,X_2=x_2,\dots X_n=x_n\right]= \frac{\theta_1 y(\text{e}^{\theta_1y}-\text{e}^{-\theta_1y})} {\text{e}^{\theta_1y}+\text{e}^{-\theta_1y}}$
where $y=\sqrt{s_1-(x_2-\mu)^2-\dots-(x_n-\mu)^2}$.  The right hand side
is an increasing function of $|\theta_1y|$, which is zero when
$\theta_1y=0$, and this completes the proof.

\nextq Note that, as $\matr A\transpose\matr A+k\matr I$ and
$\matr A\transpose\matr A$ commute,
\begin{align*}
\bhtheta-\bhtheta_k
& =\left\{(\matr A\transpose\matr A+k\matr I)-
(\matr A\transpose\matr A)\right\}
(\matr A\transpose\matr A)^{-1}
(\matr A\transpose\matr A+k\matr I)^{-1}\matr A\transpose\vect x \\
& =k(\matr A\transpose\matr A)^{-1}\bhtheta_k.
\end{align*}
The bias is
$\vect b(k)=\left\{(\matr A\transpose\matr A+k\matr I)^{-1} \matr A\transpose\matr A-\matr I\right\}\btheta,$
and the sum of the squares of the biases is
$\mathcal G(k)=\vect b(k)\transpose\vect b(k)= (\E\bhtheta_k-\btheta)\transpose(\E\bhtheta_k-\btheta).$
The variance-covariance matrix is
$\Var(\bhtheta_k)=\phi(\matr A\transpose\matr A+k\matr I)^{-1} \matr A\transpose\matr A(\matr A\transpose\matr A+k\matr I)^{-1}$
so that the sum of the variances of the regression coefficients is
\begin{align*}
\mathcal F(k) & =\E(\bhtheta_k-\E\btheta)\transpose(\bhtheta_k-\E\btheta) \\
& =\text{Trace}(\Var(\bhtheta_k)) \\
& =\phi\,\text{Trace}
\left\{(\matr A\transpose\matr A+k\matr I)^{-1}
\matr A\transpose\matr A\,
(\matr A\transpose\matr A+k\matr I)^{-1}\right\}
\end{align*}
and the residual sum of squares is
\begin{align*}
(\vect x-\matr A\bhtheta_k)\transpose(\vect x-\matr A\bhtheta_k)  \\
& =\E\{(\vect x-\matr A\bhtheta)+\matr A(\bhtheta-\bhtheta_k)\}\transpose
\{(\vect x-\matr A\bhtheta)+\matr A(\bhtheta-\bhtheta_k)\}.
\end{align*}
Because $\matr A\transpose(\vect x-\matr A\bhtheta)=\matr A\transpose\vect x- \matr A\transpose\matr A(\matr A\transpose\matr A)^{-1} \matr A\transpose\vect x=\vect 0$, this can be written as
\begin{align*}
(\vect x-\matr A\bhtheta)\transpose(\vect x-\matr A\bhtheta)
+(\bhtheta-\bhtheta_k)\transpose\matr A\transpose\matr A
(\bhtheta-\bhtheta_k) \\
k^2\,\bhtheta_k\transpose(\matr A\transpose\matr A)^{-1}\bhtheta_k
\end{align*}
while the mean square error is
\begin{align*}
\textit{MSE}_k & =\E(\bhtheta_k-\btheta)\transpose(\bhtheta_k-\btheta) \\
& =\E\{(\bhtheta_k-\E\bhtheta_k)+(\E\bhtheta_k-\btheta)\}\transpose
\{(\bhtheta_k-\E\bhtheta_k)+(\E\bhtheta_k-\btheta)\} \\
& =\E(\bhtheta_k-\E\bhtheta_k)\transpose(\bhtheta_k-\E\bhtheta_k)+
(\E\bhtheta_k-\btheta)\transpose(\E\bhtheta_k-\btheta) \\
& =\mathcal F(k) + \mathcal G(k)
\end{align*}
using the fact that $\E(\bhtheta_k-\E\bhtheta_k)=0$.

It can be shown that $\mathcal F(k)$ is a continuous monotonic
decreasing function of $k$ with $\mathcal F'(0) < 0$, while
$\mathcal G(k)$ is a continuous monotonic increasing function with
$\mathcal G(0)=\mathcal G'(0)=0$ which approaches
$\btheta\transpose\btheta$ as an upper limit as $k\to\infty$.  It
follows that there always exists a $k > 0$ such that
$\textit{MSE}_k < \textit{MSE}_0$.  In fact, this is always true when
$k < 2\phi/\btheta\transpose\btheta$ (cf.\ C.~M.~Theobold,
Generalizations of mean square error applied to ridge regression'',
\textit{Journal of the Royal Statistical Society Series} \textbf{B},
\textbf{36} (1974), 103--106).

\nextq We defined
$\matr H^{-1}=\matr\Psi^{-1}-\matr\Psi^{-1}\matr B \left(\matr B\transpose\matr\Psi^{-1}\matr B\right) \matr B\transpose\matr\Psi^{-1}$
so that
$\matr B\transpose\matr H^{-1}\matr B=\matr B\transpose\matr\Psi^{-1}\matr B -\matr B\transpose\matr\Psi^{-1}\matr B=\matr 0.$
If $\matr B$ is square and non-singular then
$\matr H^{-1}=\matr\Psi^{-1}-\matr\Psi^{-1}\matr B \matr B^{-1}\matr\Psi\left(\matr B\transpose\right)^{-1} \matr B\transpose\matr\Psi^{-1} =\matr\Psi^{-1}-\matr\Psi^{-1}=\matr 0.$

\nextq It is easily seen that
\begin{align*}
\matr B\transpose\matr\Psi^{-1} & =\left(\begin{array}{cccc}
\psi_{\alpha}^{-1} & \psi_{\alpha}^{-1} & 0 & 0 \\
0 & 0 & \psi_{\beta}^{-1} & \psi_{\beta}^{-1}
\end{array}\right), \\
\matr B\transpose\matr\Psi^{-1}\matr B & =\left(\begin{array}{cc}
2\psi_{\alpha}^{-1} & 0 \\ 0 & 2\psi_{\beta}^{-1}
\end{array}\right)
\end{align*}
so that
\begin{align*}
\matr H^{-1} & =\matr\Psi^{-1}-\matr\Psi^{-1}\matr B
\left(\matr B\transpose\matr\Psi^{-1}\matr B\right)
\matr B\transpose\matr\Psi^{-1} \\
& =\left(\begin{array}{cccc}
\half\psi_{\alpha}^{-1} & -\half\psi_{\alpha}^{-1}& 0 & 0 \\
-\half\psi_{\alpha}^{-1} & \half\psi_{\alpha}^{-1}& 0 & 0 \\
0 & 0 & \half\psi_{\beta}^{-1} & -\half\psi_{\beta}^{-1} \\
0 & 0 & -\half\psi_{\beta}^{-1} & \half\psi_{\beta}^{-1}
\end{array}\right)
\end{align*}
while
$\matr A\transpose\matr A=\left(\begin{array}{cccc} 4 & 0 & 2 & 2 \\ 0 & 4 & 2 & 2 \\ 2 & 2 & 4 & 0 \\ 2 & 2 & 0 & 4 \end{array}\right).$
Now
\begin{align*}
\matr K^{-1} & =\matr A\transpose\matr\Phi^{-1}\matr A+\matr H^{-1} \\
& =\left(\begin{array}{cccc}
4\phi+\half\psi_{\alpha}^{-1} & -\half\psi_{\alpha}^{-1} & 2\phi & 2\phi \\
-\half\psi_{\alpha}^{-1} & 4\phi+\half\psi_{\alpha}^{-1} & 2\phi & 2\phi \\
2\phi & 2\phi & 4\phi+\half\psi_{\beta}^{-1} & -\half\psi_{\beta}^{-1} \\
2\phi & 2\phi & -\half\psi_{\beta}^{-1} & 4\phi+\half\psi_{\beta}^{-1}
\end{array}\right)
\end{align*}

\nextq See D.~V.~Lindley and A.~F.~M.~Smith, Bayes estimates for the
linear model'' (with discussion), \textit{Journal of the Royal
Statistical Society Series} \textbf{B}, \textbf{34} (1971), 1--41
[reprinted in N.~Polson and G.~C.~Tiao, \textit{Bayesian Inference} (2 vols),
(\textit{The International Library of Critical Writings in Econometrics},
No.\ 7), Aldershot: Edward Elgar (1995)].

\nextq We have all $a_i=a$ so $\vect u\transpose\vect u=-mb/a$
so $1+\vect u\transpose\vect u=(a-mb)/a$ and thus
\begin{align*}
\Sigma_{ii} & =\frac{1}{a}\left(1-\frac{1}{1+\vect u\transpose\vect u}
\frac{b}{a}\right)=\frac{1}{a}\left(1-\frac{b}{a-mb}\right)
=\frac{a-(m+1)b}{a(a-mb)} \\
\Sigma_{ij} & =-\frac{(m+1)b}{a(a-mb)}
\end{align*}
where $a=n/\phi+1/\phi$ and $b=-1/r\psi$.

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq A simulation in \textsf{R}, the program for which can be seen on

\bigskip

{\small
\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/tprogs/integral.txt}
}

\bigskip

\noindent
resulted in values 1.684199, 1.516539, 1.7974, 1.921932, 1.595924, 1.573164,
1.812976, 1.880691, 1.641073, 1.770603 with a mean of 1.719450 as as opposed
to the theoretical value of $\text{e}-1=1.71828$. The theoretical variance
of a single value of $\text{e}^X$ where $X\sim\U(0,1)$ is
$\half(\text{e}-1)(3-\text{e})=0.24204$ so that the standard deviation of the
mean of 10 values is $\sqrt{0.24204/10}=0.15557$ which compares with a sample
standard deviation of 0.1372971.

A simulation in C++, the program for which can be seen on

\bigskip

\noindent
{\small
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/cprogs/integral.cpp}
}

\bigskip

\noindent
resulted in values 1.74529, 1.86877, 1.68003, 1.95945, 1.62928, 1.62953,
1.84021, 1.87704, 1.49146, 1.55213 with a mean of 1.72732 and a
sample standard deviation of 0.155317.

\nextq The easiest way to check the value of $\matr A^t$ is by
induction.  Then note that for large $t$ the matrix $\matr A^t$ is
approximately
$\left(\begin{array}{cc}\slopefrac{2}{5} & \slopefrac{3}{5} \\ \slopefrac{2}{5} & \slopefrac{3}{5}\end{array}\right)$
from which the result follows.

\nextq In this case
$Q=\E[(y_1+y_2-1)\log\eta+(y_3+y_4-1)\log(1-\eta)]$
giving
$\eta^{(t+1)}=\frac{\E(y_1\,|\,\eta^{(t)},\vect x)+x_2-1} {\E(y_1\,|\,\eta^{(t)},\vect x)+x_2+x_3+\E(y_4\,|\,\eta^{(t)},\vect x)-1}$
where $y_1\sim\B(x_1,\eta/(\eta+1))$ and $y_4\sim\B(x_4,(1-\eta)/(2-\eta))$,
so that
\begin{align*}
\eta^{(t+1)} & =\frac{x_1\eta^{(t)}/(\eta^{(t)}+1)+x_2-1}
{x_1\eta^{(t)}/(\eta^{(t)}+1)+x_2+x_3+x_4(1-\eta^{(t)})/(2-\eta^{(t)})-2} \\
& =\frac{461\eta^{(t)}/(\eta^{(t)}+1)+130-1}
{461\eta^{(t)}/(\eta^{(t)}+1)+130+161+515(1-\eta^{(t)})/(2-\eta^{(t)})-2}.
\end{align*}
Starting with $\eta^{(0)}=0.5$, successive iterations give 0.4681, 0.4461,
0.4411, 0.4394, 0.4386, 0.4385, and thereafter 0.4384.  This compares with a
value of 0.439 found by maximum likelihood in C.~A.~B.~Smith,
\textit{op. cit.}

\nextq The proof that
$p(\eta^{(t+1)}\,|\,\vect x)\geqslant p(\eta^{(t+1)}\,|\,\vect x)$
in the subsection Why the \EM\ algorithm works'' only uses the
properties of a \GEM\ algorithm.  As for the last part, if convergence
has occurred there is no further room for increasing $Q$, so we must
have reached a maximum.

\nextq Take prior for variance $S_0\chi_{\nu}^2$, that is,
$350\chi_4^2$, and prior for mean which is $\N(\theta_0,\phi_0)$ where
$\theta_0=85$ and $\phi_0=S_0/n_0(\nu-2)=350/(1\times2)=175$.  We then
have data with $n=100$, $\mean x=89$ and $S_1=(n-1)s^2=2970$.
Successive iterations using the \EM\ algorithm give the posterior mean
as 88.9793 and at all subsequent stages 88.9862.  This compares with
88.96 as found in Exercise 16 on Chapter 2.

\nextq Means for the four looms are 97.3497, 91.7870, 95.7272 and 96.8861,
overall mean is 97.4375.  Variance of observations from the same loom is
1.6250 and variance of means from different looms in the population is
5.1680.

\nextq \!\!(a) For the example on genetic linkage in Section 9.3, see

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/rprogs/dataaug.txt}

\bigskip

\noindent
or

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/cprogs/dataaug.cpp}

\bigskip

\blankline (b) For the example on chained data sampling due to Casella
and George, see a similar file with \texttt{chained} in place of
\texttt{dataaug}.

\blankline (c) For the example on the semi-conjugate prior with a normal
likelihood (using both the \EM\ algorithm and chained data augmentation,
see a similar file with \texttt{semiconj} in place of
\texttt{dataaug}.

\blankline (d) For the example on a change-point analysis of data on
coal mining disasters, see similar files with \texttt{coalnr.cpp} or
\texttt{coal.cpp} in place of \texttt{dataaug.cpp} (the difference is
that the former uses only routines in W.~H.~Press \textit{et al.},
\textit{Numerical Recipes in C} (2nd edn), Cambridge: University Press
1992, whereas the latter program, which is in many ways preferable, uses
gamma variates with non-integer parameters).

\nextq See the program referred to in part (a) of the previous answer.

\nextq See the program referred to in part (b) of the answer to the
question before last.

\nextq See the program

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/rprogs/semicon2.txt}

\bigskip

\noindent
or

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/$\sim$pml1/bayes/cprogs/semicon2.cpp}

\nextq B.~P.~Carlin and T.~A.~Louis, \textit{Bayes and Empirical Bayes
Methods for Data Analysis} (2nd edn), Chapman and Hall 2000, p.\,149,
remark that

In our dataset, each rat was weighed once a week for five consecutive
weeks.  In fact the rats were all the same age at each weighing
$x_{i1}=8$, $x_{i2}=15$, $x_{i3}=22$, $x_{i4}=29$, and $x_{i5}=36$ for
all $i$.  As a result we may simplify our computations by rewriting the
likelihood as
$Y_{ij}\sim N(\alpha_i+\beta_i(x_{ij}-\overline x)\,,\,\sigma^2), \ i=1, ...,k,\ j=1, ..., n_i,$
so that it is now reasonable to think of $\alpha_i$ and $\beta_i$ as
independent \textit{a priori}.  Thus we may set
$\Sigma=\textit{Diag}(\sigma_{\alpha}^2,\sigma_{\beta}^2)$, and replace
the Wishart prior with a product of idependent inverse gamma priors, say
$\textit{IG}(a_{\alpha},b_{\alpha})$ and
$\textit{IG}(a_{\beta},b_{\beta})$.''

This being so, it probably suffices to proceed as in the Rats' example
supplied with WinBUGS.

A more detailed description of the general set-up is to be found in
Taken from Illustration of Bayesian Inference in Normal Data Models
Using Gibbs Sampling'', Alan E.~Gelfand, Susan E.~Hills, Amy Racine-Poon,
and Adrian F.~M.~Smith, \textit{Journal of the American Statistical
Association} \textbf{85} (412) (1990), 972--985, Section 6, pp.\
978--979, which can be found at

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow
pml1/bayes/rprogs/gelfand.pdf}

\bigskip

\noindent
(\LaTeX source at \texttt{gelfand.htm}).  A program for
generating random matrices with a Wishart distribution based on the
algorithm of Odell and Feiveson described in W.~J.~Kennedy and
J.~E.~Gentle, Statistical Computing, Marcel Dekker 1980, pp.\,231--232
can be found at

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow
pml1/bayes/rprogs/wishart.txt}

\bigskip

while the relevant extract from Kennedy and Gentle is at

\bigskip

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow
pml1/bayes/rprogs/rwishart.pdf}

\nextq See the program at

\bigskip

\noindent

\nextq See the program at

\bigskip

\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/winbugs/wheat.txt}

\bigskip

\noindent
or

\bigskip

\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/rprogs/wheat.txt}

\nextq See the program at

\bigskip

\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/winbugs/logisti2.txt}

\bigskip

\noindent
or

\noindent
\texttt{http://www-users.york.ac.uk/\texttildelow pml1/bayes/rprogs/logisti2.txt}

\section {Exercises on Chapter \arabic{section}}

\startqs

\nextq For sampling with the Cauchy distribution we proceed thus:
\begin{description}
\item[\quad(a)] Substituting $x=\tan\theta$ we find
\begin{align*}
\eta &= \Pr(x > 2) = \int_2^{\infty} \frac{1}{\pi}\frac{1}{1+x^2}\dx
=\half-\tan^{-1}(2)/\pi = \tan^{-1}(\half)/\pi \\
&= 0.147\,583\,6.
\end{align*}
The variance is
$\eta(1-\eta)/n = 0.125\,802\,7/n.$
\item[\quad(b)] Use the usual change of variable rule to deduce the
density of $y$.  Then note that
$\frac{q(y_i)}{p(y_i)}=\frac{1/(\pi(1+y_i^2))}{2/y_i^2} =\frac{1}{2\pi}\frac{y_i^2}{1+y_i^2}$
and that we can take $f(y)=1$ as all values of $y$ satisfy
$y\geqslant 2$.
\item[\quad(c)] Substitute $x_i=2/y_i$ in the above to get
$\frac{1}{2\pi}\frac{4}{4+x_i^2}.$
\item[\quad(d)] On writing $x=2\tan\theta$ and
$\theta_0=\tan^{-1}(\half)$ we find
$\E\widehat\eta=\int_0^1\frac{1}{2\pi}\frac{4}{4+x^2}\dx = \int_0^1\frac{1}{2\pi}\frac{1}{\sec^2\theta}2\sec^2\theta\dtheta = \theta_0/\pi = \eta$
Similarly, noting that $\sin(2\theta_0)=4/5$ the integral for
$\E\widehat\eta^2$ transforms to
\begin{align*}
\int_0^1\frac{1}{4\pi^2}\frac{1}{\sec^4\theta}2\sec^2\theta\dtheta
&= \int_0^1\frac{1}{4\pi^2} 2\cos^2\theta\dtheta \\
&= \int_0^1\frac{1}{4\pi^2}\{1+\cos(2\theta)\}\dtheta \\
&= \left[\theta+\half\sin(2\theta)\right]_0^{\theta_0}/(4\pi)^2 \\
&= \left\{\tan^{-1}(\half)+\twofifths\right\}/(4\pi^2) \\
&= 0.021\,876\,4
\end{align*}
so that
\begin{align*}
\Var\widehat\eta = \E\widehat\eta^2 - (\E\widehat\eta)^2
= 0.000\,095\,5.
\end{align*}
\end{description}

\nextq A suitable program is
\begin{tabbing}
\hspace*{2ex}
\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\kill
\> \prog{n <- 10000} \\
\> \prog{alpha <- 2} \\
\> \prog{beta <- 3} \\
\> \prog{x <- runif(n)} \\
\> \prog{p <- function(x) dunif(x)} \\
\> \prog{q <- function(x) dbeta(x,alpha,beta)} \\
\> \prog{w <- q(x)/p(x)} \\
\> \prog{pi <- w/sum(w)} \\
\> \prog{samp <- sample(x,size=n,prob=pi,replace=T)} \\
\> \prog{print(mean(samp))} \\
\> \prog{print(var(samp))}
\end{tabbing}
A run of this program resulted in a value of the mean of 0.399 as
compared with the true value of 0.4 and a value of the mean of
0.0399 compared with a true value of 0.04.

\nextq Continue the program (avoiding ambiguity in the definition of
$\alpha$) by
\begin{tabbing}
\hspace*{2ex}
\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\kill
\> \prog{theta <- 0.1} \\
\> \prog{le <- (1-theta)*n} \\
\> \prog{lo <- 1:(n-le)} \\
\> \prog{hi <- (le+1):n} \\
\> \prog{y <- sort(samp)} \\
\> \prog{r <- y[hi]-y[lo]} \\
\> \prog{rm <- min(r)} \\
\> \prog{lom <- min(lo[r==rm])} \\
\> \prog{him <- min(hi[r==rm])} \\
\> \prog{dd <- function(x) dbeta(x,alpha,beta)} \\
\> \prog{plot(dd,xlim=c(0,1),ylim=c(0,2))} \\
\> \prog{par(new=T)} \\
\> \prog{plot(density(samp),xlim=c(0,1),ylim=c(0,2),} \\
\>  \> \prog{main="",xlab="",ylab="")} \\
\> \prog{abline(v=y[lom])} \\
\> \prog{abline(v=y[him])} \\
\> \prog{print(y[lom])} \\
\> \prog{print(y[him])}
\end{tabbing}
The lower limit resulting from a run of the program was 0.061 and the
upper limit was 0.705.  Referring to the tables in the Appendix we find
that values of $F$ corresponding to a 90\% HDR of $\log \F_{6,4}$ are
$\underline\F = 0.21$ and $\overline\F = 5.75$.   It follows that an
appropriate interval of values of $F_{4,6}$ is  $(1/\overline\F, 1/\underline\F)$, that is $(0.17, 4.76)$, so that an  appropriate
interval for $\pi$ is
$\frac{2\times 0.17}{3+2\times 0.17}\leqslant\pi\leqslant \frac{2\times 4.76}{3+2\times 4.76}$
that is $(0.10, 0.76)$ (but note that in Section 3.1 we were looking for
intervals in which the distribution of the log-odds is higher than
anywhere outside, rather than for HDRs for the beta distribution
itself).

\nextq A suitable program is
\begin{tabbing}
\hspace*{2ex}
\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\kill
\> \prog{r <- 10} \\
\> \prog{phi <- rep(NA,r)} \\
\> \prog{mu <- rep(NA,r)} \\
\> \prog{S <- rep(NA,r)} \\
\> \prog{n <- 100; xbar <- 89; SS <- (n-1)*30} \\
\> \prog{mu0 <- 85; S0 <- 350; nu0 <- 4; n0 <- 1} \\
\> \prog{phi0 <- S0/(n0*(nu0-2))} \\
\> \prog{S[1] <- S0} \\
\> \prog{nustar <- nu0 + n} \\
\> \prog{for (i in 2:r) \{} \\
\> \> \prog{phi[i] <- (1/phi0 + n*nustar/S[i-1])\^{}\{-1\}} \\
\> \> \prog{mu[i] <- phi[i]*(mu0/phi0 + n*xbar*nustar/S[i-1])} \\
\> \> \prog{S[i] <- S0 + (SS+n*xbar\^{}2) - 2*n*xbar*mu[i+1] +} \\
\> \> \> \> \> \> \> \> \> \prog{n*(mu[i]\^{}2 + phi[i])} \\
\> \> \prog{cat("i",i,"phi",phi[i],"mu",mu[i],"S",S[i],"$\backslash$n")} \\
\> \prog{\}} \\
\> \prog{mustar <- mu[r]; phistar <- phi[r]; Sstar <- S[r]} \\
\> \prog{cat("mu has mean",mustar,"and s.d.",sqrt(phistar),"$\backslash$n")} \\
\> \prog{cat("phi has mean",Sstar/(nustar-2),"and s.d.",} \\
\> \> \> \> \prog{(Sstar/(nustar-2))*sqrt(2/(nustar-4)),"$\backslash$n")} \\
\end{tabbing}
We get convergence to $\mu=88.993$ and $\phi=0.322$.

\nextq In the discrete case the Kullback-Leibler divergence is
\begin{align*}
\I(q:p) &= \sum q(x) \log\{q(x)/p(x)\} \\
&= \sum \binom{n}{x}\rho^x(1-\rho^{n-x})
\log \left\{\binom{n}{x}\rho^x(1-\rho^{n-x})\left/
\binom{n}{x}\pi^x(1-\pi^{n-x})\right.\right\} \\
&= \sum \binom{n}{x}\rho^x(1-\rho^{n-x})
\left\{x\log(\rho/\pi)+(n-x)\log[(1-\rho)/(1-\pi)]\right\} \\
&= n\rho\log(\rho/\pi) + n(1-\rho)\log[(1-\rho)/(1-\pi)]
\end{align*}
using the fact that the mean of $\B(n,\rho)$ is $n\rho$.

For $\I(q:p)=\I(p:q)$ we need
{\small
$n\rho\log(\rho/\pi) + n(1-\rho)\log[(1-\rho)/(1-\pi)] = n\pi\log(\pi/rho) + n(1-\pi)\log[(1-\pi)/(1-\rho)]$
}
so
$(\pi-\rho)\log\left\{\frac{\pi}{1-\pi}\left/ \frac{\rho}{1-\rho}\right.\right\}$
from which it is clear that this can happen if and only if $\pi=\rho$
(when $\I(q:p)=\I(p:q)=0$).

\nextq In the continuous case the Kullback-Leibler divergence is
\begin{align*}
\I(q:p) &= \int q(x) \log\{q(x)/p(x)\}\dx \\
&= \int (2\pi\psi)^{-1/2}\exp\{-\half(x-\nu)^2/\psi\}\times \\
(2\pi\psi)^{-1/2}\exp\{-\half(x-\nu)^2/\psi\}\right]\right. \\
(2\pi\phi)^{-1/2}\exp\{-\half(x-\mu)^2/\phi\}\right]\right\} \\
&= \half\log(\psi/\phi)-\E(x-\nu)^2/\phi+\E(x-\mu)^2/\phi
\end{align*}
where $x\sim\N(\nu,\phi)$.  By writing $(x-\mu)^2=\{(x-\nu)+(\nu-\mu)\}^2$
it is easily concluded that
$2\I(q:p) = \log(\phi/\psi)+(\phi-\psi)/\psi+(\nu-\mu)^2/\phi.$
In particular if $\phi=\psi$ then
$\I(q:p) = \half(\nu-\mu)^2/\phi.$

\nextq We find
\begin{align*}
\I(q:p) &= \int q(x) \log\{q(x)/p(x)\}\dx \\
&= \int_0^{\infty} \beta^{-1}\exp(-x/\beta)
\log\left\{\left(\beta^{-1}\exp(-x/\beta)\right)\left/
2(2\pi)^{-1/2}\exp(-\half x^2)\right.\right\} \\
&= \half\log(8\pi)-\log\beta - \E x/\beta +\half\E x^2
\end{align*}
where $x \sim \Ex(\beta)$.  It follows that
$\I(q:p) = \half\log(8\pi)-\log\beta - 1 + \beta^2$
and hence
\begin{align*}
\frac{\partial\I(q:p)}{\partial\beta}
&= -\frac{1}{\beta} + 2\beta \\
\frac{\partial^2\I(q:p)}{\partial\beta^2}
&=\frac{1}{\beta^2} + 2
\end{align*}
It follows that $\I(q:p)$ is a minimum when $\beta=\sqrt{2}$.

\nextq See the paper by Corduneaunu and Bishop (2001).

\nextq The model is
$(\quarter+\quarter\eta,\,\quarter\eta,\,\quarter(1-\eta),\, \quarter(1-\eta)+\quarter)$
and the values quoted are $x_1=461$, $x_2=130$, $x_3=161$ and
$x_4=515$.  An \R\ program for the \ABCREJ\ algorithm is
\begin{tabbing}
\hspace*{2ex}
\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\=\hspace{2ex}\kill
\> \prog{N <- 100000} \\
\> \prog{etastar <- c(461,130,161,515)} \\
\> \prog{n <- sum(etastar)} \\
\> \prog{eps <- 22} \\
\> \prog{d <- rep(NA,N)} \\
\> \prog{trial <- rep(NA,N)} \\
\> \prog{for (j in 1:N) \{} \\
\> \prog{etatrial <- runif(1)} \\
\>  \> \prog{inds <- sample(4,n,replace=T,p=c(0.25+etatrial/4,etatrial/4,} \\
\>  \> \> \> \> \> \> \> \prog{(1-etatrial)/4,(1-etatrial)/4+0.25))} \\
\> \> \prog{samp <- c(sum(inds==1),sum(inds==2),sum(inds==3),sum(inds==4))} \\
\> \> \prog{d <- sqrt(sum((etastar-samp)\^{}2))} \\
\> \> \prog{if (d <= eps) trial[j] <- etatrial} \\
\> \prog{\}} \\
\> \prog{eta <- trial[!is.na(trial)]} \\
\> \prog{k <- length(eta)} \\
\> \prog{m <- mean(eta)} \\
\> \prog{s <- sd(eta)} \\
\> \prog{cat("k",k,"m",m,"s",s,"$\backslash$n")}
\end{tabbing}
Similar modifications can be made for the other algorithms.

\nextq Let $M_i$ denote model $i$ ($i = 1$, 2, 3, 4) with model
parameters $\theta_i$, where $\theta_1=(\alpha,\tau)$, $\theta_2 = (\alpha,\beta,\tau)$, $\theta_3 = (\alpha,\lambda,\tau)$ and $\theta_4 = (\alpha,\beta,\lambda,\tau)$.

\medskip

\noindent
The reversible jump MCMC algorithm is as follows:
\begin{itemize}
\item Initiate the algorithm by choosing an initial model, $m$ to
start in and initial parameter values $\theta_m$.
\item Each iteration $t$:
\begin{itemize}
\item Update the model parameters $\theta_{m_{t-1}}$, where
$m_{t-1}$ is the model which the algorithm is residing in at the end of
iteration $t - 1$.
\item Propose to move to a new model $M_t'$ with parameter values
$\theta'$. Calculate the acceptance probability for the move and decide
whether to accept or reject the new model (and parameter values).
\end{itemize}
\end{itemize}
The models are nested with model 4 being the full model with sub-models
model 2 and model 3 and the simplest model 1. Each model either includes
or excludes $\beta$ and $\lambda$ (exclusion of a parameter is
equivalent to setting it equal to 0).

\medskip

\noindent
Given that the current model $j$, employ the following reversible jump:

\medskip

\noindent
Choose randomly either $\beta$ and $\lambda$, and propose to move to
model $i$ has the chosen parameter included (excluded) if the parameter
is excluded (included) in model $j$.

\medskip

\noindent
A bijection is required for transforming between the models. Therefore
for inclusion of a parameter propose $u \sim \N(0, \sigma^2)$ for
suitably chosen $\sigma^2$ and propose to set the parameter equal to
$u$.  (This works best if $x_i$ and $t_i$ have been standardised to have
mean 0 and variance 1.) Leave the other parameters unchanged. Thus the
Jacobian of the transformation is simply 1.

\medskip

\noindent
Let $L_j(\theta_j)$ denote the likelihood of $\theta_j$ given model
$M_j$.  Then the acceptance probability for the inclusion of a parameter
(either $\beta$ or $\tau$ set equal to $u$) is
$\min\left\{1,\ \frac{L_i(\theta_i)}{L_j(\theta_j)}\times \frac{\pi(\theta_i,M_i)}{\pi(\theta_j,M_j)}\times\frac{1/2}{1/2} \times\frac{1}{\exp(-\half u^2/\sigma^2)/\sqrt{2\pi\sigma^2}} \right\}.$

\end{document}

% `