\chapter{Inference for a normal population} This chapter shows how to make inferences for the mean and variance of a normal population using a conjugate prior distribution. First we need the multi-parameter version of Bayes Theorem. \section{Bayes Theorem for many parameters} Suppose that now the probability (density) function we used to describe the data depends on many parameters, that is, $f(\xvec|\thetavec)$ where $\thetavece$. After observing the data, the likelihood function for $\thetavec$ is $f(\xvec|\thetavec)$. Prior beliefs about $\thetavec$ are represented through a probability (density) function $\pi(\thetavec)$. Therefore, using Bayes Theorem, the posterior probability (density) function for $\thetavec$ is \begin{align*} \pi(\thetavec|\xvec)&=\frac{\pi(\thetavec)\,f(\xvec|\thetavec)}{f(\xvec)} \\ \intertext{where} f(\xvec)&= \begin{cases} \int_\Theta\pi(\thetavec)\,f(\xvec|\thetavec)\,d\thetavec & \text{if $\thetavec$ is continuous}, \\ \\ \sum_\Theta\pi(\thetavec)\,f(\xvec|\thetavec) & \text{if $\thetavec$ is discrete}. \end{cases} \\ \intertext{As in Chapter~1, this can be rewritten as} \pi(\thetavec|\xvec)&\propto \pi(\thetavec)\times f(\xvec|\thetavec) \\ i.e.~~\text{posterior}&\propto\text{prior}\times\text{likelihood}. \end{align*} Next we introduce a new distribution which will be useful later on. %%Hack for vspace \newpage \begin{example} \label{ex:student}~\\ If $X$ has a generalised $t_a(b,c)$ distribution (see formula sheet) then show that $Y=(X-b)/\sqrt{c}\sim t_a\equiv t_a(0,1)$. Recall the general result: if $X$ is a random variable with probability density function $f_X(x)$ and $g$ is a bijective (1--1) function then the random variable $Y=g(X)$ has probability density function \begin{equation} \label{eq:trans1} f_Y(y)=f_X\left\{g^{-1}(y)\right\}\left|\frac{d}{dy}\,g^{-1}(y)\right|. \end{equation} \framebox[17cm]{ \parbox[t]{16cm}{ \vspace{0.4cm} \huge \ding{46} \normalsize \textcolor{white}{ Here we take $Y=g(X)=(X-b)/\sqrt{c}$ from which we obtain $X=g^{-1}(Y)=b+\sqrt{c}\,Y$. Therefore using $\eqref{eq:trans1}$ we have \begin{align*} f_Y(y) &=f_X\left\{g^{-1}(y)\right\}\left|\frac{d}{dy}\,g^{-1}(y)\right|\\ &=f_Y\left(b+\sqrt{c}\,y\right)\times\sqrt{c}\\ &=\frac{\Gamma\left(\frac{a+1}{2}\right)} {\sqrt{ac\pi}\,\Gamma\left(\frac{a}{2}\right)} \left(1+\frac{y^2}{a}\right)^{-\frac{a+1}{2}}\times\sqrt{c},\quad y\in\mathbb{R}\\ &=\frac{\Gamma\left(\frac{a+1}{2}\right)} {\sqrt{a\pi}\,\Gamma\left(\frac{a}{2}\right)} \left(1+\frac{y^2}{a}\right)^{-\frac{a+1}{2}},\quad y\in\mathbb{R}. \end{align*} This is the $t_a$ density and so $Y=(X-b)/\sqrt{c}\sim t_a$.} \vspace{2cm} }} \end{example} \vspace{-0.75cm} \subsection*{Comment} Values for the density function $f_Y(y)$ and the distribution function $F_Y(y)$ can be obtained by using the \texttt{R} functions \texttt{dgt} and \texttt{pgt} in the package \texttt{nclbayes}. It is clear that $t_a(0,1)\equiv t_a$ by examining their densities. Therefore, it makes sense to think of the $t_a$ distribution as the standard $t_a$--distribution and make all calculations for the generalised $t_a(b,c)$ distribution from this standard distribution. The relationship between this standard and generalised version of the $t$-distribution is directly analogous to that between the standard normal $N(0,1)$ distribution and its more general version: the $N(b,c)$ distribution. In both cases the relationship is one of location and scale: \begin{center} \begin{tabular}{lcl} $Y\sim N(b,c)$ & $\Longrightarrow$ & $\dfrac{Y-b}{\sqrt{c}}\sim N(0,1)$\\ \\ $Y\sim t_a(b,c)$ & $\Longrightarrow$ & $\dfrac{Y-b}{\sqrt{c}}\sim t_a$.\\ \end{tabular} \end{center} \section{Prior to posterior analysis} \label{sec:nochtheory} Suppose we have a random sample from a normal distribution in which both the mean~$\mu$ and the precision~$\tau$ are unknown, that is, $X_i|\mu,\tau\sim N(\mu,1/\tau)$, $i=1,2,\ldots,n$ (independent). We shall adopt a (joint) prior distribution for $\mu$ and $\tau$ for which \begin{equation*} \mu|\tau\sim N\left(b,\frac{1}{c\tau}\right) \quad\quad\text{and}\quad\quad \tau\sim Ga(g,h) \end{equation*} for known values $b$, $c$, $g$ and $h$. This distribution has density function \begin{align} \label{eq:nogaprior} \pi(\mu,\tau)&= \pi(\mu|\tau)\pi(\tau) \notag \\ &=\left(\frac{c\tau}{2\pi}\right)^{1/2}\exp\left\{-\frac{c\tau}{2}(\mu-b)^2\right\} \times \frac{h^g\tau^{g-1}e^{-h\tau}}{\Gamma(g)}, \quad \mu\in\R, ~\tau>0 \notag \\ &\propto\tau^{g-\frac{1}{2}}\exp\left\{-\frac{\tau}{2} \left[c(\mu-b)^2+2h\right]\right\},\quad \mu\in\R, ~\tau>0. \end{align} We will use the notation $NGa(b,c,g,h)$ for this distribution. Thus we take the prior distribution \[ \mutau\sim NGa(b,c,g,h). \] Determine the posterior distribution for $\mutau$. Hint: \[ c(\mu-b)^2+n(\bar x-\mu)^2 =(c+n)\left\{\mu-\left(\frac{cb+n\bar x}{c+n}\right)\right\}^2 +\frac{nc(\bar x-b)^2}{c+n}. \] \framebox[17cm]{ \parbox[t]{16cm}{ \vspace{0.4cm} \huge \ding{46} \normalsize \textcolor{white}{ From \eqref{eq:p10}, the likelihood function is \begin{align*} f(\xvec|\mu,\tau)&=\left(\frac{\tau}{2\pi}\right)^{n/2} \exp\left[-\frac{n\tau}{2} \left\{s^2+(\bar x-\mu)^2\right\}\right]. %\end{align*} %\begin{align} \intertext{Using Bayes Theorem, the posterior density is} \pi(\mu,\tau|\xvec)&\propto\pi(\mu,\tau)\,f(\xvec|\mu,\tau) \intertext{and so, for $\mu\in\R$, $\tau>0$} \pi(\mu,\tau|\xvec)&\propto\tau^{g-\frac{1}{2}}\exp\left\{-\frac{\tau}{2} \left[c(\mu-b)^2+2h\right]\right\}\\ &\hspace*{3cm}\times \tau^{\frac{n}{2}} \exp\left[-\frac{n\tau}{2} \left\{s^2+(\bar x-\mu)^2\right\}\right] \\ &\propto\tau^{g+\frac{n}{2}-\frac{1}{2}}\exp\left\{-\frac{\tau}{2} \left[c(\mu-b)^2+n(\bar x-\mu)^2+2h+ns^2\right]\right\}\\ &\propto\tau^{g+\frac{n}{2}-\frac{1}{2}}\exp\left\{-\frac{\tau}{2} \left[(c+n)\left\{\mu-\left(\frac{cb+n\bar x}{c+n}\right)\right\}^2+\frac{nc(\bar x-b)^2}{c+n}+2h+ns^2\right]\right\} \end{align*} } }} \framebox[17cm]{ \parbox[t]{16cm}{ \vspace{0.4cm} \huge \ding{46} \normalsize \textit{...Solution to Example 1.2} \textcolor{white}{ using the hint. Let \begin{equation} \begin{split} \label{eq:nogapostpara} B&=\frac{bc+n\bar x}{c+n}, \quad\quad C=c+n,\\ G&=g+\frac{n}{2}, \quad\quad H=h+\frac{cn(\bar x-b)^2}{2(c+n)}+\frac{ns^2}{2}. \end{split} \end{equation} Then the posterior density is \begin{equation*} %\label{eq:nogaposterior} \begin{split} \pi(\mu,\tau|\xvec)&\propto\tau^{G-\frac{1}{2}}\exp\left\{-\frac{\tau}{2} \left[C(\mu-B)^2+2H\right]\right\},\\ &\hspace{7cm}\mu\in\R, \tau>0 \end{split} \end{equation*} Notice that this posterior density %$\eqref{eq:nogaposterior}$ is of the same form as the prior density $\eqref{eq:nogaprior}$. Therefore, we can conclude that the posterior distribution is \[ \left.\mutau\right|\xvec\sim NGa(B,C,G,H). \] Thus, the $NGa$ distribution is conjugate to this data model.} \vspace{10cm} }} %\end{soln} \newpage \subsection{Marginal distributions} Suppose $(\mu,\tau)^T\sim NGa(b,c,g,h)$. From the definition of the $NGa$ distribution we know that $\tau\sim Ga(g,h)$. This also means that $\sigma=1/\sqrt\tau\sim\textit{Inv-Chi}(g,h)$; see formula sheet. The (marginal) density for $\mu$ is, for $\mu\in\R$ \begin{align*} \pi(\mu)&=\int_0^\infty \pi(\mu,\tau)\,d\tau \\ &\propto\int_0^\infty \tau^{g-\frac{1}{2}}\exp\left\{-\frac{\tau}{2} \left[c(\mu-b)^2+2h\right]\right\}\,d\tau. \end{align*} Now, as the integral of a gamma density over its entire range is one, we have \[ \int_0^\infty\frac{b^a\theta^{a-1}e^{-b\theta}}{\Gamma(a)}\,d\theta=1 \quad\Longrightarrow\quad \int_0^\infty\theta^{a-1}e^{-b\theta}\,d\theta =\frac{\Gamma(a)}{b^a}. \] Therefore, for $\mu\in\R$ \begin{align*} \pi(\mu) &\propto\int_0^\infty \tau^{g+\frac{1}{2}-1}\exp\left\{-\frac{\tau}{2} \left[c(\mu-b)^2+2h\right]\right\}\,d\tau\\ &\propto\frac{\Gamma\left(g+\frac{1}{2}\right)} {\left[\left\{c(\mu-b)^2+2h\}/2\right\}\right]^{g+\frac{1}{2}}} \\ &\propto h^{-g-1/2}\left\{1+\frac{c(\mu-b)^2}{2h}\right\}^{-g-1/2}\\ &\propto\left\{1+\frac{c(\mu-b)^2}{2h}\right\}^{-\frac{2g+1}{2}}. \end{align*} Comparing this density with that of the generalised $t$--distribution (see formula sheet) gives \begin{equation} \label{eq:margprior} \mu\sim t_{2g}\left(b,\frac{h}{gc}\right). \end{equation} Thus, marginally, the prior distribution for $\mu$ is a $t$--distribution. Similar calculations can be used to determine the (marginal) posterior distributions. \subsection*{Summary of marginal distributions} The prior $\mutau\sim NGa(b,c,g,h)$ has marginal distributions \begin{itemize} \item $\mu\sim t_{2g}\left(b,\frac{h}{gc}\right)$ \medskip \item $\tau\sim Ga(g,h)$; also $\sigma=1/\sqrt\tau\sim\textit{Inv-Chi}(g,h)$. \end{itemize} \newpage The posterior $\left.\mutau\right|\xvec\sim NGa(B,C,G,H)$ has marginal distributions \begin{itemize} \item $\mu|\xvec\sim t_{2G}\left(B,\frac{H}{GC}\right)$ \medskip \item $\tau|\xvec\sim Ga(G,H)$; also $\sigma|\xvec\sim\textit{Inv-Chi}(G,H)$. \end{itemize} It can be shown that the posterior mean of $\mu$ is greater than its prior mean if and only if the sample mean (likelihood mode) is greater than its prior mean, that is, \begin{equation*} E(\mu|\xvec)>E(\mu)\quad\iff\quad \bar x >b. \end{equation*} The relationships between the prior and posterior variance of $\mu$ and mean and variance of $\tau$ and of $\sigma$ are rather more complex. \begin{example}~\\ Recall Example~\ref{ex:cavendish} on the earth's density. Previously \label{ex:cavbiv} we assumed that the measurements followed a $N(\mu,0.2^2)$ distribution, that is, the standard deviation of the measurements was known to be~$0.2\,g/cm^3$. Now we consider the case where this standard deviation is unknown and determine posterior distributions using the theory in section~\ref{sec:nochtheory}. Before we can proceed, we must specify the parameters in the $NGa(b,c,g,h)$ prior distribution for $(\mu,\tau)$. In the previous analysis, we assumed that the population measurement precision was $\tau=1/0.2^2=25$ and assumed a $N(5.41,0.4^2)$ prior distribution for the population mean, that is, $\mu|\tau=25\sim N(5.41,0.4^2)$. Choice of $b$ and $c$: the conditional prior distribution for $\mu$ is $\mu|\tau\sim N\{b,1/(c\tau)\}$ and so matching the prior distributions for~$\mu$ (when $\tau=25$) gives $b=5.41$ and $c=0.25$. Choice of $g$ and $h$: the marginal prior distribution for $\tau$ is $\tau\sim Ga(g,h)$. Previously, we assumed $\tau=25$ (with $Var(\tau)=0$) and so take this value as the prior mean: $E(\tau)=25$. Suppose we also decide that $Var(\tau)=250$. These two requirements give $g=2.5$ and $h=0.1$. Therefore, we will assume the prior distribution \[ \mutau\sim NGa(5.41,0.25,2.5,0.1). \] We have seen that if $(\mu,\tau)^T\sim NGa(b,c,g,h)$ then the marginal distribution of $\mu$ is $\mu\sim t_{2g}\bigl\{b,h/(gc)\bigr\}$. Therefore, with this choice of prior distribution, the marginal prior distribution for $\mu$ is \[ \mu\sim t_5(5.41,0.16). \] Figure~\ref{fig:cav-prior} shows the close match between the new (marginal) prior distribution for $\mu$ and that used previously. \begin{figure}[ht] \centering \includegraphics[width=0.65\textwidth]{cav-prior-mu} %\includegraphics[width=0.5\textwidth]{cav-prior-tau} %\includegraphics[width=0.5\textwidth]{cav-prior-sigma} %\caption{Marginal prior densities for $\mu$ (new version (solid) and previous version (dashed)), $\tau$ and $\sigma$} \caption{Marginal prior density for $\mu$: new version (solid) and previous version (dashed)} \label{fig:cav-prior} \end{figure} Determine the posterior distribution for $(\mu,\tau)^T$. Also determine the marginal prior distribution for $\tau$ and for $\sigma$, and the marginal posterior distribution for each of $\mu$, $\tau$ and~$\sigma$. \end{example} \framebox[17cm]{ \parbox[t]{16cm}{ \vspace{0.4cm} \huge \ding{46} \normalsize \textcolor{white}{ We can combine the information in the $NGa(5.41,0.25,2.5,0.1)$ prior distribution for $(\mu,\tau)^T$ with that in the data ($n=23$, $\bar x=5.4848$, $s=0.1882$) using the results in section~\ref{sec:nochtheory} to obtain a $NGa(B,C,G,H)$ posterior distribution, where \begin{align*} B&=\frac{bc+n\bar x}{c+n} =\frac{(5.41\times 0.25)+(23\times 5.4848)}{23.25}=5.4840,\\ C&=c+n=23.25,\\ G&=g+\frac{n}{2}=14,\\ H&=h+\frac{cn(\bar x-b)^2}{2(c+n)}+\frac{ns^2}{2} =0.1+\frac{5.75}{46.5}(5.4848-5.41)^2+11.5\times 0.1882^2=0.5080. \end{align*} The marginal prior distributions for $\tau$ and $\sigma$ are \begin{align*} \tau&\sim Ga(g,h)\equiv Ga(2.5,0.1)\\ \noalign{\medskip} \sigma&\sim\textit{Inv-Chi}(g,h)\equiv\textit{Inv-Chi}(2.5,0.1) \end{align*} Also the marginal posterior distributions for $\mu$, $\tau$ and $\sigma$ are \begin{align*} \mu|\xvec&\sim t_{2G}\left(B,\frac{H}{GC}\right)\equiv t_{28}(5.4840,0.001561)\\ \tau|\xvec&\sim Ga(G,H)\equiv Ga(14,0.5080)\\ \noalign{\medskip} \sigma|\xvec&\sim\textit{Inv-Chi}(G,H)\equiv\textit{Inv-Chi}(14,0.5080) \end{align*}} %%Hack for vspace \vspace{2cm} }} Plots of the (marginal) prior and posterior distributions of $\mu$, $\tau$ and $\sigma$ are given in Figure~\ref{fig:cav-post}. Note that the (marginal) prior and posterior distributions for $\sigma$ can be determined from that of $\tau$. We can also examine the joint prior and posterior distributions for $(\mu,\tau)^T$ via the contour plots of their densities to see if there is any change in the dependence structure; see Figure~\ref{fig:cav-post-joint}. This figure is produced by using the \texttt{R} command \texttt{NGacontour} in the \texttt{nclbayes} package as follows: \begin{verbatim} mu=seq(4.5,6.5,len=1000) tau=seq(0,71,len=1000) NGacontour(mu,tau,b,c,g,h,lty=3) NGacontour(mu,tau,B,C,G,H,add=TRUE) \end{verbatim} in which the variables \texttt{b,c,g,h,B,C,G,H} have already been set to their prior/posterior values. A careful look at the values of the contour levels plotted shows that the highest contour level plotted for the prior density is 0.024 and the lowest level for the posterior density is 0.05. From this we can conclude that the posterior distribution is far more concentrated than the prior distribution. Also the contours for the posterior distribution are much more elliptical than those for the prior distribution. This indicates a change in the dependence structure. However, the main changes shown by the figure are in the mean and variability of $\mu$ and~$\tau$. \begin{figure}[t!] \includegraphics[width=0.5\textwidth]{cav-post-mu}\\ \includegraphics[width=0.5\textwidth]{cav-post-tau} \includegraphics[width=0.5\textwidth]{cav-post-sigma} \caption{Prior (dashed) and posterior (solid) densities for $\mu$, $\tau$ and $\sigma$} \label{fig:cav-post} \end{figure} \begin{figure}[t] \centering\bigskip \includegraphics[width=0.7\textwidth]{cav-post-joint} %\includegraphics[width=0.7\textwidth]{cav-post-joint-sigma} \caption{Contour plot of the prior (dashed) and posterior (solid) densities for $(\mu,\tau)^T$.} \label{fig:cav-post-joint} \end{figure} Wikipedia tells us that the actual mean density of the earth is $5.515\,g/cm^3$. We can determine the (posterior) probability that the mean density is within 0.1 of this value as follows. We already know that $\mu|\xvec\sim t_{28}(5.484,0.001561)$ and so we can calculate \[ Pr(5.415<\mu<5.615|\xvec)=0.9529 \] using \verb$pgt(5.615,28,5.484,0.001561)-pgt(5.415,28,5.484,0.001561)$. Without the data, the only basis for determining the earth's density is via the prior distribution. Here the prior distribution is $\mu\sim t_5(5.41,0.16)$ and so the (prior) probability that the mean density is within 0.1 of the (now known) true value is \[ Pr(5.415<\mu<5.615)=0.1802, \] calculated using \verb$pgt(5.615,5,5.41,0.16)-pgt(5.415,5,5.41,0.16)$. These probability calculations demonstrate that the data have been very informative and changed our beliefs about the earth's density. \newpage \section{Confidence intervals and regions} \label{sec:confint} \begin{example}~\\ Determine the $100(1-\alpha)\%$ highest density interval (HDI) for the population mean~$\mu$ in terms of quantiles of the standard $t$-distribution. \end{example} \framebox[17cm]{ \parbox[t]{16cm}{ \vspace{0.4cm} \huge \ding{46} \normalsize \textcolor{white}{ The marginal posterior distribution is $\mu|\xvec\sim t_{2G}\left(B,\frac{H}{GC}\right)$. This is a symmetric distribution and so the HDI is an equi-tailed interval. Therefore the HDI $(\ell,u)$ for $\mu$ must satisfy \[ Pr(\mu<\ell|\xvec)=\alpha/2\qquad\text{and}\qquad Pr(\mu>u|\xvec)=\alpha/2. \] Now, given the data $\xvec$ \begin{align*} \frac{\mu-B}{\sqrt{H/(GC)}}&\sim t_{2G} \intertext{and so} Pr(\mu>u|\xvec)=\alpha/2\quad&\Rightarrow\quad Pr\left(\left.\frac{\mu-B}{\sqrt{H/(GC)}}>\frac{u-B}{\sqrt{H/(GC)}}\right|\xvec\right)=\alpha/2\\ &\Rightarrow\quad \frac{u-B}{\sqrt{H/(GC)}}=t_{2G;\alpha/2} \intertext{where $t_{2G;p}$ is the upper $p$ point of the $t_{2G}$ distribution. Therefore} u&=B+t_{2G;\alpha/2}\sqrt{\frac{H}{GC}}. \intertext{Similar calculations give} \ell&=B+t_{2G;1-\alpha/2}\sqrt{\frac{H}{GC}}=B-t_{2G;\alpha/2}\sqrt{\frac{H}{GC}} \end{align*} since the $t$ distribution is symmetric about zero. Thus the $100(1-\alpha)\%$ HDI for $\mu$ is \[ \left(B-t_{2G;\alpha/2}\sqrt{\frac{H}{GC}},~B+t_{2G;\alpha/2}\sqrt{\frac{H}{GC}}\right). \]} \vspace{2cm} }} \newpage These intervals can be calculated easily using the \verb$R$ function \texttt{qgt} in the package \texttt{nclbayes}. For example, the prior and posterior 95\% HDIs for $\mu$ can be calculated using \begin{center} \texttt{c(qgt(0.025,2*g,b,h/(g*c)),qgt(0.975,2*g,b,h/(g*c)))}\\ \texttt{c(qgt(0.025,2*G,B,H/(G*C)),qgt(0.975,2*G,B,H/(G*C)))} \end{center} Determining a highest density interval (HDI) for the population precision~$\tau$ or standard deviation~$\sigma$ is more complicated as their posterior distributions are not symmetric. The (marginal) posterior for $\tau$ is $\tau|\xvec\sim Ga(G,H)$ and the (marginal) posterior for $\sigma$ is $\sigma|\xvec\sim\textit{Inv-Chi}(G,H)$. HDIs can be found by using the \texttt{R} functions \texttt{hdiGamma} and \texttt{hdiInvchi} in the package \texttt{nclbayes}. More standard equi-tailed confidence intervals can be found using the functions \texttt{qgamma} and \texttt{qinvchi}. For example, the prior and posterior 95\% HDIs for $\tau$ can be calculated using \texttt{R} commands \texttt{hdiGamma(0.95,g,h)} and \texttt{hdiGamma(0.95,G,H)}, and those for $\sigma$ using commands \texttt{hdiInvchi(0.95,g,h)} and \texttt{hdiInvchi(0.95,G,H)}. The 95\% equi-tailed confidence intervals are calculated in a similar way to the HDIs for $\mu$ above. So for $\tau$, the prior and posterior intervals are calculated using \begin{center} \texttt{c(qgamma(0.025,g,h),qgamma(0.975,g,h))}\\ \texttt{c(qgamma(0.025,G,H),qgamma(0.975,G,H))} \end{center} and those for $\sigma$ using \begin{center} \texttt{c(qinvchi(0.025,g,h),qinvchi(0.975,g,h))}\\ \texttt{c(qinvchi(0.025,G,H),qinvchi(0.975,G,H))} \end{center} The numerical values for the prior and posterior 95\% intervals for the analysis in Example~\ref{ex:cavbiv} are given in Table~\ref{tab:cavbivint}. Notice that there is little difference between the posterior HDI and equi-tailed intervals for~$\tau$ and for~$\sigma$, whereas the prior intervals are fairly different. This is because the prior distributions are quite skewed but the posterior distributions are fairly symmetric; see Figure~\ref{fig:cav-post}. \begin{table} \begin{center} \begin{tabular}{cccl} & Prior & Posterior \\ $\mu$: & $(4.3818,6.4382)$ & $(5.4031,5.5649)$ \\ \noalign{\medskip} $\tau$: & {\d}$(1.4812,55.9573)$ & $(14.0193,42.2530)$ & $\gets$ HDI\\ & {\d}$(4.1561,64.1625)$ & $(15.0674,43.7625)$ \\ \noalign{\medskip} $\sigma$: & $(0.1062,0.4246)$ & $(0.1466,0.2505)$ & $\gets$ HDI\\ & $(0.1248,0.4905)$ & $(0.1512,0.2576)$ \end{tabular} \end{center} \caption{Prior and posterior 95\% intervals for the analysis in Example~\ref{ex:cavbiv}} \label{tab:cavbivint} \end{table} In Bayesian inference it can also be useful to determine (joint) confidence regions for several parameters, in this case, for $(\mu,\tau)^T$. In general this is a difficult problem to solve mathematically, and it is in this case. %%Hack for vspace \newpage \begin{example}~\\ Determine a joint confidence region for $(\mu,\tau)^T$. \end{example} \framebox[17cm]{ \parbox[t]{16cm}{ \vspace{0.4cm} \huge \ding{46} \normalsize \textcolor{white}{ We know that the (joint) prior distribution for these parameters is \[ \mutau\sim NGa(b,c,g,h). \] Therefore an HDI--type confidence region takes the form \begin{align*} &\left\{\mutau:~\pi(\mu,\tau)>k\right\}\\ &=\left\{\mutau:~\tau^{g-\frac{1}{2}}\exp\left\{-\frac{\tau}{2}\left[c(\mu-b)^2+2h\right]\right\}>k'\right\}\\ %\end{align*} %\begin{align*} &=\left\{\mutau:~\left(g-\frac{1}{2}\right)\log\tau -\frac{\tau}{2}\left[c(\mu-b)^2+2h\right]>k''\right\}\\ &=\left\{\mutau:~\frac{\tau c(\mu-b)^2}{2}+h\tau -\left(g-\frac{1}{2}\right)\log\tau