\documentclass[12pt,a4paper]{article} \usepackage{amsthm,amsmath,amsfonts} \usepackage{latexsym} \usepackage{fancyhdr} \usepackage{enumerate} %\usepackage{marginnote} \usepackage{graphicx} \usepackage{skull} \usepackage{hyperref} \usepackage{fancybox} \usepackage{xspace} %%From web for code highlighting:: \usepackage{listings} \usepackage{cmbright} \addtolength{\parskip}{1ex} \baselineskip = 6mm %\parskip = 2mm \textheight 23.6cm \textwidth 16.6cm \oddsidemargin 0.0cm \evensidemargin 0.0cm \topmargin -1.4cm %\headsep -.60in \setlength{\parindent}{0cm} \pagestyle{fancy} \lhead{} \chead{} \rhead{} \lfoot{} \cfoot{\thepage} \rfoot{} \rhead{} \newcommand{\pr}[2][]{\ensuremath{\mathrm{Pr}_{#1}\!\left(#2\right)}} \newenvironment{myhighlight}{\textit}{} \renewenvironment{emph}{\textbf}{} \newcommand{\Exp}[2][]{\ensuremath{\mathrm{Exp}\left(#2\right)}} \newcommand{\Bin}[2][]{\ensuremath{\mathrm{Bin}\left(#2\right)}} \newcommand{\Po}[2][]{\ensuremath{\mathrm{Po}\left(#2\right)}} \newcommand{\cov}[2][]{\ensuremath{\mathrm{Cov}\left(#2\right)}} \newcommand{\cor}[2][]{\ensuremath{\mathrm{Cor}\left(#2\right)}} \newcommand{\diag}[2][]{\ensuremath{\mathrm{diag}\left(#2\right)}} \newcommand{\given}{\ensuremath{\ |\ }} \newcommand{\stationary}{\ensuremath{\vec{\pi}}} \newcommand{\identity}{\ensuremath{\vec{I}}} \renewcommand{\vec}[1]{\underline{#1}} \newcommand{\remove}[1]{} %\newcommand{\cover}[2]{\hi{#1}} \newcommand{\cover}[2]{#2} \newcommand{\hi}[2][]{\textbf{#2}} \newcommand{\bigfont}{\fontsize{1.45cm}{1.5cm}\selectfont} \newcommand{\announce}{\setbeamercolor{structure}{fg=green!60!black}} \newcommand{\sol}{\setbeamercolor{structure}{fg=purple!80!black}} \newcommand{\normal}{\setbeamercolor{structure}{fg=blue!60!green}} \newcommand{\vsp}{\vspace{0.3cm}} \newcommand{\vspl}{\vspace{0.15cm}} \newcommand{\xvec}{\boldsymbol{x}} \newcommand{\yvec}{\vec{y}} \newcommand{\Xvec}{\boldsymbol{X}} \newcommand{\Yvec}{\vec{Y}} \newcommand{\Evec}{\vec{E}} \newcommand{\tvec}{\bodsymbol{t}} \newcommand{\Tvec}{\bodsymbol{T}} \newcommand{\mutau}{\begin{pmatrix}\mu\\\tau\end{pmatrix}} \newcommand{\thetavec}{\vec{\theta}} \newcommand{\thetavechat}{\vec{\hat\theta}} \newcommand{\thetahat}{\hat\theta} \newcommand{\xvece}{\boldsymbol{x}=(x_1,x_2,\ldots,x_n)^T} \newcommand{\Xvece}{\boldsymbol{X}=(X_1,X_2,\ldots,X_n)^T} \newcommand{\thetavece}{\vec{\theta}=(\theta_1,\theta_2,\ldots,\theta_p)^T} \newcommand{\exxvec}{$x_1,x_2,\ldots,x_n$\xspace} \newcommand{\exxvecdot}{$x_1,x_2,\ldots,x_n$.\ } \newcommand{\exXvec}{$X_1,X_2,\ldots,X_n$\ } \newcommand{\muvec}{\vec{\mu}} \newtheoremstyle{exmp}% hnamei {3pt}% hSpace abovei {3pt}% hSpace belowi {}% hBody fonti {}% hIndent amounti1 {\bf}% hTheorem head fonti {:}% hPunctuation after theorem headi {.5em}% hSpace after theorem headi2 {}% hTheorem head spec (can be left empty, meaning `normal')i \theoremstyle{exmp} \newtheorem{example}{\textcolor{purple}{\textbf{Example}}}[section] %\def \var{{\textrm{Var}}} \newcommand{\E}[2][]{\ensuremath{\mathrm{E}\left[#2\right]}} \def \R{\mathbb{R}} \def \Z{\mathbb{Z}} \def \proof{\noindent{\bf Proof: }} \def \eproof{\qed\vskip .3cm} \def \union{\cup} \def \intersect{\cap} \newcommand{\ex}[2][]{\ensuremath{\mathrm{E}_{#1}\!\left[#2\right]}} \newcommand{\var}[2][]{\ensuremath{\mathrm{Var}_{#1}\left(#2\right)}} %\renewcommand{\det}[1][]{\ensuremath{\mathrm{det}\left(#1\right)}} \newtheorem{defn}{Definition} %\newtheorem*{example}{Example} \newtheorem{result}{Result}[section] \newtheorem{algorithm}{Algorithm}[section] \usepackage{Sweave} \usepackage[usenames, dvipsnames]{xcolor} \begin{document} %%From web for code/highlighting \definecolor{mygreen}{rgb}{0,0.6,0} \definecolor{mygray}{rgb}{0.5,0.5,0.5} \definecolor{mymauve}{rgb}{0.58,0,0.82} \definecolor{light-gray}{gray}{0.95} \definecolor{LEEgreen}{rgb}{0.04,0.667,0.137} \lstset{ backgroundcolor=\color{light-gray}, % choose the background color; you must add \usepackage{color} or \usepackage{xcolor}; should come as last argument basicstyle=\footnotesize, % the size of the fonts that are used for the code breakatwhitespace=false, % sets if automatic breaks should only happen at whitespace breaklines=true, % sets automatic line breaking captionpos=b, % sets the caption-position to bottom commentstyle=\color{mygreen}, % comment style deletekeywords={...}, % if you want to delete keywords from the given language escapeinside={\%*}{*)}, % if you want to add LaTeX within your code extendedchars=false, % lets you use non-ASCII characters; for 8-bits encodings only, does not work with UTF-8 frame=single, % adds a frame around the code keepspaces=true, % keeps spaces in text, useful for keeping indentation of code (possibly needs columns=flexible) keywordstyle=\color{blue}, % keyword style language=R, % the language of the code morekeywords={*,...}, % if you want to add more keywords to the set numbers=left, % where to put the line-numbers; possible values are (none, left, right) numbersep=5pt, % how far the line-numbers are from the code numberstyle=\tiny\color{mygray}, % the style that is used for the line-numbers rulecolor=\color{black}, % if not set, the frame-color may be changed on line-breaks within not-black text (e.g. comments (green here)) showspaces=false, % show spaces everywhere adding particular underscores; it overrides 'showstringspaces' showstringspaces=false, % underline spaces within strings only showtabs=false, % show tabs within strings adding particular underscores stepnumber=1, % the step between two line-numbers. If it's 1, each line will be numbered stringstyle=\color{mymauve}, % string literal style tabsize=2, % sets default tabsize to 2 spaces title=\lstname % show the filename of files included with \lstinputlisting; also try caption instead of title } %\SweaveOpts{concordance=TRUE} \pagenumbering{gobble} \setcounter{page}{-3} % to get contents page labelled (i) %%%Practicals %\section*{Aims} %This introductory practical is split into two parts: %\begin{itemize} %\item In the first hour or so, we will focus on the basics of R, revising some of the key concepts from last year's MAS1802 material %\item In the second hour we will focus on material from this week's lecture -- using R to simulate from standard probability distributions, as well as more complex models including mixture distributions %\end{itemize} %\vspace{0.25cm} %\noindent Please read through all of this handout \textit{very carefully}, and follow the instructions \textit{meticulously}. Whenever %you are typing commands into R, make sure you do so \textit{extremely carefully}. If you get an error message, %try typing the code in again, just to make sure you've not made a spelling mistake or other simple typo. A %common error is to forget that R is case sensitive, so use upper/lower cases exactly when you're told to %do so. %\vspace{0.25cm} %\noindent For all R practicals, make sure you've got your lecture notes handy. %\section*{Getting started} %\noindent Log-in to a PC in the usual way. Now open Rstudio. If you can't remember how to do this, click on the ``Start'' button in the bottom %left-hand-corner and search for ``rstudio''. Once you've opened Rstudio you should see four panels; if there is no text editor (usually the top-left panel), click on the green button with a white plus symbol and open a new R script. Remember -- to execute a line of code in the text editor, hit \texttt{Ctrl}-\texttt{Enter}. %\vspace{0.25cm} %\noindent It is good practice to always use the text editor, rather than type into the console window (bottom- %left panel) directly. This is especially true in future practicals when you may be building large functions; it %is much easier to de-bug and edit code in the text editor than in the console window. %\vspace{0.25cm} %\noindent \textbf{Hint: To commit several lines of code in one go, highlight your block of code in the text editor and hit \texttt{Ctrl}-\texttt{Enter} just once!} %\vspace{0.25cm} %\noindent Now you've got Rstudio open, you're ready to go! Try out \textit{all} of the following exercises. %\newpage \lhead{MAS3902} \rhead{Bayesian Inference} \section*{Bayesian Inference: Assignment 4} \noindent Answers to the starred (\textbf{*}) questions should be submitted to Canvas by no later than \textcolor{red}{\bf 4pm, Friday 3rd May 2024}. Solutions to non-assessed questions will be made available in class. \subsection*{Question 1*} The following probability density function (PDF) is often used in geology, for example to model the size of meteorites: \[ f(x|\alpha,\theta) =\frac{\alpha\theta^\alpha}{x^{\theta+1}},\quad x>\theta \] where $\alpha>0$ and $\theta>0$. Suppose \exxvec are a random sample of meteorite sizes; suppose further, from previous studies, we assume that $\alpha=1$ in the PDF above. \begin{itemize} \item [(a)] Determine the Jeffreys prior for $\theta$. \item [(b)] Determine the posterior distribution for $\theta$, assuming a very large sample. \end{itemize} \subsection*{Question 2*} The salaries $X$ of a random sample of $n$ workers in a city follow a reparameterised log-normal $RLN(\mu,\alpha)$ distribution, where $(\mu,\alpha)\in\mathbb{R}^2$ and \[ f(x|\mu,\alpha)= \frac{e^{\alpha/2}}{x\sqrt{2\pi}} \exp\left\{-\frac{e^\alpha}{2}(\log x-\mu)^2\right\},\quad x>0. \] Find the Jeffreys prior for $(\mu, \alpha)^{T}$. \textcolor{gray}{ Hint: If $X\sim RLN(\mu,\alpha)$ then $E(\log X)=\mu$ and $Var(\log X)=e^{-\alpha}$. } \subsection*{Question 3} A random sample $\xvece$ is taken from a log-normal $LN(\mu,\sigma^2)$ distribution. %Find the Jeffreys prior. \begin{itemize} \item [(a)] Find the Jeffreys prior. \item [(b)] Is this a proper prior distribution? Explain your answer. \end{itemize} \newpage \subsection*{Question 4} Suppose that you have a random sample $\xvece$ from a $Ga(\alpha,\lambda)$ distribution. \begin{enumerate}[(a)] \item Determine the asymptotic posterior distribution for $\thetavec=(\alpha,\lambda)^T$. Hint: you should define the maximum likelihood estimates $\hat\alpha$ and $\hat\lambda$ in terms of the equations they must satisfy, with these involving the sample mean $\bar x$, the geometric mean $\bar x_g$ and the digamma function $\psi(x)=\Gamma'(x)/\Gamma(x)$. \item Suppose a sample of size $n=100$ gives sample mean $\bar x=3.0$ and geometric mean $\bar x_g=2.5$ leading to maximum likelihood estimates $\hat\alpha=2.8983$ and $\hat\lambda=0.9661$. Determine the asymptotic posterior distribution for $\thetavec$. Also calculate the asymptotic posterior correlation between $\alpha$ and $\lambda$. Hint: you will need to use the \texttt{R} functions \texttt{digamma} for $\psi(\cdot)$ and \texttt{trigamma} for $\psi'(\cdot)$. \end{enumerate} \newpage \section*{Solutions} \subsection*{Question 3} \begin{enumerate}[(a)] \item The likelihood function is \begin{align*} f(\xvec|\mu,\sigma)&=\prod_{i=1}^n\frac{1}{\sqrt{2\pi}\sigma x_i} \exp\left\{-\frac{1}{2\sigma^2}(\log x_i-\mu)^2\right\}\\ &\propto\sigma^{-n}\exp\left\{-\frac{1}{2\sigma^2}\sum_{i=1}^n(\log x_i-\mu)^2\right\}. \end{align*} We have \begin{align*} \log f(\xvec|\mu,\sigma)&=-n\log\sigma-\frac{1}{2\sigma^2}\sum_{i=1}^n(\log x_i-\mu)^2\\ \Longrightarrow\quad&\frac{\partial}{\partial\mu} \log f(\xvec|\mu,\sigma)=\frac{1}{\sigma^2}\sum_{i=1}^n(\log x_i-\mu)\\ &\frac{\partial}{\partial\sigma} \log f(\xvec|\mu,\sigma)=-\frac{n}{\sigma}+\frac{1}{\sigma^3}\sum_{i=1}^n(\log x_i-\mu)^2\\ \Longrightarrow\quad&\frac{\partial^2}{\partial\mu^2} \log f(\xvec|\mu,\sigma)=-\frac{n}{\sigma^2}\\ &\frac{\partial^2}{\partial\mu\partial\sigma} \log f(\xvec|\mu,\sigma)=-\frac{2}{\sigma^3}\sum_{i=1}^n(\log x_i-\mu)\\ &\frac{\partial^2}{\partial\sigma^2} \log f(\xvec|\mu,\sigma)=\frac{n}{\sigma^2}-\frac{3}{\sigma^4}\sum_{i=1}^n(\log x_i-\mu)^2. \end{align*} Now, given our knowledge of the log-normal distribution (would be provided as a hint in the exam), $\log X_i\sim N(\mu,\sigma^2)$ and so \[ E(\log X_i|\mu,\sigma)=\mu\quad\text{and}\quad E\{(\log X_i-\mu)^2|\mu,\sigma)=Var(\log X_i|\mu,\sigma)=\sigma^2. \] Therefore, the elements of Fisher's information matrix are \begin{align*} I_{11}&=E_{\Xvec|\mu,\sigma} \left[-\frac{\partial^2}{\partial\mu^2} \log f(\Xvec|\mu,\sigma)\right]=\frac{n}{\sigma^2}\\ I_{12}&=E_{\Xvec|\mu,\sigma} \left[-\frac{\partial^2}{\partial\mu\partial\sigma} \log f(\Xvec|\mu,\sigma)\right] =\frac{2n}{\sigma^3}\,E_{X|\mu,\sigma}[\log X-\mu]=0\\ I_{22}&=E_{\Xvec|\mu,\sigma} \left[-\frac{\partial^2}{\partial\sigma^2} \log f(\Xvec|\mu,\sigma)\right]=-\frac{n}{\sigma^2}+\frac{3n}{\sigma^4}\,E_{X|\mu,\sigma}[(\log X-\mu)^2]=\frac{2n}{\sigma^2}. \end{align*} Hence, Fisher's information matrix is \[ I(\mu,\sigma) =\frac{n}{\sigma^2}\begin{pmatrix} 1 & 0 \\ 0 & 2 \\ \end{pmatrix}, \] from which we obtain the Jeffreys prior density as \begin{align*} \pi(\mu,\sigma)&\propto|I(\mu,\sigma)|^{1/2} \\ &\propto\frac{\sqrt{2}n}{\sigma^2},\qquad-\infty<\mu<\infty,\quad\sigma>0 \\ &\propto\frac{1}{\sigma^2},\qquad-\infty<\mu<\infty,\quad\sigma>0. \end{align*} \item This is not a proper distribution as $\int\pi(\mu,\sigma)\,d\mu d\sigma$ is a divergent integral. Also, $\mu$ and $\sigma$ are independent as the joint density can be factorised as \[ \pi(\mu,\sigma)=\pi(\mu)\,\pi(\sigma) \] where \[ \pi(\mu)\propto 1,\quad -\infty<\mu<\infty \qquad\text{and}\qquad \pi(\sigma)\propto\frac{1}{\sigma^2},\quad\sigma>0. \] \end{enumerate} \subsection*{Question 4} The likelihood function is \begin{align*} f(\vec{x}|\alpha,\lambda) &=\prod_{i=1}^n \displaystyle{\lambda^\alpha x_i^{\alpha-1}e^{-\lambda x_i}}{\Gamma(\alpha)} =\displaystyle{\lambda^{n\alpha}\bar x_g^{n(\alpha-1)}e^{-n\bar x\lambda}} {\Gamma(\alpha)^n}, \intertext{where $\bar x_g=\sqrt[n]{\prod x_i}$, and so the log-likelihood function is} \log f(\vec{x}|\alpha,\lambda) &=n\alpha\log\lambda +n(\alpha-1)\log\bar x_g -n\bar x\lambda -n\log\Gamma(\alpha). \intertext{Therefore} \frac{\partial}{\partial\alpha}\log f(\vec{x}|\alpha,\lambda) &=n\log\lambda+n\log\bar x_g-n\psi(\alpha)\\ \frac{\partial}{\partial\lambda}\log f(\vec{x}|\alpha,\lambda)&=\frac{n\alpha}{\lambda}-n\bar x \intertext{and so the maximum likelihood estimates $(\hat\alpha,\hat\lambda)$ satisfy} \log\hat\lambda+\log\bar x_g-\psi(\hat\alpha)&=0\qquad\text{and}\qquad \frac{\hat\alpha}{\hat\lambda}-\bar x=0, \intertext{that is,} \log\hat\lambda+\log\bar x_g-\psi(\hat\lambda\bar x)&=0\qquad\text{and}\qquad \hat\alpha=\hat\lambda\bar x. \end{align*} The second derivatives are \begin{align*} \frac{\partial^2}{\partial\alpha^2}\log f(\vec{x}|\alpha,\lambda) &=-n\psi'(\alpha)\\ \frac{\partial^2}{\partial\lambda^2}\log f(\vec{x}|\alpha,\lambda) &=-\frac{n\alpha}{\lambda^2}\\ \frac{\partial^2}{\partial\alpha\partial\lambda}\log f(\vec{x}|\alpha,\lambda) &=\frac{n}{\lambda}. \end{align*} Therefore \begin{align*} J_{11}(\hat\alpha,\hat\lambda)&=n\psi'(\hat\alpha),\qquad J_{12}(\hat\alpha,\hat\lambda)=-n/\hat\lambda,\qquad J_{22}(\hat\alpha,\hat\lambda)=n\hat\alpha/\hat\lambda^2 \intertext{and so} J(\hat\alpha,\hat\lambda)&= \begin{pmatrix} n\psi'(\hat\alpha) & -n/\hat\lambda \\ -n/\hat\lambda & n\hat\alpha/\hat\lambda^2 \\ \end{pmatrix} =n\begin{pmatrix} \psi'(\hat\alpha) & -1/\hat\lambda \\ -1/\hat\lambda & \hat\alpha/\hat\lambda^2 \\ \end{pmatrix} \end{align*} whence \begin{align*} J(\hat\alpha,\hat\lambda)^{-1}&= \displaystyle{1}{n\left\{\displaystyle{\hat\alpha\psi'(\hat\alpha)}{\hat\lambda^2}-\displaystyle{1} {\hat\lambda^2}\right\}} \begin{pmatrix} \hat\alpha/\hat\lambda^2 & 1/\hat\lambda \\ 1/\hat\lambda & \psi'(\hat\alpha) \\ \end{pmatrix} =\frac{1}{n\{\hat\alpha\psi'(\hat\alpha)-1\}} \begin{pmatrix} \hat\alpha & \hat\lambda \\ \hat\lambda & \hat\lambda^2\psi'(\hat\alpha) \\ \end{pmatrix}. \end{align*} Therefore, the asymptotic posterior distribution distribution (as $n\to\infty$) is \[ \left. \begin{pmatrix}\alpha \\ \lambda \\ \end{pmatrix}\right|\xvec\sim N\left\{\begin{pmatrix}\hat\alpha \\ \hat\lambda \\ \end{pmatrix},~ \frac{1}{n\{\hat\alpha\psi'(\hat\alpha)-1\}} \begin{pmatrix} \hat\alpha & \hat\lambda \\ \hat\lambda & \hat\lambda^2\psi'(\hat\alpha) \\ \end{pmatrix} \right\}. \] (ii) We have $n=100$, $\bar x=3.0$ and $\bar x_g=2.5$ leading to maximum likelihood estimates $\hat\alpha=2.8983$ and $\hat\lambda=0.9661$. Note that the value of $\hat\lambda$ (and hence that of $\hat\alpha$) can be found using the R code \begin{verbatim} zerofn=function(x){(log(x)+log(2.5)-digamma(3*x))^2} optimise(zerofn,interval=c(0,5)) \end{verbatim} Now $\psi'(\hat\alpha)=0.4112474$ and so \begin{align*} \frac{1}{n\{\hat\alpha\psi'(\hat\alpha)-1\}} \begin{pmatrix} \hat\alpha & \hat\lambda \\ \hat\lambda & \hat\lambda^2\psi'(\hat\alpha) \\ \end{pmatrix} =\frac{1}{19.1918} \begin{pmatrix} 2.8983 & 0.9661 \\ 0.9661 & 0.3838 \\ \end{pmatrix} =\begin{pmatrix} 0.1510 & 0.0503 \\ 0.0503 & 0.0200 \\ \end{pmatrix}. \end{align*} Therefore, the asymptotic posterior distribution distribution (as $n\to\infty$) is \[ \left. \begin{pmatrix}\alpha \\ \lambda \\ \end{pmatrix}\right|\xvec\sim N\left\{\begin{pmatrix}2.8983 \\0.9661 \\ \end{pmatrix},~ \begin{pmatrix} 0.1510 & 0.0503 \\ 0.0503 & 0.0200 \\ \end{pmatrix} \right\}. \] The asymptotic posterior correlation between $\alpha$ and $\lambda$ is \[ Corr(\alpha,\lambda|\xvec)=\frac{Cov(\alpha,\lambda|\xvec)} {SD(\alpha|\xvec),SD(\lambda|\xvec)} =\frac{0.0505}{\sqrt{0.1510\times 0.0200}}=0.9189, \] that is, there is very strong positive posterior correlation between the parameters. %\end{document} \subsection*{Question 1} \begin{itemize} \item [(a)] The Jeffreys prior is given by \[ \pi(\theta) \propto \sqrt{I(\theta)}, \] where \[ I(\theta) = E_{X|\theta}\left[-\frac{\partial^{2}}{\partial \theta^{2}} \log f(\boldsymbol{x}|\theta)\right]. \] Now \begin{eqnarray*} f(\boldsymbol{x}|\theta) &=& \prod_{i=1}^{n}\frac{\theta}{x_{i}^{\theta+1}} = \theta^{n} \left(\prod_{i=1}^{n}x_{i}^{-(\theta+1)}\right), \quad \text{and} \\ \ell = \log f(\boldsymbol{x}|\theta) &=&n\log \theta -(\theta+1)\sum_{i=1}^{n}\log x_{i} = n\log\theta - (\theta+1)n \bar{x}_{\text{log}}.\\ && \text{\textcolor{red}{[2 marks, no need to express in terms of $\bar{x}_{\text{log}}$.]}} \end{eqnarray*} Then \begin{eqnarray*} \frac{\partial \ell}{\partial \theta} &=& \frac{n}{\theta} - n\bar{x}_{\text{log}}, \quad \text{and} \quad \text{\textcolor{red}{[1 mark]}} \\ \frac{\partial^{2} \ell}{\partial \theta^{2}} &=& -\frac{n}{\theta^{2}}. \text{\textcolor{red}{[1 mark]} } \end{eqnarray*} Then $\displaystyle E_{X|\theta}\left[-\frac{\partial^{2}\ell}{\partial \theta^{2}}\right] = \frac{n}{\theta^{2}}$, and so \[ \pi(\theta) \propto \sqrt{\frac{n}{\theta^{2}}} \propto \frac{1}{\theta}, \quad \theta < x. \text{\textcolor{red}{[1 mark]} } \] \item [(b)] The asymptotic posterior distribution (as $n\to\infty$) is \begin{equation*} \theta|\xvec\sim N\left(\hat\theta,J(\hat\theta)^{-1}\right), \end{equation*} where \begin{equation*} J(\theta)=-\frac{\partial^2}{\partial\theta^2} \log f(\xvec|\theta)=\frac{n}{\theta^2}. \end{equation*} Now \begin{eqnarray*} \frac{\partial}{\partial\theta} \log f(\xvec|\theta)=0 \quad\quad&\Longrightarrow&\quad\quad \frac{n}{\hat\theta}-n\bar{x}_{\text{log}}=0 \\ &\Longrightarrow&\quad\quad \hat\theta=1\left/\,\bar{x}_{\text{log}}\right. \text{\textcolor{red}{[1 mark]}} \\ &\Longrightarrow&\quad\quad J(\hat\theta)^{-1}=\frac{1}{n\bar{x}_{\text{log}^{2}}}. \end{eqnarray*} Therefore, for large $n$, the posterior distribution for $\theta$ is \begin{equation*} \theta|\xvec\sim N \left(\frac{1}{\bar{x}_{\text{log}}}, \frac{1}{n \bar{x}_{\text{log}}^{2}}\right) \quad\quad\text{approximately}. \end{equation*} \textcolor{red}{[2 marks, must state that this is an approximate results as $n \to \infty$. Also - it is not a requirement for students to show their results in terms of $\bar{x}_{\text{log}}$; for example, it is fine to say that $\hat{\theta} = n/\sum \log x_{i}$.]} \end{itemize} \subsection*{Question 2} The likelihood function is \begin{align*} f(\xvec|\mu,\alpha)&=\prod_{i=1}^n\frac{e^{\alpha/2}}{x_i\sqrt{2\pi}} \exp\left\{-\frac{e^\alpha}{2}(\log x_i-\mu)^2\right\}\\ &\propto e^{n\alpha/2}\exp\left\{-\frac{e^\alpha}{2}\sum_{i=1}^n(\log x_i-\mu)^2\right\}. \quad \text{\textcolor{red}{[1 mark]}} \end{align*} We have \begin{align*} \log f(\xvec|\mu,\alpha)&=\frac{n\alpha}{2}-\frac{e^\alpha}{2}\sum_{i=1}^n(\log x_i-\mu)^2+ const\\ \Longrightarrow\quad&\frac{\partial}{\partial\mu} \log f(\xvec|\mu,\alpha)=e^{\alpha}\sum_{i=1}^n(\log x_i-\mu) \quad \text{\textcolor{red}{[1 mark]}}\\ &\frac{\partial}{\partial\alpha} \log f(\xvec|\mu,\alpha)=\frac{n}{2}-\frac{e^\alpha}{2}\sum_{i=1}^n(\log x_i-\mu)^2 \quad \text{\textcolor{red}{[1 mark]}}\\ \Longrightarrow\quad&\frac{\partial^2}{\partial\mu^2} \log f(\xvec|\mu,\alpha)=-ne^{\alpha} \quad \text{\textcolor{red}{[1 mark]}} \\ &\frac{\partial^2}{\partial\mu\partial\alpha} \log f(\xvec|\mu,\alpha)=e^{\alpha}\sum_{i=1}^n(\log x_i-\mu) \quad \text{\textcolor{red}{[1 mark]}}\\ &\frac{\partial^2}{\partial\alpha^2} \log f(\xvec|\mu,\alpha)=-\frac{e^{\alpha}}{2}\sum_{i=1}^n(\log x_i-\mu)^2. \quad \text{\textcolor{red}{[1 mark]}} \end{align*} From the hint \[ E(\log X_i|\mu,\alpha)=\mu\quad\text{and}\quad E\{(\log X_i-\mu)^2|\mu,\alpha)=Var(\log X_i|\mu,\alpha)=e^{-\alpha}. \] Therefore, the elements of Fisher's information matrix are \begin{align*} I_{11}&=E_{\Xvec|\mu,\alpha} \left[-\frac{\partial^2}{\partial\mu^2} \log f(\Xvec|\mu,\alpha)\right]=ne^\alpha\\ I_{12}&=E_{\Xvec|\mu,\alpha} \left[-\frac{\partial^2}{\partial\mu\partial\alpha} \log f(\Xvec|\mu,\alpha)\right] =-ne^\alpha\,E_{X|\mu,\alpha}[\log X-\mu]=0\\ I_{22}&=E_{\Xvec|\mu,\alpha} \left[-\frac{\partial^2}{\partial\alpha^2} \log f(\Xvec|\mu,\alpha)\right]=\frac{ne^{\alpha}}{2}E_{X|\mu,\alpha}[(\log X-\mu)^2]=\frac{n}{2}. \end{align*} Hence, Fisher's information matrix is \[ I(\mu,\alpha) =n\begin{pmatrix} e^\alpha & 0 \\ 0 & 1/2 \\ \end{pmatrix} \quad \text{\textcolor{red}{[4 marks]}}, \] and so Jeffreys prior density is, for $\mu\in\mathbb{R}$, $\alpha\in\mathbb{R}$ \begin{align*} \pi(\mu,\alpha)&\propto|I(\mu,\alpha)|^{1/2} \\ &\propto\sqrt{\frac{n^2e^\alpha}{2}}\\ &\propto e^{\alpha/2}. \quad \text{\textcolor{red}{[2 marks]}} \end{align*} \end{document}