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\rhead{Jimin Ding\\Scribe:Yuchen Han, Bei Wu, Hiro Ono}
\lhead{Math 5062 Spring 2018}


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\section{Lecture 7: Bayesian Prospective of MRIE}
\underline{Motivating example}: location invariant estimator\\
Suppose the p.d.f. of $X$ satisfies $f(x-\theta), \theta \in \mathbb{R}, x \in\mathbb{R}$, under loss function $L(\theta, a)=(\theta-a)^2$. \\
Our goal is to estimate $\theta$.\\
Consider $\delta_0(x)=x$, then $\delta(x)$ is invariant if and only if $\delta(x)=\delta_0(x)-u=x-u.$\\
$\delta^*(x)=x-u^*$ is MRIE if and only if
$$\delta^*(x)=\arg \min_{\delta \in \mathbb{A}}R(\theta,\delta)=\arg \min 
E_\theta[(x-u)^2]=\arg \min E_0[(x-u)^2].$$
So $\delta^*=E(x|\theta)=E_{\theta=0}(x)=0.$ Let's look at the function to be 
minimized.
\begin{align*}
 E_0[(x-u)^2] & = \int (x'-u)^2f_0(x') dx' = \int (x' +\delta(x) -x)^2f_0(x') dx', \quad \mbox{let }x'=x-\theta, \\
&=  \int (x-\theta +\delta(x) -x)^2f_0(x-\theta) d\theta\\
&=  \int (\delta(x) -\theta)^2f_0(x-\theta) d\theta\\
&=  \int (\delta(x) -\theta)^2\Pi(\theta|x) d\theta\\
&\propto r(\Pi, \delta), \quad \mbox{Bayes risk w.r.t. }\Pi.
\end{align*}
where $\Pi$ is the generalized Bayes posterior under uniform prior on $\mathbb{R}$ (improper prior),
$$\Pi(\theta|x)= \frac{\Pi(\theta)f(x|\theta)}{\int \Pi(\theta)f(x|\theta) d \theta} =\frac{f(x|\theta)}{\int f(x|\theta) d \theta} =\frac{f_0(x-\theta)}{\int f(x|\theta) d \theta}f_0(x-\theta).$$
Hence the generalized bayes rule that minimizes the above Bayes risk is
\begin{align*}
\delta^*(x)&=E(\theta|x)=\int \theta\Pi(\theta|x) d\theta \\
&= \int \theta f_0(x-\theta)d\theta, \quad \mbox{let }x'=x-\theta, \\
&= \int (x-x') f_0(x')dx' \\
&= x\int f_0(x')dx' - \int x' f_0(x')dx'\\
&= x-E(x|\theta=0)=x
\end{align*}
The generalized Bayes rule w.r.t. uniform prior is sames as MRIE.

\begin{theorem}[Eaton 1989]
If $\bar{G}$ acts transitive on $\Theta$, the MRIE is given by
$$\delta^*(x)=\arg \min _{\delta\in\mathbb{A}}\int _{\bar{G}} L(\bar{g}\theta, \delta) f(x|\bar{g}\theta_0) d \mu_r(\bar{g}),$$
where $\mu_r$ is the right-invariant measure on $\bar{G}$ and $\theta_0$ is any 
point in $\Theta$.\\
(Note: as $\bar{g}$ ranges over $\bar{G}$, $\bar{g}\theta_0$ 
ranges over entire 
$\Theta$.)
\end{theorem} 

\begin{definition}
Right-invariant prior corresponding to $\bar{G}$ on $\Theta$ is
$$\forall A \in \sigma(\Theta), \quad \Pi_r(A)=\mu_r\{\bar{g}:\bar{g}\theta_0\in A\}=\int \mathbf{1}(\bar{g}\theta_0\in A) d\mu_r(\bar{g}),$$
where $\theta_0$ is any point in $\Theta$, $\mu_r$ is the right-invariant measure on $\bar{G}$.
\end{definition}
Interpretation: $\Pi_r$ is the distribution of $\bar{g}\theta_0$, where $\bar{g}$ is randomly selected from $\bar{G}$.

\begin{corollary}
If $\bar{G}$ is transitive. The MRIE is given by $$\delta^*(x) = \arg \min 
_{\delta\in\mathbb{A}}\int _{ \Theta}  L(\theta, \delta) d \Pi_r(\theta|x), $$ 
where $\Pi_r(\theta|x)$ is the posterior of $\theta $ with p.d.f ( w.r.t $ 
\Pi_r(\theta)) $ $$\frac{d \Pi_r(\theta|x)}{d \Pi_r(\theta)} = 
\frac{f(x|\theta)}{\int f(x|\theta)d\Pi_r(\theta) }$$ $$  (d \Pi_r(\theta|x) = 
\frac{f(x|\theta)d\Pi_r(\theta)}{\int f(x|\theta)d\Pi_r(\theta) }  )$$
\end{corollary}
\Remark This corollary shows that the MRIE $\delta^*$ can be viewed as a 
generalized Bayes estimator w.r.t 
prior $\Pi_r$, since  $$\int _{ \Theta}  L(\theta, \delta) d 
\Pi_r(\theta|x) =\frac{1}{C_x} \int _{ \Theta}  L(\theta, \delta)  f(x|\theta) d 
\Pi_r(\theta)  = \frac{1}{C_x }r(\Pi_r, \delta),$$ where $C_x= \int 
f(x|\theta)d\Pi_r(\theta)$ is independent of $\theta$. 


\medskip

\noindent \textbf{Finding the Right-invariant Measure $\mu_r$ on 
$\bar{G}$ }\\

A measure $\mu_r$ is right-invariant iff
$\forall \bar{g_0} \in \bar{G} $ and measurable $A \subset \bar{G}$, $$\mu_r(A\bar{g_0}^{-1}) = \mu_r(A) $$ In other words, if view $\mu_r$ as a probablity measure $$P(\bar{g} \in A) = P(\bar{g}\circ \bar{g_0} \in A)$$

\Example
\begin{enumerate}
\item Additive groups: $\bar{G} = \{\bar{g}:\bar{g}(\theta) = \theta+c, c\in 
\mathbb{R} \} $.\\
$\bar{g_0}(\theta) = \theta + c_0$, then $ \bar{g} \circ \bar{g_0} = \theta + (c+ c_0) $. $\mu_r$ is right-invariant if $\mu_r(A-c_0) = \mu_r(A), \forall c_0 \in \mathbb{R}, \forall A \in \mathcal{B}_{\mathbb{R}} $\\
Lebesgue measure is invariant on $\mathbb{R}$. The induced prior is given by the distribution of $\theta_0 +c$, where $c\sim \mu_r = \mu_L $ (lebesgue measure)\\
$\therefore \Pi_r = \mu_r = \mu_L$
\item Multiplicative group: $\bar{G} = \{\bar{g}:\bar{g}(\theta) = c \theta, c 
> 0\} $.\\
$\bar{g_0}(\theta) = c_0\theta$, then $ \bar{g} \circ \bar{g_0} = c\cdot c_0\theta  $. $\bar{G} $ is isomorphic on $\mathbb{R}^+$. $\mu_r$ is right-invariant if $$\mu_r(A/c_0) = \mu_r(A), \forall c_0 \in \mathbb{R}^+, \forall A \subset \mathcal{B}_{\mathbb{R^+}}$$
Lebesgue measure on $\mathbb{R}^+$ is not right- invariant. Consider the measure $\mu_r$ with the density $1/c$ w.r.t lebesgue measure $\mu_L$.
$\forall B=(b_1,b_2) \in  \mathbb{R}^+,$ let $c'=c \cdot c_0 $
Then $$\mu_r(B/c_0) = \int_{b1/c_0}^{b2/c_0}  
\frac{1}{c} d \mu_L(c) = \int_{b1}^{b2}  
\frac{c_0}{c'} d \mu_L(\frac{c'}{c_0}) = \int_{b1}^{b2}  
\frac{1}{c'} d \mu_L(c')=\mu_r(B) $$
$\therefore \mu_r$ is the right-invariant measure. The induced 
right-invariant prior $\Pi_r = \mu_r$
\end{enumerate}



\subsection{Related Reading}
\begin{enumerate}
 \item Sh  P253-255 %\cite{shao03}
 \item LC Chapter 4%\cite{bd01}
\end{enumerate}

\end{document}


