## A Theoretical Result

tsakanikasnickos on Thursday, November 20 2014, 07:55 PM
0

Show that a Bayes estimator is a function of the sufficient statistic.

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• Replied by tsakanikasnickos on Wednesday, February 04 2015, 05:29 PM
Let $$\displaystyle \underset{\sim}{X} =( X_{1}, \dots , X_{n} )$$ be a random sample from a density $$\displaystyle f(x|\theta)$$, where $$\theta$$ is a value of the random variable $$\Theta$$ with prior density $$\displaystyle \pi_{\Theta}$$. Let $$\displaystyle d^{*}$$ be a Bayes estimator of $$\theta$$ and let $$\mathcal{L}(\theta,\cdot)$$ be the loss function for the estimation of $$\theta$$. By definition, $$\displaystyle d^{*}$$ is the statistic that minimizes the Bayes risk function. Equivalently, it can be easily shown that $$\displaystyle d^{*}$$ is the statistic that minimizes the posterior risk function $$\displaystyle E_{\Theta | \underset{\sim}{X} }\left( \mathcal{L}(\theta,\cdot) \right)$$. We have that

\begin{align*}
\displaystyle
E_{\Theta | \underset{\sim}{X} } \left( \mathcal{L}(\theta,d) \right) &= \int_{\Theta} \mathcal{L}(\theta,d) \pi_{\Theta | \underset{\sim}{X} }(\theta | \underset{\sim}{X} ) \mathrm{d}\theta \\
&= \int_{\Theta} \mathcal{L}(\theta,d) \pi_{\Theta}(\theta) f(\underset{\sim}{X} | \theta) \frac{1}{c(\underset{\sim}{X} )} \mathrm{d}\theta \; \; \; (1)
\end{align*}

where $$\displaystyle c = c(\underset{\sim}{X} )$$ is a constant. By Neyman-Fischer's factorization theorem, if $$\displaystyle T=T(\underset{\sim}{X} )$$ is the sufficient statistic, then
$\displaystyle f(\underset{\sim}{X} | \theta) = g\left( T(\underset{\sim}{X} ) , \theta \right) h(\underset{\sim}{X} ) \; \; \; (2)$
We now observe, due to (1) and (2), that, in order to minimize the quantity
$E_{\Theta | \underset{\sim}{X} } \left( \mathcal{L}(\theta,d) \right)$with respect to $$d$$, we have to minimize the quantity
$\int_{\Theta} \mathcal{L}(\theta,d) \pi_{\Theta}(\theta) g\left( T(\underset{\sim}{X} ) , \theta \right) \mathrm{d}\theta$with respect to $$d$$, which is a function of $$\displaystyle T=T(\underset{\sim}{X} ) \text{ and } d$$. We thus conclude that the Bayes estimator $$\displaystyle d^{*}$$ of $$\theta$$ is a function of the sufficient statistic $$T(\underset{\sim}{X} )$$, since we already mentioned that $$\displaystyle d^{*} = \min_{d} E_{\Theta | \underset{\sim}{X} }\left( \mathcal{L}(\theta,d) \right)$$.

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