# Bayes estimate with weighted square error loss

First, let $T(x)$ be an estimator of $g(\theta)$ and assume we have a square error loss function defined as $$L[g(\theta),T(x)]=[g(\theta)-T(x)]^2$$ Then the posterior expected risk of $T$ is $$\rho_T(x)=\int_{\Theta}[g(\theta)-T(x)]^2\pi(\theta|x)d\theta$$ Since we know the fact that $E(Y-b)^2$ is minimized by $b=E(Y)$, thus $$T(x)=E[g(\theta)|X=x]$$ minimizes $\rho_T(x)$.

Now we want to generalize this idea to a weighted square error loss defined as $$L^*[g(\theta),T(x)]=w(\theta)[g(\theta)-T(x)]^2$$ The solution says that the Bayes estimate now is $$T^*(x)=\frac{E[g(\theta)w(\theta)|X=x]}{E[w(\theta)|X=x]}$$ I was wondering why is this? Since the posterior expected risk is now as $$\rho_T^*(x)=\int_{\Theta}w(\theta)[g(\theta)-T(x)]^2\pi(\theta|x)d\theta=E\{w(\theta)[g(\theta)-T(x)]^2|X=x \}$$ Then why this $T^*(x)$ minimize our $\rho_T^*(x)$?

Thanks~

• This is Corollary 2.5.2 (page 78) in my book, The Bayesian Choice. – Xi'an Jul 21 '18 at 14:45

Well, the simplest way to look at this phenomenon is to group the posterior density in the integrand with the weighting function (since they are both functions of $\theta$) to form a product function:
$$\rho_T^* (x) = \int_\Theta [g(\theta) - T(x)]^2 \Big( w(\theta) \pi(\theta|x) \Big) d\theta.$$
Now, this product can be considered to be the kernel of a new density (which might be an improper density for some weighting functions), which we can write as $\pi^*(\theta |x) \propto w(\theta) \pi(\theta|x)$. Via analogy to the first result we then have:
\begin{equation} \begin{aligned} T^*(x) &= \int_\Theta g(\theta) \pi^*(\theta|x) d\theta \\[6pt] &= \int_\Theta g(\theta) \Big( \frac{w(\theta) \pi(\theta|x)}{\int_\Theta w(\theta’) \pi(\theta’|x) d\theta’} \Big) d\theta \\[6pt] &= \frac{ \int_\Theta g(\theta) w(\theta) \pi(\theta|x) d\theta}{\int_\Theta w(\theta’) \pi(\theta’|x) d\theta’} \\[6pt] &= \frac{\mathbb{E}(g(\theta) w(\theta)|X=x)}{\mathbb{E}(w(\theta)|X=x)}. \\[6pt] \end{aligned} \end{equation}