I have been working through a wide variety of problems involving Bayes risk and loss functions and I couldn't immediately solve the following
From "The Bayesian Choice",
Consider
$x \sim N(\theta,1)$ ,
$\theta \sim N(0,1)$
and the loss function
$L(\theta, \delta)= e^{3\theta^{2}/2}(\theta-\delta)^{2}$
Then show that the Bayes estimator is $\delta^{\pi}(x)=2x$
My thoughts:
If it was the usual quadratic loss, the Bayes estimator would simply be the posterior mean, the posterior is
$\theta | x \sim N(\frac{x}{2},\frac{1}{2})$
However it is not usual quadratic loss.
If $e^{3\theta^{2}/2}\pi(\theta|x)$ itself was a distribution then we could take the posterior mean of that to be the Bayes estimator. But it is not obvious that it is a distribution.
So,
$$\delta^{\pi}(x)=argmin_{\delta} \int_{-\infty}^{\infty} e^{3\theta^{2}/2}(\theta-\delta)^{2}\pi(\theta|x) d\theta$$
ie
$$\delta^{\pi}(x)=\frac{E(w(\theta)\theta|X]}{E[w(\theta)|X]}$$
where $w(\theta)$ is the non negative weight.
So how would we see the result to be true from this?
Thanks