# Bayes estimator with weighted Loss

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

It thus uses $$\exp\{3\theta^2/4\}$$ as a weight.
You are correct in stating that the solution is $$\delta^{\pi}(x)=\arg\min_{\delta} \int_{-\infty}^{\infty} e^{3\theta^{2}/4}(\theta-\delta)^{2}\pi(\theta|x) \,\text{d}\theta$$ From there, differentiating in $$\delta$$ leads to the equation $$\int_{-\infty}^{\infty} 2e^{3\theta^{2}/2}(\theta-\delta^\pi(x))\pi(\theta|x) \,\text{d}\theta=0$$ that is $$\delta^\pi(x)\int_{-\infty}^{\infty} e^{3\theta^{2}/4}\pi(\theta|x) \,\text{d}\theta=\int_{-\infty}^{\infty} e^{3\theta^{2}/4}\theta\pi(\theta|x) \,\text{d}\theta$$ i.e. $$\delta^\pi(x)=\dfrac{\int_{-\infty}^{\infty} \theta e^{3\theta^{2}/4}\pi(\theta|x) \,\text{d}\theta}{\int_{-\infty}^{\infty} e^{3\theta^{2}/4}\pi(\theta|x) \,\text{d}\theta}=\frac{\mathbb{E}(w(\theta)\theta|X=x]}{\mathbb{E}[w(\theta)|X=x]}$$with $$w(\theta)=\exp\{3\theta^2/4\}$$ (Corollary 2.5.2, page 78). This is also the posterior mean associated with the new prior $$\pi^*(\theta)=\exp\{3\theta^2/4\}\pi(\theta)=\exp\{3\theta^2/4\}\exp\{-2\theta^2/4\}=\exp\{\theta^2/4\}$$ which leads to the posterior$$\pi^*(\theta)\propto \exp\{\theta^2/4\}\exp\{-(\theta-x)^2/2\}\propto \exp\{-\theta^2/4+2\times2\times\theta x/4\}\propto \exp\{-(\theta-2x)^2/4\}$$which concludes question a).
Note: there is no Bayes estimator associated with the weight $$\exp\{3\theta^2/2\}$$ as the posterior loss is always infinite.