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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

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Warning: The correct wording of the exercise is as follows:

enter image description here

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.

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