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",


$x \sim N(\theta,1)$ ,

$\theta \sim N(0,1)$

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.


$$\delta^{\pi}(x)=argmin_{\delta} \int_{-\infty}^{\infty} e^{3\theta^{2}/2}(\theta-\delta)^{2}\pi(\theta|x) d\theta$$



where $w(\theta)$ is the non negative weight.

So how would we see the result to be true from this?



1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.