# Fastest way to solve Bayes estimator problem

The below problem is from an old PhD qualifying exam in our department. My own solution below is time-consuming and quite possibly wrong. It also relies on recognizing a less common distribution, so I wonder: is there a faster way to solve this problem, and how is that done? I suspect there are applicable theorems about Bayes estimators that I don't know about or have forgotten.

Problem

Given $\theta >0$, let $Y_1,\dots, Y_n$ be iid from the distribution with density

$$f(y\mid \theta ) = I_{y>0}c\sqrt{\theta}\exp\{-\theta y^2 / 2\},$$ where $c=(2\pi)^{-1/2}$. Suppose the loss function for estimating $f(0\mid \theta) = c\sqrt{\theta}$ is given by $L(\delta,c\sqrt{\theta}):=(\delta-c\sqrt{\theta})^2/(c\sqrt{\theta})^2$ and the prior on $\theta$ is the gamma distribution with parameters $\alpha>1,\beta>0$.

My solution

Let $\gamma = f(0\mid\theta)$. Then $$f(\mathbf{y}\mid\gamma)= \gamma^n \exp\{-\gamma^2 \sum_i y_i^2/(2c^2)\}$$ and by a density tranform with Jacobian $2\gamma$ we get

$$p(\gamma)\propto \gamma^{2\alpha - 1}\exp\{ -\beta \gamma^2 / c^2\},$$

so that

$$p(\gamma\mid \mathbf{y}) \propto \gamma^{n+2\alpha-1}\exp{\{-\gamma^2[\beta/c^2 + \mathbf{y'y}/(2c^2)] \}}.$$ Now minimizing $\mathbb E L(\delta, \gamma)$ under this posterior w.r.t. $\delta$ is equivalent to minimizing $\mathbb E (\delta-\gamma)^2$ (i.e. results in the same objective function) under the posterior

$$p(\gamma\mid \mathbf{y}) \propto \gamma^{n+2\alpha-3}\exp{\{-\gamma^2[\beta/c^2 + \mathbf{y'y}/(2c^2)] \}}.$$

We know that setting $\delta = \mathbb E (\gamma \mid \mathbf y)$ solves such a problem. We also recognize the latest posterior as a generalized Gamma with parameters (following Wikipedia's notation) $d=n+2\alpha - 2, p = 2, 1/a = \sqrt{\beta/c^2 + \mathbf{y'y}/(2c^2)}$. The mean of the generalized gamma is $a\Gamma((d+1)/p)/\Gamma(d/p)$, and we are done.

• Sorry, I can't help solve the question, but out of interest what kind of PhD program was this for? Was it in mathematics, or statistics? Just asking as an undergraduate. Commented Jul 8, 2015 at 15:14
• @ChrisC it's from a stats program. Don't think one gets away with this abuse of notation in a maths program, but I'm not sure:).
– KOE
Commented Jul 8, 2015 at 15:17
• Ha, thanks! I definitely have some brushing up to do. Hope you find an answer! Commented Jul 8, 2015 at 15:38

Found a quicker way:

We want to minimize $$r(\delta):=E\left(\frac{c\sqrt \theta - \delta}{c\sqrt \theta}\right)^2=E\left(\frac{c^2\theta - 2c\sqrt \theta \delta+\delta ^2}{c^2 \theta}\right)$$ under the posterior distribution. For the no-data problem we get $$r(\delta) = 1 - \delta 2c^{-1} E\theta^{-1/2} + \delta^2 c^{-2} E\theta^{-1},$$

which is minimized at

$$\delta = c\frac{E\theta^{-1/2}}{E \theta^{-1}}.$$

For a gamma distributed random variable, $X\sim Gamma(a,b)$, we have, for parameter values such that the integral exists: $$EX^{-c} = \frac{b^a}{\Gamma(a)}\int x^{a-1-c}e^{-bx}=\frac{b^a\Gamma(a-c)}{b^{a-c}\Gamma(a)}.$$

Since the posterior is also gamma, viz. $$p(\theta \mid \mathbf y)\propto \theta^{\alpha + n/2}e^{-\theta (\beta + \sum_i y_i^2 / 2)},$$

the required moments follow directly from this and gives $\delta(\mathbf y).$