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.


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.

  • $\begingroup$ 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. $\endgroup$
    – Chris C
    Commented Jul 8, 2015 at 15:14
  • $\begingroup$ @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:). $\endgroup$
    – KOE
    Commented Jul 8, 2015 at 15:17
  • $\begingroup$ Ha, thanks! I definitely have some brushing up to do. Hope you find an answer! $\endgroup$
    – Chris C
    Commented Jul 8, 2015 at 15:38

1 Answer 1


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 $\delta(\mathbf y).$


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