I'm not even sure if the title of the question makes sense...

I have the solution to the following question in front of me, but can not make sense of it:

Let $X_1, \ldots, X_n$ be a random sample from a distribution $\mathrm{N}(0, \theta)$.
Is $\hat{\theta}_{MLE}$ a UMVUE of $\theta$?

(HINT: $\frac{X^2}{\theta} \sim \chi_{1}^2$, and $\mathrm{E}(\frac{X^2}{\theta})=1$, $\mathrm{Var}(\chi_{k}^2) = 2k$.)

I fairly easily found $\hat{\theta}_{MLE} = \frac{1}{n} \sum X_i^2$; and further that this is unbiased, since $\mathrm{E}(\hat{\theta}_{MLE})=\mathrm{E}(X_{1}^{2})=\theta$. Now, upon computing the relevant derivative to find the CRLB I have arrived at the following expectation: $$ \mathrm{E} \left[ \frac{\partial}{\partial\theta} \mathrm{ln} \ f(X; \theta) \right]^2 = \mathrm{E} \left[- \frac{1}{2\theta} + \frac{\mathrm{X}^2}{2\theta^2} \right]^2 = \frac{1}{4\theta^2} \times \mathrm{E} \left[ \frac{\mathrm{X}^2}{\theta} - 1 \right]^2 $$

I am certain that it is here I need to apply the hint given to me in the question, but I have lost my way and would greatly appreciate a pointer in the right direction.

  • $\begingroup$ Please add the self-study tag. Another hint: $Var(Y)=E[Y-E(Y)]^2$ $\endgroup$ Commented Jun 18, 2017 at 11:16
  • $\begingroup$ @ChristophHanck thanks for hint; but I'm really not getting it. I'm just losing it at the whiteboard in the hall... $\endgroup$
    – c SS
    Commented Jun 18, 2017 at 12:13
  • $\begingroup$ Try writing $Y=X^2/\theta$ and then apply Christoph Hanck's hint. $\endgroup$ Commented Jun 18, 2017 at 12:23
  • $\begingroup$ ah! Ok. Thanks very much to you both @Cristoph & Gordon: $\textrm{E}(\frac{X^2}{\theta}-1)^2 = \textrm{Var}(\frac{X^2}{\theta}-1) + \left[\textrm{E}(\frac{X^2}{\theta}-1)\right]^2 = \textrm{Var}(\frac{X^2}{\theta}) + \left(1-1\right)^2 = \textrm{Var}(\frac{X^2}{\theta})$ $\endgroup$
    – c SS
    Commented Jun 18, 2017 at 13:32

1 Answer 1


As pointed out to me; I should have used the definition of the variance to proceed. Below is the rest of the answer.

Using the hint $\textrm{E}\left(\frac{X^2}{\theta}\right) = 1$, we have: $$ \textrm{E}\left(\frac{X^2}{\theta}-1\right)^2 = \textrm{Var}\left(\frac{X^2}{\theta}-1\right) + \left[\textrm{E}(\frac{X^2}{\theta}-1)\right]^2 = \textrm{Var}\left(\frac{X^2}{\theta}\right) + \left(1-1\right)^2 = \textrm{Var}\left(\frac{X^2}{\theta}\right) $$ and using the hint $\frac{X^2}{\theta} \sim \chi_{1}^{2}$, with $\textrm{Var}\left(\chi_{k}^{2}\right)=2k$, we have $\frac{1}{4\theta^2} \textrm{Var}\left(\frac{x^2}{\theta}\right)=\frac{1}{2\theta^2}$. Hence, we find:

$$ \textrm{CRLB}\left(\theta\right) = \frac{1}{n \frac{1}{2\theta^2}}=\frac{2 \theta^2}{n}. $$

Now, since $\textrm{Var}(\hat{\theta}_{MLE})=\textrm{Var}\left(\frac{1}{n} \sum X_{i}^{2}\right)= \frac{1}{n^2} \cdot n \cdot \textrm{Var} \left( \sum X_{1}^{2} \right)= \frac{1}{n} \cdot \theta^2 \cdot \textrm{Var}\left( \frac{X_1^2}{\theta} \right) = \frac{2 \theta^2}{n}$

we have shown that $\textrm{CRLB}(\theta) = \textrm{Var}(\hat\theta_{MLE}); $ and may conclude that the estimator is an UMVU estimator of $\theta$.


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