Question: Suppose $X_1, \cdots, X_n$ are $iid$ normal random variables with unknown mean $\mu$ and known variance $\sigma^2$. Find the UMVUE for $\Phi(\mu)$, where $\Phi(\cdot)$ is the cdf of a standard normal random variable.

I used to guess the desired UMVUE is or similar to $\Phi(\bar X)$ since it is a function of complete and sufficient statistic. However, it is not really unbiased (see here). Could anyone shed light on how to find the UMVUE, please? Thank you!


1 Answer 1


Here is a solution for $\sigma^2 < n$. You can work this out from the result provided in the link you gave. From the link you gave we have the following. For $X \sim N(\mu, \sigma^2)$:

$$E(\Phi(X)) = \frac{\mu}{\sqrt{1 + \sigma^2}}$$

Finding the UMVUE will follow from this result. Note that $\bar{X}_n \sim N(\mu, \sigma^2/n)$ so that

$$E(\Phi(\bar{X}_n)) = \frac{\mu}{\sqrt{1 + \frac{\sigma^2}{n}}}$$

Then for any $a \in \mathbb{R}$ such that $a \neq 0$, $a \bar{X}_n \sim N(a\mu, a^2\sigma^2/n)$ so that

$$E(\Phi(a\bar{X}_n)) = \frac{a\mu}{\sqrt{1 + \frac{a^2 \sigma^2}{n}}}$$

Now we want to find $a$ such that

$$\frac{a\mu}{\sqrt{1 + \frac{a^2 \sigma^2}{n}}} = \mu$$

which solves for $a = \frac{1}{\sqrt{1 - \frac{\sigma^2}{n} }}$. Thus, the UMVUE is $\Phi \left(\frac{\bar{X}_n}{\sqrt{1 - \frac{\sigma^2}{n} }} \right)$ since it's a function of the complete sufficient statistic $\bar{X}_n$. You can further see that this statistic will only be valid if $1 - \frac{\sigma^2}{n} > 0$ which means $\sigma^2 < n$.

What should we do if $\sigma^2$ is greater than $n$ then? Since $\sigma^2$ is known we could always increase $n$ to be larger than $\sigma^2$. I don't have any great ideas for what to do if $n$ can't be increased though. The above result does eliminate the choice of $\Phi (a \bar{X}_n )$ as the UMVUE when $\sigma^2 > n$ though.

  • $\begingroup$ Thanks, there is a small error in the third equation where it should be $a\bar X$ in $\Phi(\cdot)$. $\endgroup$
    – LaTeXFan
    Jun 9, 2014 at 11:39
  • $\begingroup$ @20826 - fixed the error $\endgroup$ Jun 9, 2014 at 20:20

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.