# Find the UMVUE for $\Phi(\mu)$

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!

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.

• Thanks, there is a small error in the third equation where it should be $a\bar X$ in $\Phi(\cdot)$. – LaTeXFan Jun 9 '14 at 11:39
• @20826 - fixed the error – Samuel Benidt Jun 9 '14 at 20:20