# MLE for the .95 percentile of the normal distribution

The question is: let $X_1, ..., X_n \sim N(\mu, \sigma^2)$. Let $\tau$ be the .95 percentile, i.e. $P(X<\tau)$ = .95. What is the MLE of $\tau$?_

What I have tried:

$P(X<\tau) = P(Z < \frac{\tau - \mu}{\sigma}) = \Phi(\frac{\tau - \mu}{\sigma})$ where Z denotes a Standard Normal random variable, and $\Phi$ is the CDF for the normal distribution.

I think I need to invoke the delta method for this, since $\tau$ is a function of the parameters for the normal distribution, but I don't quite see how I can use this method for finding a quantile. I also don't see how I need to invoke the maximum likelihood. I do know, that for a normal distribution the MLE's are:

$\hat{\mu} = \sum_{i=1}^n X_i/n$ ,the sample mean and $\hat{\sigma} = \sqrt{\sum_{i=1}^n(X_i - \mu)^2/n}$. How would I need those? Any help is appreciated!

Edit: the question is one from Larry Wasserman's All of Statistics that I'm working through, practising for an exam I have this week. And I suppose that $\tau$ is almost equal to two $\sigma$, I believe it is ~$1.96 \sigma$, looking at a normal table?

• This appears to be a routine textbook-style question. How does it arise? – Glen_b Jan 12 '15 at 13:14
• Can you write $\tau$ in terms of $\mu$ and $\sigma$? How would you calculate the MLE of that function of $\mu$ and $\sigma$? You may need to recall some properties of ML estimation. – Glen_b Jan 12 '15 at 13:17

The sample mean and variance are sufficient statistics for the Normal distribution. All normal inference goes through them. Basically pretend that the mle's for $\mu$ and $\sigma$ are the true values and get the quantile from there.
Another way to look at it is to reflect that the MLE of a function is the function of the MLE. For example, the MLE of $\sigma$ is the quare root of the MLE for $\sigma^2$. The MLE of a quantile is the quantile of the distribution whose parameters are the MLE's. I'm not suggesting this as a quickie workaround: this really will be the MLE.
$$\hat{\tau} = \hat{\mu} + 1.96 \, \hat{\sigma}$$