Confidence interval for the 95th percentile of the normal distribution

Let $$X_1, .., X_n \sim Normal(\mu, \sigma^2)$$.

Let $$\tau$$ be the 95th percentile of this distribution. Thus,

$$P(X_i < \tau) = 0.95$$.

What is the $$1 - \alpha$$ confidence interval for $$\tau$$?

I know how to get the maximum likelihood estimator for $$\tau$$; I would invoke the equivariance principle and plug in the MLEs for $$\mu$$ and $$\sigma$$.

$$\hat{\tau} = \bar{X} + S \Phi^{-1}(0.95)$$.

However, I'm struggling to estimate the standard error for it. It likely involves Fisher's information matrix, but I'm stuck at this point.

For normal distribution, $$\bar X$$ and $$S$$ are independent. So $$\mathrm{Var}(\hat \tau) = \mathrm{Var}(\bar X) +\mathrm{Var}(S \Phi^{-1}(0.95)) = \frac {\sigma^2}n + (\Phi^{-1}(0.95))^2 \mathrm{Var}(S)$$

$$\sqrt {n-1} S/\sigma$$ follows chi distribution with $$n-1$$ degree of freedom. Its variance is $$\frac {2[\Gamma(\frac {n-1}2)[\Gamma(1+ \frac {n-1}2)-[\Gamma(\frac {n}2)]}{\Gamma(\frac {n-1}2) } = V$$. So the variance of $$S$$ is $$\frac {\sigma^2}{n-1}V$$.

So $$\mathrm{Var}(\hat \tau) = \frac {\sigma^2}n + (\Phi^{-1}(0.95))^2\frac {\sigma^2}{n-1}V$$

The square root of variance is the standard error.

• How do you know that the asymptotic normal distribution has this variance for the MLE? Shouldn't we use the inverse of the Fisher information matrix to estimate the variance instead?
– MSE
Nov 5, 2018 at 2:26
• Nov 5, 2018 at 2:31
• If you doubt about why S has relation with chi distribution, you can find the answer at the beginning of the linked article. For a statistics, if we have its exact variance, we should not use the Fisher information. Of cause, maybe they are the same. Nov 5, 2018 at 3:30
• OK - I did not think about getting an exact distribution of S. Thanks for the solution, @user158565.
– MSE
Nov 27, 2018 at 18:50