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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – MSE Nov 5 '18 at 2:26
  • $\begingroup$ eqn 9 on mathworld.wolfram.com/ChiDistribution.html $\endgroup$ – user158565 Nov 5 '18 at 2:31
  • $\begingroup$ 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. $\endgroup$ – user158565 Nov 5 '18 at 3:30
  • $\begingroup$ OK - I did not think about getting an exact distribution of S. Thanks for the solution, @user158565. $\endgroup$ – MSE Nov 27 '18 at 18:50

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