# Proof regarding chi-squared confidence intervals

$X_1, ... X_n$ are an iid sample from a normal population whose mean and variance are unknown, and let $S^2 = \frac{1}{n}\sum_{i=1}^n (X_i-\bar{X})^2$. Suppose $1 - \alpha = P(\frac{\sqrt nS}{\sqrt a} \leq \sigma \leq \frac{\sqrt n S}{\sqrt b})$ such that $[\frac{\sqrt nS}{\sqrt a}, \frac{\sqrt n S}{\sqrt b}]$ is a $1 - \alpha$ confidence interval of $\sigma$. Show that the values of a and b that minimises the length of the CI satisfies $a^{n/2}e^{-a/2} - b^{n/2}e^{-b/2} = 0$.

What I have: I know that $\frac{nS^2}{\sigma^2} \sim \chi^2_{df = n-1}$. The equation is reminiscient of the $\chi^2$ pdf, and the furthest I've got so far is:

$$\frac{1}{2^{{(n-1)}/{2}} \Gamma(\frac{n-1}{2})}\int_a^b x^{\frac{n-1}{2}-1} e^{-\frac{x}{2}} dx = 1 - \alpha$$

One thing I've considered is to take partial derivatives wrt b on both sides, using the Fundamental Theorem of Calculus on the right, which allows me to show that $b^{n/2}e^{-b/2} = 0$; then do the same process, taking derivatives wrt a to show that $a^{n/2}e^{-a/2} = 0$. Obviously, 0-0=0, but I feel the answer can't be that obvious, so there must be something wrong with this answer.

Can someone help me with this? Thanks a lot!

• Use Lagrange multipliers to optimize $b-a$ subject to the integral equaling $1-\alpha$. (Arguably the length isn't as relevant as the ratio $b/a$, but one could use the same method to address that case, too.) – whuber Nov 28 '16 at 18:38
• S^2 is incorrectly defined. It should have n-1 in the denominator rather than n. With that change we also have in the formula for the chi square with n-1 degrees, n-1 replaces n. – Michael R. Chernick Nov 28 '16 at 23:22
• @whuber: Thanks! Minimising the difference between the bounds of the confidence interval worked for me. – WavesWashSands Nov 29 '16 at 9:55
• @MichaelChernick: It's just the (idiosyncratic?) way my professor defines S^2... I don't know why he does it either. – WavesWashSands Nov 29 '16 at 9:56
• pace @Michael, there's nothing incorrect at all about $S^2$. It's a perfectly fine statistic. Indeed, many well-respected statistics textbooks by excellent statistician use this formula throughout. – whuber Nov 29 '16 at 18:51