# Shortest Confidence interval for $\sigma^2$

I have a $X_1, \cdots , X_n$ random samples from a $N(\mu,\sigma^2)$. $\mu$ is known, but $\sigma^2$ is unknown. I would like to know how to go about constructing a $(1-\alpha)$100% shortest confidence interval for $\sigma^2$.

I know how how to construct that of a known variance, but unknown mean. I'm bit confused as how to proceed when the situation is reversed.

Thank you.

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The shortest interval can be found by solving for $a$ and $b$ such that $a^2 f(a) = b^2 f(b)$ subject to $\int_{a}^b f(x) \,\mathrm{d}x = \alpha$ where $f$ is the density of a chi-square distribution with $n$ degrees of freedom. No closed form exists, so people in practice tend to use an equal-tails confidence interval, which is the one that @Sivaram gave. However, the equal-tails interval is not the shortest possible. – cardinal Jun 23 '11 at 13:10
The integral in my previous comment should obviously be equal to $1-\alpha$. Apologies for the typo. – cardinal Jun 23 '11 at 13:29

First note that $\mathbb{E}\left[(X-\mu)^2 \right] = \sigma^2$

An unbiased estimator of $\sigma^2$ is given by$$\hat{s}^2 = \displaystyle \frac{\displaystyle \sum_{k=1}^{n} \left( X_i - \mu \right)^2}{n}$$ Now we have that $\displaystyle \frac{n \hat{s}^2}{\sigma^2}$ is a $\chi^2_n$ i.e. a $\chi^2$ random variable with $n$ degrees of freedom (since $X_i$ are normally distributed random variables)

Find $a$ and $b$ such that $\displaystyle 1- \frac{\alpha}{2} = P\left(\chi_n^2 \leq b \right)$ and $\displaystyle \frac{\alpha}{2} = P\left(\chi_n^2 \leq a \right)$

Hence, we have $$\displaystyle P \left(a \leq \chi_n^2 \leq b \right) = 1 - \alpha$$ $$\displaystyle P \left(a \leq \frac{n \hat{s}^2}{\sigma^2} \leq b \right) = 1 - \alpha$$ $$\displaystyle P \left(\frac{n \hat{s}^2}{b} \leq \sigma^2 \leq \frac{n \hat{s}^2}{a} \right) = 1 - \alpha$$

Hence, the desired $1-\alpha$ confidence interval is given by $$\left(\frac{n \hat{s}^2}{b},\frac{n \hat{s}^2}{a} \right)$$ where $a$ and $b$ are given by the $\displaystyle F_{\chi}(a) = \frac{\alpha}{2}$ and $\displaystyle F_{\chi}(b) = 1 - \frac{\alpha}{2}$ where $F_{\chi}$ denotes the cumulative distribution function.

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thanks for your answer. How do we show that the given interval is the shortest? – Mike Jun 23 '11 at 8:16
@Mike: You may want to look at this. en.wikipedia.org/wiki/Minimum-variance_unbiased_estimator – user17762 Jun 23 '11 at 8:19
Is it the shortest becasue $\hat{s}^2$ is the MVUE? – Mike Jun 23 '11 at 8:42
(-1) This is not the shortest interval. Indeed, no closed form for thresholds $a$ and $b$ exist for giving the shortest interval. Try an example, like, say, $n = 10$, $\alpha = 0.05$. For more information, see, e.g., p. 383 of Mood, Graybill and Boes, 3rd. ed. (1974). – cardinal Jun 23 '11 at 13:05

It is described here how to construct confidence intervals for the variance in the normal model when $\mu$ is known (as well as when $\mu$ is unknown).

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This is probably not what you are looking for, but maybe it can help you... – Shai Covo Jun 23 '11 at 8:03
@Shai: It helped, but I didn't understand what they meant by "use the pivotal quantity $V$ to show the given 1-$r$ confidence interval... – Mike Jun 23 '11 at 8:06
I'll consider your question shortly. – Shai Covo Jun 23 '11 at 8:38
Ok. Thanks....:) – Mike Jun 23 '11 at 8:43
Consider the problem backwards (I use $d^2$ instead of $\sigma^2$). You want to show that ${\rm P}(\frac{{nW^2 }}{{\nu _{n.1 - r/2} }} \le d^2 \le \frac{{nW^2 }}{{\nu _{n.r/2} }}) = 1 - r$, that is ${\rm P}(\nu _{n,r/2} \le \frac{{nW^2 }}{{d^2 }} \le \nu _{n,1 - r/2} ) = 1 - r$. Now note that $V = nW^2 /d^2$ has the $\chi ^2$ distribution with $n$ degrees of freedom... – Shai Covo Jun 23 '11 at 8:53