Understanding the Confidence Interval for Variance $\sigma^2$

We are given a confidence interval $I_{a,b} = \left[ \frac{\sum (X_i - \bar{X})^2}{b}, \frac{\sum (X_i - \bar{X})^2}{a} \right]$ for the unknown variance $\sigma^2$ of $n$ i.i.d. $N(\mu, \sigma^2)$ variables. Let $V = \sum (X_i - \bar{X})^2$. The quantity $\frac{V}{\sigma^2}$ follows a chi-square distribution with $n-1$ degrees of freedom, $\chi^2_{n-1}$.

A $(1-\alpha)\times 100\%$ confidence interval requires $P\left(\frac{V}{b} \le \sigma^2 \le \frac{V}{a}\right) = 1 - \alpha$. This is equivalent to $P\left(a \le \frac{V}{\sigma^2} \le b\right) = 1 - \alpha$. Using the cumulative distribution function $G_{n-1}$ of the $\chi^2_{n-1}$ distribution, this condition becomes $G_{n-1}(b) - G_{n-1}(a) = 1 - \alpha$. For a $95\%$ interval, $1 - \alpha = 0.95$. Thus, we need $G_{n-1}(b) - G_{n-1}(a) = 0.95$. We analyze each option based on this framework.

Analyzing Option A: Possibility of $ab=1$

Option A states it's possible to have a $95\%$ confidence interval where $ab = 1$. This implies $b = 1/a$. The condition becomes $G_{n-1}(1/a) - G_{n-1}(a) = 0.95$. For the interval bounds to be valid, we require $b > a > 0$. If $ab = 1$, then $1/a > a$, which means $a^2 < 1$. Since $a > 0$, we must have $0 < a < 1$, which implies $b = 1/a > 1$.

Let $f(a) = G_{n-1}(1/a) - G_{n-1}(a)$ for $a \in (0, 1)$. The range of the CDF $G_{n-1}(x)$ is $(0, 1)$.

• As $a \to 0^+$, $1/a \to \infty$. $G_{n-1}(a) \to 0$ and $G_{n-1}(1/a) \to 1$. Thus, $\lim_{a \to 0^+} f(a) = 1$.
• As $a \to 1^-$, $1/a \to 1^+$. $G_{n-1}(a) \to G_{n-1}(1)$ and $G_{n-1}(1/a) \to G_{n-1}(1)$. Thus, $\lim_{a \to 1^-} f(a) = 0$.

Since $f(a)$ is a continuous function on $(0, 1)$ ranging from $1$ to $0$, by the Intermediate Value Theorem, there must exist an $a \in (0, 1)$ such that $f(a) = 0.95$. Therefore, Option A is true.

Analyzing Option B: Symmetric Interval vs. Shortest Interval

Option B considers the case where $G_{n-1}(a) = 0.025$ and $1 - G_{n-1}(b) = 0.025$. This means $G_{n-1}(a) = 0.025$ and $G_{n-1}(b) = 0.975$. This defines a $95\%$ confidence interval ($0.975 - 0.025 = 0.95$). This corresponds to an interval symmetric in probability mass excluded from the tails ($2.5\%$ in each tail).

However, the length of the interval is $L = V(\frac{1}{a} - \frac{1}{b})$. The expected length is $E[L] = (n-1)\sigma^2 (\frac{1}{a} - \frac{1}{b})$. The shortest interval minimizes this expected length. Minimizing $\frac{1}{a} - \frac{1}{b}$ subject to $G_{n-1}(b) - G_{n-1}(a) = 0.95$ does not generally yield the interval symmetric in probability tails for a skewed distribution like chi-square. Thus, the symmetric interval is not necessarily the shortest. Option B is false.

Analyzing Option C: Condition for Shortest Interval

The shortest confidence interval is found by minimizing the expected length $E[L] = (n-1)\sigma^2 (\frac{1}{a} - \frac{1}{b})$ subject to the constraint $G_{n-1}(b) - G_{n-1}(a) = 0.95$. Using optimization techniques (like Lagrange multipliers), the condition derived for the shortest interval is $(n+1) \log(b/a) = b-a$. The condition stated in Option C is $b - a = (n-3) \log \frac{b}{a}$. These two conditions are different.

For example, if $n=4$, the shortest interval condition is $5 \log(b/a) = b-a$, while Option C's condition is $b-a = 1 \log(b/a)$. These are not equivalent. Therefore, Option C is false.

Analyzing Option D: Expected Length Calculation

Option D concerns the expected length of the $95\%$ confidence interval $I_{a,b}$ where $G_{n-1}(b) - G_{n-1}(a) = 0.95$.

The interval is $I_{a,b} = \left[ \frac{V}{b}, \frac{V}{a} \right]$, where $V = \sum (X_i - \bar{X})^2$. The length of the interval is $L = \frac{V}{a} - \frac{V}{b} = V \left( \frac{1}{a} - \frac{1}{b} \right)$.

The expected length is $E[L] = E\left[ V \left( \frac{1}{a} - \frac{1}{b} \right) \right]$. Since $a$ and $b$ are constants, this is $E[L] = \left( \frac{1}{a} - \frac{1}{b} \right) E[V]$.

We know $\frac{V}{\sigma^2} \sim \chi^2_{n-1}$. The expected value of a $\chi^2_k$ distribution is $k$. So, $E\left[ \frac{V}{\sigma^2} \right] = n-1$. This implies $E[V] = (n-1)\sigma^2$.

Substituting $E[V]$ back into the expression for $E[L]$:

This matches the expression given in Option D. Therefore, Option D is true.

Conclusion: True Statements

Based on the analysis, Option A and Option D are true statements regarding the confidence interval for $\sigma^2$. Option B is false because the symmetric interval is not necessarily the shortest. Option C is false because the condition provided does not match the derived condition for the shortest interval.