the 100(1-$\alpha$) confidence interval for $\sigma^2$

Question

Let $X_1, X_2$ be random samples from N($\mu, \sigma^2$). Let $V=X_1-X_2$.
I have to find out the 100(1-$\alpha$) confidence interval for $\sigma^2$.
I know that V~N(0,2$\sigma^2$).
Also I found out that $V^2$/2$\sigma^2$ follows the chi-square distribuiton.
This is where I'm stuck and I have no idea how to find out the 100(1-$\alpha$) confidence interval for $\sigma^2$ using these facts. Any suggestions please?

Answer 1

You already know that $V^2/(2\sigma^2)$ is a pivotal quantity that is chi-squared with $1$ degree of freedom. Therefore, $$\Pr\left[\chi_{1,\alpha/2}^2 < \frac{V^2}{2\sigma^2} < \chi_{1,1-\alpha/2}^2 \right] = 1 - \alpha,$$ where $\chi_{1,\alpha/2}^2$ and $\chi_{1,1-\alpha/2}^2$ are the $\alpha/2$ and $1 - \alpha/2$ quantiles of the chi-squared distribution with $1$ degree of freedom. Thus $$\Pr\left[\frac{V^2}{2\chi_{1,1-\alpha/2}^2} < \sigma^2 < \frac{V^2}{2\chi_{1,\alpha/2}^2}\right] = 1 - \alpha.$$ So your lower and upper confidence limits for an equal-tailed $100(1-\alpha)\%$ confidence interval will be $$L = \frac{(X_1 - X_2)^2}{2\chi_{1,1-\alpha/2}^2}, \quad U = \frac{(X_1 - X_2)^2}{2\chi_{1,\alpha/2}^2}.$$

Let's now look at an example. Suppose I generated two normal random variables with mean $\mu = 1$, $\sigma^2 = 2$. I got $$X_1 = 1.64735, \quad X_2 = -0.393634.$$ I want the equal-tailed $95\%$ confidence interval for $\sigma^2$ from this sample. (Although we know the true value of $\sigma^2$ in this case, for illustrative purposes, we seek to compute the interval estimate so that we can compare the result against the true variance.)

Then for a chi-squared distribution with $1$ degree of freedom, the $5^{\rm th}$ and $95^{\rm th}$ percentiles are $$\chi_{1,0.025}^2 \approx 0.000982069, \quad \chi_{1,0.975}^2 \approx 5.02389.$$ Then we compuute $$L \approx \frac{(1.64735 - (-0.393634))^2}{2(5.02389)} \approx 0.414581, \quad U \approx \frac{(1.64735 - (-0.393634))^2}{2(0.000982069)} \approx 2120.84.$$

This interval is huge. The reason is because the sample consists of essentially two realizations, $X_1$ and $X_2$. There is very little information from which to construct an interval estimate.

What would happen if instead we had a set of paired differences, say $$V_i = X_{i,1} - X_{i,2}, \quad i \in \{1, 2, \ldots, n\}?$$ Then $$\frac{V^2}{2\sigma^2} = \frac{1}{2\sigma^2} \sum_{i=1}^n V_i^2$$ would be chi-squared with $n$ degrees of freedom, and the corresponding confidence interval would look like $$L = \frac{V^2}{2\chi_{n,1-\alpha/2}^2}, \quad U = \frac{V^2}{2\chi_{n,\alpha/2}^2},$$ where the change from the previous $n = 1$ case is that the quantiles in the denominator are for the chi-squared distribution with $n$ degrees of freedom.