Suppose $X_1, ..., X_n, Y_1, ..., Y_m$ are independent random variables and $X_1, ..., X_n$ i.i.d.∼$\mathcal{N}(\mu_1, \sigma_1^2)$ and $Y_1, ..., Y_m$i.i.d.∼$\mathcal{N}(\mu_2, \sigma_2^2)$, with $\mu_1, \mu_2 \in \mathbb{R}$ known. For $\alpha \in (0, 1)$, how do I construct an exact $(1 − \alpha)$-confidence interval for the quantity $\frac{\sigma_1^2}{\sigma_2^2}$?
I am confused on how to construct confidence intervals, and I hope an explanation for this problem can help me on the subject. Thanks.
In the case where the means are unknown, the usual point estimate for the variance ratio is simply $s_1^2/s_2^2$ where $$s_1^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar x)^2, \quad s_2^2 = \frac{1}{m-1} \sum_{i=1}^m (y_i - \bar y)^2, \tag{1}$$ the ratio of the unbiased sample variances. Then since $$\frac{(n-1) s_1^2}{\sigma_1^2} \sim \chi_{n-1}^2 \tag{2}$$ and similarly for $Y$, the pivotal quantity $$\frac{\sigma_1^2}{\sigma_2^2} \cdot \frac{s_2^2}{s_1^2} \sim F(m-1, n-1) \tag{3}$$ is $F$-ratio distributed with $m-1$ and $n-1$ degrees of freedom, respectively. Thus the $100(1-\alpha)\%$ confidence interval is $$\left[\frac{1}{F^*_{\alpha/2}(n-1,m-1)} \frac{s_1^2}{s_2^2}, \frac{1}{F^*_{\alpha/2}(m-1,n-1)} \frac{s_1^2}{s_2^2}\right] \tag{4}$$ where $F^*_{\alpha/2}$ is the upper $\alpha/2$ quantile of the $F$-ratio distribution, namely $\Pr[F > F^*_{\alpha/2}] = \alpha/2$.
However, when $\mu_1, \mu_2$ are known, then the estimators $(1)$ are not efficient, and the resulting interval $(4)$ may have a larger than nominal coverage probability. Instead, one may consider the alternative unbiased estimators $$\varsigma_1^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \mu_1)^2, \quad \varsigma_2^2 = \frac{1}{m} \sum_{i=1}^m (y_i - \mu_2)^2. \tag{5}$$ In such a case, I believe (but have not bothered to prove it) that the pivotal quantity in $(3)$ becomes $$\frac{\sigma_1^2}{\sigma_2^2} \cdot \frac{\varsigma_2^2}{\varsigma_1^2} \sim F(m,n)$$ and the corresponding interval estimate becomes $$\left[\frac{1}{F^*_{\alpha/2}(n,m)} \frac{\varsigma_1^2}{\varsigma_2^2}, \frac{1}{F^*_{\alpha/2}(m,n)} \frac{\varsigma_1^2}{\varsigma_2^2}\right]. \tag{6}$$
At the moment I do not have the time to verify this result through simulation but it should be quite straightforward, so I will leave such details to the reader.