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I am given the question: "What is the probability that both the mean is in its confidence interval for confidence level a and the variance is in its confidence interval for confidence level a?"

Note that these CIs are the t-distribution interval for the mean and the chisq-distribution interval for the variance. I am very confused by how to begin as I am under the impression that CIs are not probabilistic.

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    $\begingroup$ Welcome to CV! Re "not probabilistic:" CIs are random intervals by definition. You had better start, then, by reviewing our most popular posts on confidence intervals. Your question is not answerable in general, though, because that probability depends on the procedures you use to compute those CIs as well as your assumptions about the underlying distribution. It can also depend on the specific circumstances: that is, even having all this information might not permit you to compute the answer. $\endgroup$
    – whuber ♦
    Commented Apr 24, 2023 at 22:44
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    $\begingroup$ What would be nice if the event of the confidence interval for the mean covering the population mean and the event of the confidence interval for the variance covering the population variance were independent events, so you could just multiply the confidence levels as in $0.95^2=0.9025$. Sadly this is not the case even when sampling from a normal distribution; simulation suggests the joint probability tends to be higher than this. $\endgroup$
    – Henry
    Commented Apr 24, 2023 at 23:18

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