Confidence interval for the standard deviation of a Normal distribution with known mean

Question

Suppose $\textbf{Y} = (Y_{1}, ... , Y_{n})$ is a random sample from the $N(\mu, \sigma_{0}^{2})$ distribution where $\mathrm{E}(Y_{i}) = \mu$ is unknown but $\mathrm{SD}(Y_{i}) = \sigma_{0}$ is known.

It's possible to construct a confidence interval for the mean of a Normal distribution with known standard deviation using the pivotal quantity:

$$Q = \frac{\bar{Y} - \mu}{\sigma_{0} / \sqrt{n}}$$

Then, suppose $\mathrm{E}(Y_{i}) = \mu_{0}$ is known but $\mathrm{SD}(Y_{i}) = \sigma$ is unknown, why can't we just use the following pivotal quantity to construct a confidence interval for the standard deviation with known mean?

$$Q = \frac{\bar{Y} - \mu_{0}}{\sigma / \sqrt{n}}$$

Answer 1

After looking around for a while without finding anything satisfactory, this is the best answer that seems to make sense to me:

Notice that the sampling distribution of the unknown $\sigma$ is not Normal, so $Q = \frac{\bar{Y} - \mu_{0}}{\sigma / \sqrt{n}}$ does not actually follow $N(0, 1)$, thus it cannot be a pivotal quantity, at least not a pivotal quantity with a Normal distribution.

The case for unknown $\mu$ however, is different because the sampling distribution of $\mu$ is Normal by Central Limit Theorem, so we know $Q = \frac{\bar{Y} - \mu}{\sigma_{0} / \sqrt{n}}$ follows $N(0, 1)$, which makes it a pivotal quantity.

This is why to find the confidence interval for $\sigma$, we have to use the pivotal quantity $$\frac{(n-1)S^2}{\sigma^2},$$ which follows a $\chi^2$ distribution with $n-1$ degrees of freedom.