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?

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

  • $\begingroup$ What's G? Please make your notation more explicit. $\endgroup$ – Glen_b Jul 4 '14 at 7:45
  • $\begingroup$ I think G is for Gaussian. $\endgroup$ – soakley Jul 4 '14 at 12:51
  • $\begingroup$ Sorry, G is for Gaussian. I will change it to Normal in case it confuses anyone. $\endgroup$ – Vincent Wen Jul 4 '14 at 13:38

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.