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In the set-up for the classical CLT we have that

$$\frac{\sqrt{n}}{\sigma}(\bar{X}_n-\mu)\to^d N(0,1)$$

as $n\to \infty$, which gives rise to the $1-\alpha$ asymptotic confidence interval formula for $\bar{X}_n$:

$$[\mu-\frac{\sigma}{\sqrt{n}}z_{\alpha/2},\mu+\frac{\sigma}{\sqrt{n}}z_{\alpha/2}]$$

In practice $\mu$ and $\sigma$ are unknown, and so we replace those quantities with the sample mean $\hat{\mu}$ and the sample standard deviation $\hat{\sigma}$ respectively.

My question is what is the rigorous justification for this?

Since the sample standard deviation $\hat{\sigma}$ is a consistent estimator for $\sigma$, an application of Slutsky theorem gives

$$\frac{\sqrt{n}}{\hat{\sigma}}(\bar{X}_n-\mu)\to^d N(0,1)$$

so I understand that replacing $\sigma$ with $\hat{\sigma}$ is justified. However the result

$$\frac{\sqrt{n}}{\hat{\sigma}}(\bar{X}_n-\hat{\mu})\to^d N(0,1)$$ is clearly false, and in fact

$$P(\bar{X}_n \in [\hat{\mu}-\frac{\hat{\sigma}}{\sqrt{n}}z_{\alpha/2},\hat{\mu}+\frac{\hat{\sigma}}{\sqrt{n}}z_{\alpha/2}])=1$$ for all $n$.

Am I missing someting?

Alain-Philippe Fortin
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  • The part that is "clearly false" and its sequel are irrelevant, because the defining condition for a confidence interval is that it cover the true mean, not the estimated one. – whuber Sep 27 '20 at 12:07
  • @whuber Yes you are right. I messed up the definition of a confidence interval... – Alain-Philippe Fortin Sep 27 '20 at 12:49
  • Your first "confidence" interval actually is a probability interval, see here. I can't get what application of Slutsky theorem you are speaking of, and $\sqrt{n}(\overline{X}n-\mu)/\hat{\sigma}\sim t{n-1}$. The part that is "clearly false" does not make sense to me, because $\hat\mu=\overline{X}_n$. – Sergio Sep 27 '20 at 13:11

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As whuber said, the definition of a confidence interval is that it must cover the true mean $\mu$, not the sample mean $\bar{X}_n$.

Alain-Philippe Fortin
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