1
$\begingroup$

I am learning some new statistics concepts, and I am having a bit of difficulty discerning certain notation. How does $\hat{\sigma}$ differ from $\sigma$ in the description below:

CLT: If $X_1,...,X_n \sim \text{IID Dist}(\mu, \sigma^2)$ then: $$Z_n = \frac{\sqrt{n} (\bar{X}_n - \mu)}{\sigma} \overset{\text{dist}}{\longrightarrow} Z \sim \text{N}(0, 1).$$

Theorem: If $\hat{\sigma}_n^2 \overset{p}{\longrightarrow} \sigma^2$ then: $$Z_n = \frac{\sqrt{n} (\bar{X}_n - \mu)}{\hat{\sigma}_n} \overset{\text{dist}}{\longrightarrow} Z \sim \text{N}(0, 1).$$

$\endgroup$
1
  • $\begingroup$ $\hat\sigma^2$ is the variance estimated from the data. It serves as best guess for the unknown variance $\sigma^2$. $\endgroup$
    – Michael M
    Commented Jan 13, 2020 at 7:51

1 Answer 1

2
$\begingroup$

In statistics, we put hats on parameters to indicate that we have an estimator of the parameter (which is a function of the observed data). Thus, the notation $\hat{\sigma}_n^2$ refers to an estimator of the unknown parameter $\sigma^2$, using $n$ data points. The theorem in your question says that if you have any variance estimator that converges in probability to the true variance (i.e., is weakly consistent) then you can replace the true variance parameter in the $Z$ statistic with that estimator, and you still get the same convergence to the normal distribution. (Incidentally, both the original statement of the CLT and the subsequent theorem are incomplete. Both results should include the condition of finite variance ---i.e., $\sigma^2 < \infty$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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