I'm currently doing a self-study of Wasserman's All of Statistics.

Theorem 5.10 states that for $X_1, ..., X_n$ IID with mean $\mu$ and variance $\sigma^2$, we have $\frac{\sqrt{n}(\bar{X}_n - \mu)}{S_n} \xrightarrow{D} N(0,1)$, where $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$ and $S^2_n = \frac{1}{n - 1}\sum_{i=1}^n(X_i - \bar{X}_n)^2$. This statement seems to be the same statement as the Central Limit Theorem but with $\sigma$ replaced by $S_n$.

My questions are: is this theorem correct, and how would one prove this? No proof is given and I cannot manage to find this result elsewhere.

I want to emphasize that I'm wondering whether this statement is true as stated and not wondering whether it is useful to approximate $\sigma^2$ with $S^2_n$ for other purposes.


1 Answer 1


Yes it is a result of Slutsky's Theorem since $S_n \rightarrow_p \sigma$


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