Does the Central Limit Theorem hold if you replace the true variance with the sample variance?

Question

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.

Answer 1

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

Answer 2

You appear to be interested in a quantity like $$ \mathcal{T}=\frac{\sqrt T(\bar{y}-\mu)}{s}, $$ with $s$ the sample standard deviation of $y_t$, $t=1,\ldots,T$.

By the CLT, $\sqrt T(\bar{y}-\mu)\to_d \mathcal{N}(0,\sigma^2)$ when (e.g.) the $y_i$ are iid with mean $\mu$ and variance $\sigma^2$.

By the WLLN, $s^2\to_p\sigma^2$.

By the continuous mapping theorem, $s\to_p\sigma$.

By Slutzky's theorem, $\mathcal{T}\to_d \mathcal{N}(0,1)$.