Does $\frac{\overline X - \mu}{S/\sqrt n}$ converge to $N(0,1)$?

Question

Suppose that $X_1,...,X_n,...$ are i.i.d. with $EX_i = \mu < \infty$ and $Var X_i = \sigma^2 < \infty$. Now, I know that by the Central Limit Theorem that

$$ \frac{\overline X - \mu}{\sigma/\sqrt n} \rightharpoonup N(0,1). $$

Now, suppose that instead of the actual $\sigma^2$, I use the sample standard deviation $S^2 = \sum^n_{i=1}\frac{(X_i-\overline X)^2}{n-1}$. Thus, is it true that $$ \frac{\overline X - \mu}{S/\sqrt n} \rightharpoonup N(0,1). $$

My intuition is that this is true, but surprisingly I was not able to find easily anywhere. Sorry if this is a silly question. [

I think you want $S / \sqrt n$ in the denominator, not just $S$, right? — Aaron Montgomery, Aug 31 '21 at 19:34
If you know the mean, why use $\sum^n_{i=1}\frac{(X_i-\overline X)^2}{n-1}$ rather than $\sum^n_{i=1}\frac{(X_i-\mu)^2}{n}$? — Henry, Aug 31 '21 at 21:50

score 3 · Accepted Answer · answered Aug 31 '21 at 19:49

The sample variance $S^2$ is consistent for $\sigma^2$.

By the Mann-Wald (continuous mapping) theorem, since the function $g: \mathbb{R}_{\geq 0} \to \mathbb{R}$ defined by $g(x) = \sqrt{x}$ is continuous, it follows that $g(S^2) = S$ is consistent for $g(\sigma^2) = \sigma$.

Write $$\dfrac{\overline{X} - \mu}{S/\sqrt{n}} = \dfrac{\sigma}{S} \cdot \dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}}\text{.}$$ By the CLT, $\dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}} \overset{d}{\to} \mathcal{N}(0, 1)$.

By Mann-Wald again, we have that $h: \mathbb{R}_{> 0} \to \mathbb{R}$ given by $h(x) = \dfrac{\sigma}{x}$ is continuous, hence since $S$ is consistent for $\sigma$, $h(S) = \dfrac{\sigma}{S} $ is consistent for $h(\sigma) = \dfrac{\sigma}{\sigma} = 1$.

By Slutsky's Theorem, $$\dfrac{\overline{X} - \mu}{S/\sqrt{n}} = \dfrac{\sigma}{S} \cdot \dfrac{\overline{X} - \mu}{\sigma/\sqrt{n}} \overset{d}{\to} 1 \cdot \mathcal{N}(0, 1) = \mathcal{N}(0, 1)\text{.}$$

score 3 · Answer 2 · edited Sep 01 '21 at 08:19

Yes, this is the Slutsky's theorem https://en.wikipedia.org/wiki/Slutsky%27s_theorem.

$$\frac{\bar{X}-\mu}{S/\sqrt{n}}=\frac{\bar{X}-\mu}{\sigma/ \sqrt{n}}\frac{\sigma}{S}\rightarrow\mathcal{N}(0,1)\quad\text{(in distribution)}$$

because $\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\rightarrow\mathcal{N}(0,1)$ in distribution and $\frac{\sigma}{S}\rightarrow 1$ in probability. This happens with every consistent estimator.

Does $\frac{\overline X - \mu}{S/\sqrt n}$ converge to $N(0,1)$?

2 Answers2

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