dividing sample variance with variance

Earlier today i came across something that i could not quite understand so i figured maybe someone here sits on enough knowlege to help me out. The basic question is that one should decide what $a_s$ should be in the expression below such that the entire expression converges in distribution towards $Y$ where practiacally nothing is known about $Y$.

$$\frac {\overline{X} - \mu}{a_s \cdot s}$$

I first figured that one could simply set $a_s = \frac 1{\sqrt n}$ and use the central limit theorem ad say that it converges into a $N(0,1)$ distribution.

However my professor chose to extend the expression isntead such that:

$$\frac {\overline{X} - \mu}{a_s \cdot s} =\frac {\overline{X} - \mu}{\frac s {\sqrt{n}}} = \frac {\overline{X} - \mu}{\frac \sigma {\sqrt{n}}} \cdot \frac 1 {\sqrt{\frac {s^2(n-1)}{\sigma^2 (n-1)}}}$$

and then went on to deduct the final distribution of $Y$.

What i dont understand is first of all why the last two expressions in the equation are equal, and why the sample variance cant be used in the central limit theorem.

• If you take $s^2$ and $\sigma^2$ outside the squared root, the answer to your first question is pretty straightforward. Then, you just have convergence to a Student's $t$ rather than a Normal distribution. However, by increasing $n$, you also increase the $t$'s degrees of freedom, thing leading you to actually converge to a Normal distribution. – Federico Tedeschi Jun 1 '18 at 9:05
• i see, i guess i was overthinking it hehe, not so hard at all :) thank you – Janono Jun 1 '18 at 9:11
• @FedericoTedeschi Perhaps you should write that into an answer? It's very helpful! – surelyourejoking Jun 1 '18 at 11:43

If you take $s^2$ and $\sigma^2$ outside the squared root (by exploiting the fact tha $\sqrt{X^2}=X$), the answer to your first question is pretty straightforward. Then, you just have convergence to a Student's $t$ with $n-1$ degrees of freedom rather than to a Standard Normal distribution. Notice that, by increasing $n$, you also increase the $t$'s degrees of freedom. In particular, $n \rightarrow \infty$ implies $(n-1) \rightarrow \infty$ and, for the degrees of freedom tending to infinity, the Student's $t$ tend to a Standard Normal distribution $(N(0,1))$. To sum up:
1) In case you have observations from a Normal distribution, the distribution of the standardized sample mean (estimating the unknown variance with the unbiased version of the sample variance) is a Student's $t$,that converges to a $N(0,1)$.
2) In case you have i.i.d. variables with a finite variance, you can exploit the Central Limit Theorem and you'll have convergence to a Student's $t$, that implies again convergence to a $N(0,1)$.