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Let $\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i$, and $S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \overline{X})^2$.

It is well known that if the $X_i$ are IID normals, say $X_i \sim N(\mu, \sigma)$, then $\frac{\overline{X} - \mu}{S / \sqrt{n}}$ is $t$-distributed with $n-1$ degrees of freedom. In turn, as $n$ goes to infinity, this will converge in distribution to a $N(0, 1)$ variable.

I have two questions (I have not tried to prove the statements, and couldn't easily find precise references):

1) Is it true that if the $X_i$ are IID (with mean $\mu$, say, and finite variance), not necessarily normal, then $\frac{\overline{X} - \mu}{S / \sqrt{n}}$ converges in distribution to a $N(0, 1)$ variable?

2) Is it true that if $X_i$ are independent, not necessarily identically distributed or normal (but with some suitable growth bound on the variances), then $\frac{\overline{X} - \frac{1}{n} \sum_{i=1}^n E[X_i]}{S / \sqrt{n}}$ converges in distribution to a $N(0, 1)$ variable?


Seems like the answer to the first question is here: Is there a theorem that says that $\sqrt{n}\frac{\bar{X} - \mu}{S}$ converges in distribution to a normal as $n$ goes to infinity?

I suppose I can arrive at the same conclusion for the second question by imposing Lyapunov's central limit theorem.

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    $\begingroup$ These questions make no sense, because "$n$" cannot possibly appear in any formula for what the sequence converges to. $\endgroup$ – whuber Dec 28 '16 at 19:57
  • $\begingroup$ The allegedly nonsensical part or the question can be split in two questions: Does $\frac{\overline{X} - \mu}{S / \sqrt{n}}$ converge to normal? When is the distribution of $\frac{\overline{X} - \mu}{S / \sqrt{n}}$ well approximated by a t-student? In fact, the whole question seems equivalent to "When can we use t-test with a non normal variable?" which makes a lot of sense. $\endgroup$ – Pere Dec 28 '16 at 20:07
  • $\begingroup$ Agreed, @whuber. I re-formulated the question to what I really need to know. Thanks. $\endgroup$ – Christian Dec 28 '16 at 21:12
  • $\begingroup$ You might want to consider coupling Slutsky's theorem with whichever version of the CLT will work for known variance (or set of variances) as suited for your question. Write the appropriate fraction as a CLT-relevant term in the numerator, and a $s/\sigma$ type term in the denominator, apply Slutsky to the ratio X/Y where Y -> 1 and apply CLT.to X. Just be careful to set it up correctly for the different variances version $\endgroup$ – Glen_b Jan 16 '17 at 6:15

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