Question: Given a random sample $X_1 ... X_n$ from a cdf $F$, derive the asymptotic distribution of the the sample mean, $\overline{X}$.

I am not sure what is being requested here. Does one just need to apply the Central Limit Theorem?

In that case

$$ \sqrt{n}\frac{\overline{X}-\mu}{\sigma}\to_{(d)}\to Z \sim \mathcal{N}(0,1) $$

In that case I rewrite

$$ \sqrt{n}\frac{\overline{X}-\mu}{\sigma} = \frac{\sqrt{n}}{\sigma}\overline{X} + \left(-\frac{\mu\sqrt{n}}{\sigma}\right) $$

So I end up with

$$ \overline{X} \sim \mathcal{N}\left(-\frac{\mu\sqrt{n}}{\sigma} , \frac{n}{\sigma}\right) $$

Does this make sense?


No, the answer should be $\mathcal N(\mu, \sigma^2/n)$. Of course this is assuming the distribution has a finite variance.

  • $\begingroup$ Can you please explain how you arrived to that? $\endgroup$ – user473969 Sep 11 '17 at 19:43
  • $\begingroup$ If $\frac{\sqrt{n}}{\sigma} (\overline{X}-\mu) \approx Z$, then $\overline{X} \approx \frac{\sigma}{\sqrt{n}} Z + \mu$. Multiplying a mean-zero normal random variable by a positive constant multiplies the variance by the square of that constant; adding a constant to the random variable adds that constant to the mean, without changing the variance. $\endgroup$ – Robert Israel Sep 11 '17 at 19:48
  • $\begingroup$ Thanks I get it now $\endgroup$ – user473969 Sep 11 '17 at 19:49
  • $\begingroup$ Shouldn't it be $\mathcal N(\mu,\sigma^2/\sqrt n)$ ? $\endgroup$ – Prasun Biswas Sep 11 '17 at 19:51
  • $\begingroup$ @PrasunBiswas The standard deviation is $\sigma/\sqrt{n}$; the variance is $\sigma^2/n$. $\endgroup$ – Robert Israel Sep 12 '17 at 5:31

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