I keep reading in the documents about the asymptotic distribution of the CLT. I am learning things by myself so try to figure it out on my own and don't have a teacher to ask the question to.
So it's my understanding that the variance of the sample mean is:
$Var(\bar X) = {{ \sigma^2 } \over { n }}.$
I also know that you can write this variance as $E[(\bar X - \mu)^2]$ where $\bar X$ is the sample mean and $\mu$ is the population mean. you can definitely plot the first equation (if you know the population variance $\sigma^2$) and yes you would get a curve which is going very close to the x-axis (without even touching it); this my understanding of what they "asymptotic" word means.
I am trying to understand if this is what it means when I find in text books that the CLT has an asymptotic distribution? Could someone explained a bit more what it means when one says that "the Central Limit Theorem provides the asymptotic distribution of $\sqrt{N}(\bar X − \mu)$"?
Thank you very much.
The fact that the variance of $\bar X_n$ goes to zero when $n\to\infty$ (like $\sigma^2/n$) indicates that $\bar X_n$ looks more and more like the constant random variable equal to $E[\bar X_n]=\mu$. The CLT is a second order expansion of $\bar X_n$ around $\mu$. The surprise, when one first sees this result, is that the convergence of $\bar X_n-\mu$ suitably normalized happens in distribution (but not in probability, even less almost surely).
At the end of the day, the CLT is a relatively weak statement saying, first, that the correct normalization is $\sqrt{n}$ (not a surprise if one wants the variance to converge to a positive finite limit) and, second, that $Z_n=\sqrt{n}(\bar X_n-\mu)$ converges in distribution to a normal random variable $Z$ with mean $0$ and variance $\sigma^2$. This means that, for every real number $z$, $$ P[Z_n\leqslant z]\to P[Z\leqslant z]. $$ Since one knows the distribution of $Z$, one knows that the value of the RHS is $$ P[Z\leqslant z]=\int_{-\infty}^{x/\sigma}\frac1{\sqrt{2\pi}}\mathrm e^{-t^2/2}\mathrm dt. $$ Note that the almost sure behaviour of $Z_n$ is another matter, since $\limsup\limits_{n\to+\infty}Z_n=+\infty$ and $\liminf\limits_{n\to+\infty}Z_n=-\infty$, almost surely (a keyword here being law of iterated logarithm).
To sum up, the fact that the variance is $\sigma^2/n$ suggests the normalization in the CLT but the CLT itself is a much more encompassing result. For example, the fact that a unique distribution (the normal one) captures the behaviour of every $\bar X_n$ (provided square integrability) should be truly bewildering.