Central limit theorem question

I am thinking about the rate of convergence in central limit theorem (CLT) for different distributions. Let's assume we have a set of i.i.d random variables, $X_1,X_2,\ldots$ which follow an unknown distribution, $F$ with finite second moment. Then using the CLT we have $$\frac {\sum_{i=1}^{k}X_i-k\mu}{\sigma \sqrt k} = \frac {(1/k)\sum_{i=1}^{k}X_i-\mu}{\sigma /\sqrt k}\rightarrow N(0,1)$$

when $k\rightarrow \infty$. My question is about $k$. Is there a general rate of convergence for $k$ for a certain amount of error? For example if $F \sim N(a,b)$, then for any $k\geq 1$ the distribution of the scaled and centered sample mean is standard normal, without the need to invoke the CLT (so in a sense, here the CLT "holds with error zero"). I would be more happy if you refer me to a reference or a paper.

• My understanding is that CLT is satisfied only when k goes to infinity and therefore there is no general rate of convergence. Your example only works because you're like sampling and comparing with normals. While I'm not 100% sure, my answer would be "no, the theorem is only valid asymptotic". The error has already been specified in the question by yourself. – SmallChess Apr 6 '15 at 12:04
• Thanks @StudentT yes I think so. But I guess there might be some research on for example exponential family distributions or ... – TPArrow Apr 6 '15 at 12:12
• I do not get the question: $k$ is an integer that goes to infinity, there is no rate involved, except that the proper normalisation of the sum is $\sqrt{k}$. – Xi'an Apr 6 '15 at 13:13
• The expression in the question representing what the CLT says, is wrong, from various aspects. One should distinguish between the assertion related to a limiting distribution, from the approximate result used when we have a large but finite sample. But even in such a case, the variance (or standard deviation) shown is wrong. – Alecos Papadopoulos Apr 6 '15 at 13:51
• Note that you can exchange notation $\lim_{k\to \infty}x_k=a$ with notation $x_k\to a$ as $k\to\infty$ if and only if $a$ is constant which does not depend on $k$. Furthermore if you want to apply CLT you need to center the your sum and use normalisation. The classical CLT for iid case is stated as $\sqrt{n}(\bar X-\mu)\to N(0,\sigma^2)$, where $\bar X=\frac{1}{n}\sum_{i=1}^nX_i$, $EX_i=\mu$, $VarX_i=\sigma^2$. The centering and normalisation by $\sqrt{n}$ is important, without the it is easy to demonstrate the failure of CLT. – mpiktas Apr 6 '15 at 14:07

Is there a general rate of convergence for $k$ for a certain amount of error?
It is my impression that you are searching for literature on things like the Berry-Esseen bounds: for given $k$ can we bound the error? Alternatively, for given desired approximation error, what sample size do we need? Etc