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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – SmallChess
    Commented Apr 6, 2015 at 12:04
  • $\begingroup$ Thanks @StudentT yes I think so. But I guess there might be some research on for example exponential family distributions or ... $\endgroup$
    – TPArrow
    Commented Apr 6, 2015 at 12:12
  • 3
    $\begingroup$ 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 $\sqrt{k}$. $\endgroup$
    – Xi'an
    Commented Apr 6, 2015 at 13:13
  • 3
    $\begingroup$ 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. $\endgroup$ Commented Apr 6, 2015 at 13:51
  • 2
    $\begingroup$ 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. $\endgroup$
    – mpiktas
    Commented Apr 6, 2015 at 14:07

1 Answer 1


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

There are Uniform bounds (for all values in the support), and local bounds (that depend on a specific value of the support). Some results relate to specific distributions.

Chapter 11 "Accuracy of Central Limit Theorems" from A. Dasgupta's Asymptotic Theory of Statistics and Probability is a good place to start, and it contains also more literature references.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.