The central limit theorem as I am familiar with it applies to the limiting (rescaled) distribution of $n$ convolutions of a single probability distribution as $n$ goes to infinity, or equivalently, to distribution one gets from taking a sum of $n$ random variables each with a single fixed distribution. That is, it is a theorem about the (limiting as $n\to \infty$) probability distribution of $A_1 + A_2 + ... + A_n$ where each term has a fixed distribution $P$.

I am asking about a theorem about the limiting probability distribution of $A_1 + A_2 + ... + A_n$ where $A_1$ has probability distribution $P_1$, $A_2$ has probability distribution $P_2$, $A_3$ has probability distribution $P_3$, etc.

Also, is there a theorem for the case where each distribution isn't fixed, but is selected at random with probability determined by a measure $\mu$?

Is there such a general theorem, where the limit isn't necessarily gaussian, the limit can be reconstructed from $\mu$, and the convergence is pretty strong?

  • $\begingroup$ Could you give more details for your second question? How do you define selection of distribution at random with probability determined by measure $\mu$? $\endgroup$ – mpiktas Mar 28 '11 at 20:20
  • $\begingroup$ I've introduced latex and changed some sentences, please check if this is ok. $\endgroup$ – mpiktas Mar 28 '11 at 20:26
  • $\begingroup$ what constitutes "pretty strong" convergence? Are you asking about bounds? $\endgroup$ – mpiktas Mar 28 '11 at 20:27
  • $\begingroup$ I suppose I was really just trying to avoid a theorem that gives convergence in such a weak sense that, while technically true, it has little application, e.g. pointwise converence almost everywhere. $\endgroup$ – John Robertson Mar 30 '11 at 20:42
  • $\begingroup$ if you are talking about convergence of distribution functions, then convergence pointwise almost everywhere (on the continuity points to be more precise) is called weak convergence or convergence in distribution and is the most applied type of convergence, since all the assymptotic statistics are based on it. $\endgroup$ – mpiktas Mar 30 '11 at 20:59

The theorem 3.1 in this book answers your first question. The key restriction in central limit theorem is not different distributions but the independence. The result is a very nice one, since it says that for interesting sums of independent random variables the limiting distribution has to have certain property, namely infinite divisibility. The classical central limit theorem (with iid variables with finite variances) is then only a very special case of this theorem.

Note that this is a very general answer to very general question. Given the nature of your distributions more precise answer can be given. For example if the distributions satisfy Lindeberg's condition then the limiting distribution is necessary normal (if we exclude let us say non-interesting cases).

  • $\begingroup$ What do you mean by interesting and non-interesting cases? $\endgroup$ – Ashok Mar 29 '11 at 5:10
  • $\begingroup$ @Ashok, interesting cases are where condition of infinite smallness is satisfied. Naturally you can consider that the interesting cases are where this condition is not satisfied. $\endgroup$ – mpiktas Mar 29 '11 at 8:07
  • $\begingroup$ Very interesting. Is the theorem you cite effectively the theorem loosely summarized in the wikipedia article you cited. I don't have access to that book, especially at a retail of $250. :) $\endgroup$ – John Robertson Mar 30 '11 at 20:54
  • $\begingroup$ The wikipedia summary is quite loose, as there must clearly be more required or there are requirements on the members of the triangular array that aren't stated there, otherwise you could take two such triangular arrays with different limits and interchange every other row with the result that now neither one converges (even though the wikipedia article sounds like it says they should) $\endgroup$ – John Robertson Mar 30 '11 at 20:57
  • $\begingroup$ @John, the access in the google books is not enough? You might get access to the book via library you know, it is old-fashioned way, but it works :) $\endgroup$ – mpiktas Mar 30 '11 at 21:02

mpiktas gave a very good technical answer.

There's a nice simulation of this here:


You can manipulate the distribution at the top of this demo to show a combination of different distributions (such as bimodal), and the distribution of sample means will still be normal.


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