# Number of Gaussian mixture components needed to approximate any distribution

I remember reading an actual proven number of components, that can approximate any distribution. Somehow I think it was 18. Can someone point me to a book/article stating something of the sort?

Might've been in the following book, but I can't find the paragraph anymore.

G. McLachlan and D. Peel. Finite Mixture Models. Wiley Series in Probability and Statistics. Wiley, 2000. ISBN 9780471006268. URL http://books.google.cz/books?id=YXqflwEACAAJ.

I know how to set the number of components and all the surrounding theory, am looking for the stated lemma only.

• +1 because I am really curious. Having a particular number like that seems almost magical so I am curious to see it published. :) It would seem to render redundant a lot of work done in infinite Gaussian Mixture models... – usεr11852 Jun 6 '15 at 9:02
• @usεr11852 Don't hype your trousers off as I might just be remembering wrong ;) – mreq Jun 6 '15 at 9:08
• Without additional conditions (perhaps including some rather odd definition of 'approximate'), it seems trivial to disprove by counterexample. Consider an equal mixture of $k$ uniforms, each with range 0.01, and with the mean of the ith component $=i\,,\:i=1,\ldots,k$. – Glen_b Jun 7 '15 at 6:11
• @Glen_b: Obviously what you (and Xi'an) say is correct. The whole interesting part is what were those assumptions/conditions to begin with (aside assumptions like $k =18$, $\frac{N_i}{N_j} \rightarrow \infty, \forall i<19$ and $j>18$, etc, etc.) Awkward results are always fun! – usεr11852 Jun 7 '15 at 8:22

## 1 Answer

I am afraid this is an absurd question: there is no magical number and no upper bound on the number of components in a Gaussian mixture for approximating (in which sense?) any distribution. Just think of the Gaussian mixture with 19 components...