[Math] Mixtures of Gaussian distributions dense in distributions

pr.probabilityst.statistics

It seems that a mixture of Gaussians can approach any probability distribution, as the number of mixture components approaches infinity. Is this true? And if so, is it precise and correct to say that 'the Gaussian distribution is a dense subset of any distribution function'?

If not, how would you phrase this?

I have never seen anything about this in the literature about infinite mixtures of Gaussians.

Thanks very much!

Best Answer

One way to say this is: Given any random variable $X$, there is a sequence of random variables $X_n$ whose distributions are finite mixtures of Gaussians, such that $X_n \Rightarrow X$ (i.e. $X_n$ converges to $X$ in distribution, or weakly). I don't believe you need to consider infinite mixtures per se.

It is true. First, note that any constant random variable can certainly be approximated in distribution by Gaussians (just let the variance tend to 0, or maybe you already consider constants to be Gaussian). But any random variable can be approximated in distribution by a mixture of constants. (Approximate the cdf of $X$ by step functions.)

In functional analysis language, one might say that in the Banach space $\mathcal{M}(\mathbb{R})$ of finite signed (or complex) measures on $\mathbb{R}$, the weak-* closed convex hull of the Gaussian probability measures contains all the probability measures. Equivalently, the convex combinations (i.e. mixtures) of the Gaussian probability measures are weak-* dense in the probability measures.

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