## Completeness of an “infinite mixture of gaussians” representation

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Is there a complete "infinite mixture of gaussians representation" for densities? What I mean is, is there, for any reasonably big class of densities $\phi(x)$ I can come up with a function $c(\mu, \tau)$ such that:

$\displaystyle\phi(x) = \int d\mu d\tau\; c(\mu, \tau) \exp\left(-\frac{\tau (x-\mu)^2}{2}\right)$??

($\tau = \frac{1}{\sigma}$ is the inverse of the variance and the normalization of the gaussian was absorbed in $c(\mu,\tau)$)

I'm using the word "complete" in the sense theoretical physicists talk about the completeness of a basis for a vector space. I'm not sure if it's an adequate use of the term for mathematicians (probably not).

If yes, does it generalize for multivariate distributions? Also, is there a known "inverse integral transform" for this representation? Something like:

$\tilde{\phi}(\Sigma, \mu) = \int dx \; \tilde{\mathcal{N}}(\Sigma, \mu | x) \phi(x)$

$\phi(x) = \int d\Sigma \mathcal{N}(x | \Sigma, \mu) \tilde{\phi}(\Sigma,\mu)$

where $\Sigma$ and $\mu$ are the covariance matrix and the vector of means.

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## 1 Answer

How about this: the set of finite mixtures of (non-degenerate) Gaussians is weakly dense in the space of probability measures on $\mathbb{R}$. Proof: the set of finite mixtures of constants is certainly weakly dense. But a constant can be approximated as a Gaussian with small variance.

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So, the gaussians have even more "degrees of freedom" than necessary to represent all densities?  I guess this kills my hope for an inversion formula...  – Rorsa Sep 9 2010 at 23:38
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