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A finite Gaussian mixture with $k$ components has a probability distribution function $p(y|\mu_1,...,\mu_k, \sigma_1, ..., \sigma_k, \pi_1, ..., \pi_k)=\sum_{j=1}^{k} \pi_j\mathcal{N}(\mu_j, \sigma_j^2)$ where $\mu_i$'s are means, $\sigma_j$'s are variances, and $\pi_j$'s are mixing parameters for gaussian distribution $\mathcal{N}(\cdot,\cdot)$.

I'd like to generalize this into an infinite Gaussian mixture with infinitely many mixing parameters being also Gaussian. This can be interpreted as 'applying Gaussian smoothing to a Gaussian distribution.' Is this still Gaussian? I think it is, but I have hard time to prove it. Any idea?

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What do you mean by "infinitely many mixing parameters being also Gaussian" ? Is your question whether the resulting mixture distribution itself a Gaussian? For a specific, even infinite MOG this is not strictly true, since say you have $\pi_1,..,\pi_n,..$ with $\sum_{j=1}^{\infty} \pi_j = 1$. Assume negatively that this is always a Gaussian, then so would a (normalized) mixture with only $\pi_2, .., \pi_n, ..$. And then your distribution is a mixture of two Gaussians. If you take some limit form in which all $\pi_j$'s tend to zero, then the resulting limit distribution may be Gaussian. –  Or Zuk Feb 9 '11 at 22:48
    
@Federico Could you precise what distribution(s) of the random parameter $(\mu,\sigma^2)$ you have in mind? Obviously the distribution of $\sigma^2$ cannot be (nondegenerate) Gaussian since $\sigma^2$ is almost surely positive. (Note that if $\sigma^2$ is in fact deterministic and $\mu$ is Gaussian $N(m,v)$, then $y$ is Gaussian $N(m,v+\sigma^2)$.) –  Did Mar 3 '11 at 23:09

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What do you mean by "is this still Gaussian?", a mixture of two gaussians with different means is not a gaussian (it is a bimodal distribution, so cannot be gaussian). Moreover, any distribution can be approximated (in a weak sense) by a mixture of gaussians, so by mixing gaussians one can get lots of different distributions.

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