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We have a GMM model $\sum_{i=1}^N \alpha_k \mathcal{N}(x; \mu_k, \sigma_k^2I)$. Assuming that GMM is a single gaussian distribution $\mathcal{N}(x; \mu_\theta, \sigma_\theta^2I)$. It means that we have $$ \mathcal{N}(x; \mu_\theta, \sigma_\theta^2I) = \sum_{i=1}^N \alpha_k \mathcal{N}(x; \mu_k, \sigma_k^2I) $$ I know the $\alpha_k, \mu_k, \sigma_k$, how could I estimate the $\mu_\theta, \sigma_\theta$? Thanks!

I try: $X_i \sim \mathcal{N}(x; \mu_k, \sigma_k^2I), X \sim \mathcal{N}(x; \mu_\theta, \sigma_\theta^2I)$, could I rewrite the fomula? $$\mathbb{E}[X] = \sum \alpha_k \mathbb{E}[X_k] = \sum \alpha_k \mu_k$$ I don't know correct or not. Or maybe Moment Matching?

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In general, a mixture of Gaussian distributions is not itself Gaussian. So what you are doing is a Gaussian approximation to a true non Gaussian distribution.

Since a Gaussian distribution is completely determined by its mean vector and variance-covariance matrix, the task is simply that of computing these two moments for the Gaussian mixture.

  • Your formula for the mean of the mixture is correct. The link below provides a proof of this as well.
  • The formula for the is given in the accepted answer for Calculation of the Covariance of Gaussian Mixtures. It involves a weighted average of the component covariance matrices and also a weighted average of the deviation of the component mean vectors from the grand (mixture) mean vector.
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  • $\begingroup$ Thanks for your reply. But if I want to calculate(simplify) this mixture of Gaussian distributions, what should I do? Or under what conditions can it be simplified? $\endgroup$ Commented Jul 28 at 12:18
  • $\begingroup$ If this is difficult for u, u could recommend some paper or reference textbook. Thanks very much. $\endgroup$ Commented Jul 28 at 12:49
  • $\begingroup$ It is not clear to me what you mean by "calculate(simplify)". $\endgroup$ Commented Jul 28 at 14:27
  • $\begingroup$ Thanks for your reply. For example, Simplified expression without $\sum$ symbols. I asked this question primarily to simplify the GMM, but since I'm unsure how to approach it, I had to retain the Gaussian distribution assumption for now. If you have any relevant materials or ideas about simplifying GMM, please share them with me. Thank you very much! $\endgroup$ Commented Jul 29 at 15:05

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