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In my machine learning notes for Gaussian Mixture Models, I see the following statement:

It is common to confuse a mixture of Gaussian with a sum of Gaussians. Note that, a sum of Gaussian RVs is another Gaussian, and therefore unimodel.

This is only true if the sum of Gaussian RVs is a sum of independent Gaussian RVs right? If there is any dependency, does this guarantee that the sum of gaussians is NOT gaussian?

roulette01
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  • Even if a pair of normally distributed random variables is correlated, their sum will still be normally distributed. This Wikipedia article provides a fairly concise explanation. – Tiago C. Botelho Sep 19 '20 at 14:39
  • @TiagoC.Botelho Wait, so does this mean the sum of dependent gaussian random variables is also gaussian? – roulette01 Sep 19 '20 at 14:43
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    Precisely! Indeed, if they are independent, we can say even more about their sum: if $X \sim \mathcal{N}(\mu_X, \sigma^{2}_X)$ and $Y \sim \mathcal{N}(\mu_Y, \sigma^{2}_Y)$ are independent, then their sum $X + Y$ is distributed as $\mathcal{N}(\mu_X + \mu_Y, \sigma^{2}_X + \sigma^{2}_Y)$. A similar result holds if they are correlated, but we have to account for this correlation when we compute the variance of their sum. – Tiago C. Botelho Sep 19 '20 at 14:47
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    @Tiago: it is not true that a sum of dependent Gaussians is Gaussian. The correct statement is that if the joint distribution of $X$ and $Y$ is Gaussian then their sum is Gaussian. – Qiaochu Yuan Sep 19 '20 at 18:11
  • @QiaochuYuan I was just about to link https://stats.stackexchange.com/questions/125808/sum-of-gaussian-is-gaussian. Is there a proof for why joint normal => sum is normal? (I think it holds in the reverse direction too)? – roulette01 Sep 19 '20 at 18:12
  • @QiaochuYuan: you are absolutely right. Thank you for the heads up. My apologies to the original poster. – Tiago C. Botelho Sep 19 '20 at 20:39

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