In my problem, both random variables have zero mean, are univariate, and are independent. They may have different variances. If they happen to have the same variance, of course the mixture is Gaussian with same variance regardless of the weights in the mixture. But what if the variances are different?
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If the variances are different it's not a Gaussian. Gaussian mixtures are identifiable, which means that mixtures of Gaussians are generally different from each other and from a single Gaussian, unless parameters are exactly equal or some mixture proportions are zero. See Teicher, H. (1961). Identifiability of mixtures. Ann. Math. Statist. 32, 244–248. If you draw a density of a mixture of two Gaussians with same mean and strongly different variances you can see quite clearly that this is not a Gaussian.
answered Oct 31, 2022 at 18:15
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$\begingroup$ I don't think Teicher's paper directly applies here: it doesn't handle the case where the standard deviations vary. $\endgroup$– whuber ♦Commented Sep 5, 2023 at 14:19