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Consider $f_i(x) = \frac{1}{\sqrt{2\pi}\sigma_{i}}e^{-\frac{1}{2}\left(\frac{x-\mu_{i}}{\sigma_{i}}\right)^{2}},$ $i=1,2$. Define another density by $$f(x) \equiv wf_1(x)+(1-w)f_2(x).$$ Is $f(x)$ also a Normal density? i.e. is it true that $f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}$ for some $\mu$ and $\sigma$?

My candidate is $\mu = w\mu_1 + (1-w)\mu_2$ and $\sigma = w\sigma_1 + (1-w)\sigma_2$, but have not so far managed to prove or disprove this. I was wondering whether some algebraic manipulation could settle this.

Giorgi Chavchanidze
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    Assuming $0<w<1$, and that the two normal densities are really different, the answer is NO, see for instance https://stats.stackexchange.com/questions/594180/is-the-mixture-of-two-gaussians-with-same-mean-also-gaussian/594181#594181 – kjetil b halvorsen Nov 08 '23 at 04:54
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    Welcome to CrossValidated! This is a good question (+1), for which we already have a very good answer at the proposed duplicate. Yes, this is unintuitive. Try playing around a bit with simulations and a Kolmogorov-Smirnov test against your "candidate resulting normal"; you will see that while the simulates look bell-shaped, they are not normal. – Stephan Kolassa Nov 08 '23 at 07:02

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