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Given a distribution with PDF equal to a*N1 + (1-a)*N2, where N1 and N2 are the PDF's of normal distributions with some mean and variance, is the sum distribution also a normal distribution? The total probability is equal to 1 clearly, but is it normal?

kjetil b halvorsen
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Joe Blake
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  • If they are independent or bivariate normal random variables the answer is yes. There are examples where two univariate normals are not bivariate normal & linear combinations are not normal. This is true with any linear combination of the variables but not with the densities as you pose the problem where you have mixtures of normals. & are not normal unless the densities have the same mean & variance. – Michael R. Chernick Dec 07 '19 at 04:47
  • How might one know for sure / prove that the sum of densities are not normal (unless they have the same mean and variance)? – Joe Blake Dec 07 '19 at 04:52
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    Hi: Note that there's a big difference between a linear combination of two iid normal random variables and a linear combination of two normal densities. The former is always normal and, according to Michael, it sounds like the latter is only when the densities have the same mean and variance. Thanks Michael for explanation. – mlofton Dec 07 '19 at 04:55
  • @JoeBlake Look at the form of the density of the mixture. It is not the form of a normal distribution. But if it is of the form a $f_1(x)$ + (1-a)$f_1(x)$ where 0<a<1. Then a little algebra shows that it is equal to $f_1(x)$. That will not happen if the densities are not the same. – Michael R. Chernick Dec 07 '19 at 05:03
  • Possible duplicates, certainly related: 1. https://stats.stackexchange.com/questions/309154/intuition-for-why-sum-of-gaussian-rvs-is-different-from-gaussian-mixture/

    $\qquad$ $\qquad$.$:$ .$:$ . $:$ 2. https://stats.stackexchange.com/questions/359900/does-assumption-of-normality-of-each-mixture-components-implies-that-each-margin/

     – Glen_b Dec 07 '19 at 12:55

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