We know that the sum of two independent normal random variables is again a normal random variable. But is the reverse right? If $X$ and $Y$ are independent random variables satisfying $X+Y$~$N(\mu,\sigma^2)$ for some $\mu$ and $\sigma$, can we conclude that both $X$ and $Y$ obey normal distribution? or under some conditions added on $p_X$ and $p_Y$ (the density functions of $X$ and $Y$)?
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$\begingroup$ what if $X=constant$ ? $\endgroup$– Carlo BeenakkerCommented Mar 30, 2017 at 13:29
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$\begingroup$ @CarloBeenakker: People often consider a constant to be $N(\mu,0)$. $\endgroup$– Nate EldredgeCommented Mar 30, 2017 at 14:15
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$\begingroup$ Which people??? $\endgroup$– wolfiesCommented Apr 8, 2017 at 17:52
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2 Answers
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Yes, and the same holds for Poisson, and for mixtures of Gauss and Poisson. All these are special cases of the general question: if $X_j$ are independent and we know the distribution of their sum, what can be said about the distributions of the $X_j$. This general question is addressed in the book Linnik, Ostrovskii, Decomposition of random variables and vectors, AMS 1977 (translation from the Russian).
answered Mar 30, 2017 at 17:24
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Yes, they are normally distributed. This is the Lévy-Cramér theorem.
answered Mar 30, 2017 at 14:35