15
$\begingroup$

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 $Y$)?

$\endgroup$
  • $\begingroup$ what if $X=constant$ ? $\endgroup$ – Carlo Beenakker Mar 30 '17 at 13:29
  • $\begingroup$ @CarloBeenakker: People often consider a constant to be $N(\mu,0)$. $\endgroup$ – Nate Eldredge Mar 30 '17 at 14:15
  • $\begingroup$ Which people??? $\endgroup$ – wolfies Apr 8 '17 at 17:52
6
$\begingroup$

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).

$\endgroup$
18
$\begingroup$

Yes, they are normally distributed. This is the Lévy-Cramér theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.