I was reading this question. It is about notation but I would like to ask something regarding the sum of two normally distributed random variables. If $X$ is a normally distributed random variable with mean $\mu_{X}$ and variance $\sigma_{X}^{2}$, and $Y$ is a normally distributed random variable with mean $\mu_{Y}$ and variance $\sigma_{Y}^{2}$, what kind of normal distribution does $X+Y$ follow when

  • $X$ and $Y$ are independent,
  • $X$ and $Y$ are uncorrelated,
  • $X$ are $Y$ are correlated?

How does one give a proof for each case?

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    $\begingroup$ Hi dada and welcome to the site! Did you look at the Wikipedia page concerning this problem? It contains proof of all listed cases. If yes, in what way did you find it wanting? $\endgroup$ Jul 21, 2015 at 11:02

2 Answers 2


Regardless of whether $X$ and $Y$ are normal or not, it is true (whenever the various expectations exist) that \begin{align} \mu_{X+Y} &= \mu_X + \mu_Y\\ \sigma_{X+Y}^2 &= \sigma_{X}^2 + \sigma_{Y}^2 + 2\operatorname{cov}(X,Y) \end{align} where $\operatorname{cov}(X,Y)=0$ whenever $X$ and $Y$ are independent or uncorrelated. The only issue is whether $X+Y$ is normal or not and the answer to this is that $X+Y$ is normal when $X$ and $Y$ are jointly normal (including, as a special case, when $X$ and $Y$ are independent random variables). To forestall the inevitable follow-up question,

No, $X$ and $Y$ being uncorrelated normal random variables does not suffice to assert normality of $X+Y$. If $X$ and $Y$ are jointly normal, then they also are marginally normal. If they are jointly normal as well as uncorrelated, then they are marginally normal (as stated in the previous sentence) and they are independent as well. But, regardless of whether they are independent or dependent, correlated or uncorrelated, the sum of jointly normal random variables has a normal distribution with mean and variance as given above.

In a comment following this answer, ssdecontrol raised another question: is joint normality just a sufficient condition to assert normality of $X+Y$, or is it necessary as well?

  • Is it possible to find marginally normal $X$ and $Y$ that are not jointly normal but their sum $X+Y$ is normal? This question was asked in the comments below by Moderator Glen_b. This is indeed possible, and I have given an example in an answer to this question.

  • Is it possible to find $X$ and $Y$ such that they are not jointly normal but their sum $X+Y$ is normal? Here, we do not insist on $X$ and $Y$ being marginally normal. The answer is Yes, and an example is given by kjetil b halvorsen. Another, perhaps simpler, answer is as follows. Consider $U$ and $V$ be independent standard normal random variables and $W$ a discrete random variable taking on each of the values $+1$ and $-1$ with probability $\frac 12$. Then, $X = U+W$ and $Y=V-W$ are not marginally normal (they have identical mixture Gaussian density $\frac{\phi(t+1)+\phi(t-1)}{2}$), and so are not jointly normal either. But their sum $X+Y = U+V$ is a $N(0,2)$ random variable.

  • $\begingroup$ +1, especially for the important clarification. Is joint normality necessary, or just sufficient? $\endgroup$ Dec 26, 2015 at 23:11
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    $\begingroup$ @ssdecontrol: It is not necessary. $\endgroup$ Dec 26, 2015 at 23:25
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    $\begingroup$ @ssdecontrol See this answer of mine for a reference to another interesting result: If $X$ and $Y$ are independent, and their sum $X+Y$ is normal, then $X$ and $Y$ must themselves be normal too. $\endgroup$ Dec 27, 2015 at 0:15
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    $\begingroup$ @kjetil do you have an example where you don't have joint normality (joint normality including the cases of independence and perfect correlation), but where the sum is normal and the margins are normal? $\endgroup$
    – Glen_b
    Dec 27, 2015 at 5:48
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    $\begingroup$ @Dilip For the one you attribute to me, try this: Let $U,V$ be iid $N(0,1)$. In the first quadrant (i.e. $U>0,V>0$) let $X=\min(U,V)$ and $Y = \max(U,V)$. For the other quadrants, rotate this mapping about the origin, like this (the purple represents regions with doubled probability and the white regions are ones with no probability). Then by symmetry both $X$ and $Y$ are standard normal but are clearly not bivariate normal, and $X+Y = U+V$ which is $\sim N(0,2)$ $\endgroup$
    – Glen_b
    Jan 4, 2016 at 12:04

Let $Z = X + Y$ be, when X and Y are independent you can prove it using the moment-generating function

$ M_{X,Y}(t) = E(\exp(t(X+Y)) \\ M_{X,Y}(t)= E(\exp\{tX\}\exp\{tY\}))\\ M_{X,Y}(t) = M_x(t)M_Y(t) \\ M_{X,Y}(t) = \exp\{ (\mu_X + \mu_Y)t + \frac{\sigma^2_X + \sigma^2_Y}{2}t^2 \} $

Then, $Z \sim N( \mu_X + \mu_Y, \sigma^2_X + \sigma^2_Y )$


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