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Let $X\sim \mathcal N(\mu_X,\sigma_X^2),\ Y\sim \mathcal N(\mu_Y,\sigma_Y^2)$ two normal random variables and $a,b\in \mathbb R$.

If $X,Y$ are independent, then $$aX+bY\sim \mathcal N(a\mu_X+b\mu_Y,a^2\sigma_X^2+b^2\sigma_Y^2)$$ If $X,Y$ are jointly normally distributed with correlation $\text{corr}(X,Y)=\rho$, then $$aX+bY\sim \mathcal N(a\mu_X+b\mu_Y, a^2\sigma_X^2+b^2\sigma_Y^2 + 2ab\rho\sigma_X\sigma_Y)$$

But what if $X,Y$ are neither independent nor jointly normally distributed, what can we say about the distribution of their sum? And what would be an example of a sum of normal random variables that is not normally distributed?

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  • $\begingroup$ Please do not modify the question after an answer is posted, making the answer look off-topic. // Sums of normal random variables that are not normally distributed are all over the site. $\endgroup$
    – Did
    Apr 21, 2014 at 15:21

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Actually, your second statement is false. This would only be true if $X,Y$ are jointly normal, which you do not assume. And if you're assuming joint normality then uncorrelated=independent.

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  • $\begingroup$ Ah, I see, should I edit my question to correct the 2nd statement then? But what would be an example of $X,Y$ that are correlated with $X+Y$ not normally distributed? $\endgroup$
    – Phil-ZXX
    Apr 21, 2014 at 14:27
  • $\begingroup$ The following should work. Let $a(x)= I(|x|\leq 1) \bigl( 2 \text{sgn}(x) -1 \bigr)$ where $\text{sgn}(x)=1$ if $x>0$ and $=-1$ if $x<0$. Then consider $f(x,y)=\phi(x)\phi(y) \bigl( 1+ a(x) a(y) \bigr)$. $\endgroup$
    – JPi
    Apr 21, 2014 at 14:47
  • $\begingroup$ Sorry, that's an example of them being uncorrelated and not normally distributed! For correlated and not normally distributed there are many more possibilities; have a look at copulas if that interests you. One standard example is the distribution with joint cdf $\Phi(x) \Phi(y) \bigl( 1- 0.5 (1-\Phi(x))(1-\Phi(y))\bigr)$, where $\Phi$ is the standard normal cdf. $\endgroup$
    – JPi
    Apr 21, 2014 at 14:56
  • $\begingroup$ Thanks a lot. Regarding your first comment, what does the notation of $I$ mean? Is it the normal indicator function? $\endgroup$
    – Phil-ZXX
    Apr 21, 2014 at 15:13
  • $\begingroup$ Sorry, indicator function; sets argument equal to one if true, zero if false. $\endgroup$
    – JPi
    Apr 21, 2014 at 17:04

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