2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '14 at 15:21
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '14 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 '14 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 '14 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 '14 at 15:13
  • $\begingroup$ Sorry, indicator function; sets argument equal to one if true, zero if false. $\endgroup$ – JPi Apr 21 '14 at 17:04

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.