# Sum of correlated normal random variables

Suppose I have two correlated random variables, that were generated in the following way: \begin{align*} X_1 &\sim \mathcal{N}(0,1)\\ X_1' &\sim \mathcal{N}(0,1)\\ X_2 &= \rho X_1+\sqrt{1-\rho^2}\cdot X_1'\\ Y_1 &= \mu_1+\sigma_1 X_1\\ Y_2 &= \mu_2+\sigma_2 X_2. \end{align*}

Now, is it true that $Y_1+Y_2$ (or, more generally $\alpha_1 Y_1+\alpha_2Y_2$) normally distributed? (I can easily calculate the mean and the variance of $\alpha_1 Y_1+\alpha_2Y_2$, but I am not sure about the distribution...)

EDIT: just to clarify, $X_1$ and $X_1'$ are independent.

• You mean $X_2 \sim \rho X_1 +\sqrt{1-\rho^2} X_1'$?
– ntt
Apr 14, 2017 at 15:37
• Sure, fixed, thanks.
– Bach
Apr 14, 2017 at 15:37
• Do you mean that $X_2 \mathbf{\,=\,} \rho X_1+\sqrt{1-\rho^2}\cdot X_1'$?
– NCh
Apr 14, 2017 at 15:40
• Sure, fixed as well, thanks :)
– Bach
Apr 14, 2017 at 20:36

Take $X_1 \sim \mathcal N(0,1)$ and let $Z$ be a r.v. with $P(Z =-1)=P(Z=1)=1/2$ and $X_1, Z$ are independent. Denote $X_1'=Z\cdot X_1$, then $X_1' \sim \mathcal N(0,1)$. However $(X_1, X_1')$ is not Gaussian as $X_1+X_1'$ is not normal. Then $X_2$ should not Gaussian, perhaps $X_2$ will be Gaussian with some specific values of $\rho$. We have $Y_1$ is Gaussian, $Y_2$ is not Gaussian then $Y_1+Y_2$ is probably not Gaussian in general.

$\alpha_1 Y_1 + \alpha_2 Y_2$ is a linear combination of $X_1$ and $X_1^\prime$ - that is $\alpha_1 Y_1 + \alpha_2 Y_2 = \beta X_1 + \beta^\prime X_1^\prime$ for some $\beta, \beta^\prime$ that are a bit of a pain to calculate. Linear combinations of independent normal random variables are normal; there are several proofs of this (nontrivial, but well-known) fact. So the answer to your question is yes.

• Right, I should have seen this. Thanks!
– Bach
Apr 14, 2017 at 15:42
• @Michael Lugo Here $X_1$ and $X_1'$ are correlated, how can you deduce $\alpha_1 Y_1 + \alpha_2 Y_2$ is normal?
– ntt
Apr 14, 2017 at 15:50
• Bach: as @ntt has pointed out, this only holds if $X_1$ and $X_1^\prime$ are independent. I'd assumed from your notation that you intended independence there, but if that's not the case then my answer is incorrect. Apr 14, 2017 at 15:51
• Yes, I meant that they are independent. Sorry for not being clear about that.
– Bach
Apr 14, 2017 at 16:00
• @ntt Uncorreletedness of two r.v. with given normal marginals does not imply independence. .
– NCh
Apr 14, 2017 at 16:08