# Sum of dependent normal random variables

Let ${\bf X} =(X_1,\ldots,X_n)'$ be a vector of random variables that may be dependent and let ${\bf a}=(a_1,\ldots,a_n)'$ and ${\bf b}=(b_1,\ldots,b_n)'$ be nonrandom vectors with $a_i \neq 0$ and $b_i \neq 0$ for $i=1,\ldots,n$.

Assume ${\bf a' \bf X} =\sum_{i=1}^n a_i X_i \sim N(0,\sigma_a^2)$ and ${\bf b' \bf X} =\sum_{i=1}^n b_i X_i \sim N(0,\sigma_b^2)$ be normal random variables. Is ${(\bf a' + \bf b') \bf X}$ also a normal random variable?

• From thinking about Did's answer: If you can show that the two sums are jointly normal, then you are in business. – fabee Jul 26 '14 at 13:30
• $a'X$ and $b'X$ may not be independent. Think of $b_i=-a_i$. – Yves Daoust Jul 26 '14 at 13:34

Canonical (counter)example: Assume that $\xi$ is standard normal and that $\eta=\sigma\xi$, where $\sigma=\pm1$ is symmetric Bernoulli and independent of $\xi$. Then $\eta$ is standard normal but $\xi+\eta$ is not normal since $P(\xi+\eta=0)=P(\sigma=-1)=\frac12$ while $P(\zeta=0)$ is $0$ or $1$ for every normal random variable $\zeta$. (This argument proves that the vector $(\xi,\eta)$ is not normal.)
Variant of the same: $X=(\xi,\xi,\sigma\xi,\sigma\xi)$, $a=(1,1,1,-1)$, $b=(1,-1,1,1)$.
• Yes.    – Did Jul 26 '14 at 13:31
• Just to confirm that I'm understanding correctly: If $W_1$ and $W_2$ are normal then $W_1+W_2$ is not necessarily normal (this I understand) so you are letting $W_1={\bf a}'{\bf X}$ and $W_2={\bf b}'{\bf X}$ and that is the proof... Does it matter that all the components of $\bf a$ are nonzero (and nonrandom so you can't let $a_1=U =\pm 1$ for example) and all the components of $\bf b$ are nonzero? Does it matter that the terms in $W_1$ and $W_2$ are just different linear combinations of the components of $\bf X$? – user103828 Jul 26 '14 at 14:11
• As you say, the idea of the counterexample as I presented it is to consider $X=(X_1,UX_1)$, $a=(1,0)$ and $b=(0,1)$, then $a'X$ and $b'X$ are normal and $(a+b)'X$ is not. To get $a$ and $b$ with nonzero entries (unfortunately, I missed this condition in your question, did you add it later on?), one must work a little more... Any idea on your side? – Did Jul 26 '14 at 14:35