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?

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

Not always--otherwise every sum of normal random variables would be normal, and this ain't so.

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: $X=(\xi,\xi,\sigma\xi,\sigma\xi)$, $a=(1,1,1,-1)$, $b=(1,-1,1,1)$.

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    $\begingroup$ Can you give an example? I always thought that summing two normal variables (dependent or not) yields a normal variable again. $\endgroup$ – fabee Jul 26 '14 at 13:26
  • $\begingroup$ Ahh, I guess I see. You mean each could be marginally normal but jointly they could be non-normal. What I just wrote above just holds for jointly normal random variables. $\endgroup$ – fabee Jul 26 '14 at 13:28
  • $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ – Did Jul 26 '14 at 13:31
  • $\begingroup$ 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 $\bf X$? $\endgroup$ – user103828 Jul 26 '14 at 14:11
  • $\begingroup$ 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? $\endgroup$ – Did Jul 26 '14 at 14:35

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