Why are linear combinations of independent standard normal random variables also normally distributed?
My professor has given a list of questions that will not be appearing on my test, with this being one of them. I still feel this is extremely important to understand.
How can I prove the following
If $X$ and $Y$ are independent, standard normal random variables, then the linear combination $aX+bY,\;\forall a,b>0$ is also normally distributed.
If I am not mistaken, I believe I can find the distribution of the linear combination
If we let $Z=aX+bY$, knowing $X,Y \sim N(0,1)$, we can find the expectation and variance as $$\mathbb{E}(Z)=\mathbb{E}(aX+bY)=a\mathbb{E}(X)+b\mathbb{E}(Y)=0$$ $$Var(Z)=Var(aX+bY)=a^2Var(X)+b^2Var(Y)=a^2+b^2$$ $$$$ Thus, $Z \sim N(0,a^2+b^2)$.
I just don't think this proves that linear combination is normally distributed. I tried looking in some reference books that my professor reserved at the library, but they all just state the fact and I can't figure out how to prove it.
$\endgroup$ 21 Answer
$\begingroup$The most direct way is to look at the characteristic function.
The characteristic function of an r.v. characterizes its probability distribution completely. If you can show that the characteristic function of the linear combination is that of a normal r.v., you are done.
$\endgroup$ 5
