Let X and Y be two $ \mathcal{N}(0, 1) $ distributions. I have to prove that for $(a,b)\in \mathbb{R}^2 $, $ aX + bY $ is equal to $\mathcal{N}(0, a^2 + b^2)$.
I'm trying to do this using the characteristic function of a Gaussian distribution. $$ \phi_{aX + bY}(t) = \int_{\mathbb{R}}{ \mathbb{e}^{it(ax+by)}{\frac{1}{2} \mathbb{e}^{-\frac{x^2}{2}}} dx} $$
I don't really know what to do since by changing the variable I can't replace both x and y. Any sugestions ?
Let $Z=aX+bY$. The characteristic function of $Z$ is:
$\phi_Z(t)=E\{e^{itZ}\}=E\{e^{it(aX+bY)}\}=E\{e^{i(at)X}e^{i(bt)Y)}\}$
EDIT (Sloppy mistake...) If X and Y are independent:
$\phi_Z(t)=E\{e^{i(at)X}\}E\{e^{i(bt)Y)}\}=\phi (at) \phi (bt)$,
where $\phi(w)=e^{-\frac{w^2}{2}}$ is the characteristic function of the normal distribution. So,
$\phi_Z(t)=e^{-\frac{1}{2}(at)^2}e^{-\frac{1}{2}(bt)^2}=e^{-\frac{1}{2}(a^2+b^2)t^2}$,
which is the characteristic function of the normal distribution $\mathcal{N}(0,a^2+b^2)$.