Joint probability distribution of two normal random variables If X and Y are two independent normal random variables with mean 0 and standard deviation $ \sqrt{2}$ respectively.Prove that Z=X+Y is also a normal random variable with mean 0 and standard deviation 2. 
 A: Calculate the characteristic function of $Z$
$$\phi_Z(t)=E[e^{itZ}]=E[e^{it(X+Y)}]$$
$X$ and $Y$ are independent, therefore
$$E[e^{it(X+Y)}]=E[e^{itX}]E[e^{itY}]$$
We know that 
$$E[e^{itX}]=e^{-t^2}=E[e^{itY}]$$
therefore
$$\phi_Z(t)=e^{-2t^2}$$
or
$$\phi_Z(t)=e^{it*0-\frac{1}{2}(2^2)t^2}$$
which means that $Z$ is normally distributed with mean $0$ and standard deviation $2$.
EDIT : 
You can prove it by using the cdf definition 
$$F_Z(z)=\mathbb{P}(Z \leq z)=\mathbb{P}(X+Y \leq z)$$
$$\mathbb{P}(X+Y \leq z)=\int_{\mathbb{R}^2}{1_{x+y \leq z}}f_X(x)f_Y(y)dxdy$$
Or
$$\mathbb{P}(X+Y \leq z)=\int_{\mathbb{R}}{f_X(x) \left( \int_{-\infty}^{z-x}f_y(y)dy\right)dx}$$
Using the cdf definition of $Y$, and the fact that $X$ and $Y$ have the same distribution, we have
$$\mathbb{P}(X+Y \leq z)=\int_{\mathbb{R}}{f_X(x) F_X(z-x)dx}$$
Differentiating wrt $z$, we have the density function of $Z$
$$f_Z(z)=\int_{\mathbb{R}}{f_X(x) f_X(z-x)dx}$$ 
$$f_Z(z)=\left(\frac{1}{2\sqrt{2 \pi}}e^{-\frac{1}{2 *4}z^2}\right)\int_{\mathbb{R}}{\left(\frac{2\sqrt{2 \pi}}{1}e^{\frac{1}{2 *4}z^2}\right)f_X(x) f_X(z-x)dx}$$
We have 
$$f_X(x)=\frac{1}{\sqrt{2}\sqrt{2 \pi}}e^{-\frac{1}{2 *2}x^2}$$
The integral $\int_{\mathbb{R}}{\left(\frac{2\sqrt{2 \pi}}{1}e^{\frac{1}{2 *4}z^2}\right)f_X(x) f_X(z-x)dx}=1$ , you just factorize to have the integral on a density function with standard deviation $1$ and mean $z$
