Prove that the average of iid Gaussian random variables is Gaussian 
Given $x_1, \ldots, x_N$, independent and all distributed as a
  Gaussian with mean $\mu$ and variance $\sigma^2$. Then, the average
  $$\bar{x} = \frac{1}{N}\sum_{i=1}^Nx_i$$ is distributed as a Gaussian
  with mean $\mu$ and variance $\frac{\sigma^2}{N}.$

This is a very well-known result. Anyway, I'm looking around to find a proof for this and I'm not having luck.
 A: Let $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^{2}}{2}}$
 . Then if $$\hat{g}(\xi)=\int_{\mathbb{R}}g(x)e^{-ix\xi}dx,$$
 we have that $\hat{f}(x)=e^{-\frac{x^{2}}{2}}$
  - that is up to a constant it is a fixed point of the Fourier transform. Let $\overline{X}=\sum_{i=1}^{N}X_{i}$
  and suppose that each $X_{i}$
  has mean zero and variance 1
 . Then the probability density function of $\overline{X}$
  is a convolution $$p_{\overline{X}}(x)=p_{X_{1}}*\cdots*p_{X_{N}}(Nx),$$
 and since the Fourier transform diagonalizes the convolution operator we have that $$\hat{p}_{\overline{X}}(x)=\prod_{i=1}^{N}\hat{p}_{X_{i}}(x)=e^{-\frac{Nx^{2}}{2}},$$
 and so taking the inverse transform $$p(x)=\frac{1}{\sqrt{2\pi N}}e^{-\frac{x^{2}}{2N}},$$
 which is a Gaussian of mean zero and variance $N$
 . Your case where the variance is $\sigma$ can be obtained by scaling.
A: We can also use some known properties to prove this.
Let X1, X2, ..., Xn be independent and identically distributed random variables. Then:
1) E( a + bX1 ) = a + bE(X1)   
2) E( X1 + X2 + ... + Xn ) = E(X1) + E(X2) + ... + E(Xn) 
3) Var( X1 + X2 + ... + Xn ) = Var(X1) + Var(X2) + ... + Var(Xn)
4) Var( a + bX1 ) = b2 Var(X1)
Note that the random variables do not have to be independent or identically distributed for 1), 2) and 4).
We then have: 
E(X)  = E($\frac{X1}{n}$ + $\frac{X2}{n}$ + ... + $\frac{Xn}{n}$) = E($\frac{X1}{n}$) + E($\frac{X2}{n}$) + ... + E($\frac{Xn}{n}$)  = $\frac{1}{n}$ E(X1) + $\frac{1}{n}$ E(X2) + ... + $\frac{1}{n}$ E(Xn)  = $\frac{μ}{n}$ * n  = μ
Similarly for the variance: 
Var(X) = Var($\frac{X1}{n}$ + $\frac{X2}{n}$ + ... + $\frac{Xn}{n}$)  = Var($\frac{X1}{n}$) + Var($\frac{X2}{n}$) + ... + Var($\frac{Xn}{n}$)  = $\frac{1}{n^2}$ Var(X1) + $\frac{1}{n^2}$ Var(X2) + ... + $\frac{1}{n^2}$ Var(Xn)  = $\frac{σ^2}{n^2}$ * n = $\frac{σ^2}{n}$
