Independence of Noise at Each DFT Output My math may be a little rusty, so I would like confirmation or correction or my calculations here.

Given white noise samples, $x_i$, which are IID and zero-mean, and variance $\sigma^2_x$. I want to show that the outputs of any pair of DFT filters is also IID.

Knowing covariance of the outputs should be enough to draw a conclusion. So I examine the following:

Given the outputs $X_k$ and $X_{k'}$ defined as

$$X_k=\Sigma_{n=0}^{N-1}x_ne^{-2\pi ikn/N}$$

$$X_{k'}=\Sigma_{m=0}^{N-1}x_me^{-2\pi i{k'}m/N}$$

We can calculate the expected value of $X_k X^*_{k'}$ 

$$\begin{eqnarray*}
E(X_k X^*_{k'})
&amp;=&amp;E\left(\sum_{n=0}^{N-1}x_ne^{-2\pi ikn/N}\sum_{m=0}^{N-1}x_me^{2\pi i{k'}m/N}\right)\\
&amp;=&amp;E\left(\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}x_n x^*_m e^{-2\pi i(kn-{k'}m)/N}\right)\\
&amp;=&amp;\sum_{n=0}^{N-1}\sum_{m=0}^{N-1}E\left(x_n x^*_m \right) e^{-2\pi i(kn-{k'}m)/N}\\
\end{eqnarray*}$$

We know that $E(x_k x^*_{k'})=\sigma^2_x$ when $k={k'}$ and $E(x_k x^*_{k'})=0$ when $k\neq {k'}$. So we can simplify:

$$\begin{eqnarray*}
E(X_k X^*_{k'})
&amp;=&amp;\sigma^2_x\sum_{n=0}^{N-1}e^{-2\pi i(k-{k'})n/N}\\
&amp;=&amp;N\sigma^2_x|_{k={k'}}
\end{eqnarray*}$$

But if I'm right so far, I still am fuzzy on the case where $k \neq {k'}$. I expect it to equate to zero. Can someone help me out with that part?
Let $m=k-k'$. Then the sum in your final equation becomes

$$\sum_{n=0}^{N-1}e^{-2\pi imn/N}=\frac{1-e^{-2\pi im}}{1-e^{-2\pi im/N}},\quad m\neq 0\tag{1}$$

where I've used the formula for the geometric series. Noting that $e^{-2\pi im}=1$ for any integer $m$ shows that the sum in $(1)$ equals zero.

[Note that your sums should only have $N$ elements, i.e., their indices should go from $0$ to $N-1$ (instead of $N$).]