I got stuck when calculating the variance of this sum of complex random variables. The sum is:
$S = \sum\limits_{k = 1}^N {\sum\limits_{m = 1}^M {{a_{k,m}}{X_m}{e^{ - j\pi \left( {m - k} \right)}}} } $,
where ${a_{k,m}}$ are the complex constants, and $X_m$ are the independently complex random variable. The real and imaginary parts of $X_m$ are independent each other having zero mean and variance $\sigma^2_X$. So, X has zero mean and variance $2\sigma^2_X$.
How to calculate the variance of the real and imaginary parts of $S$.
Thank you.
Here is my attempt:
\begin{array}{l} \sigma _S^2 = E\left\{ {{{\left| S \right|}^2}} \right\}\\ = \sum\limits_{{k_1},{k_2}} {\sum\limits_{{m_1},{m_2}} {{a_{{k_1},{m_1}}}{{\left( {{a_{{k_2},{m_2}}}} \right)}^ * }} } E\left\{ {{X_{{m_1}}}{{\left( {{X_{{m_2}}}} \right)}^ * }} \right\}{e^{ - j\pi \left( {{m_1} - {k_1}} \right)}}{e^{ + j\pi \left( {{m_2} - {k_2}} \right)}}\\ = \sum\limits_{{k_1},{k_2}} {\sum\limits_m {{a_{{k_1},{m}}}{{\left( {{a_{{k_2},{m}}}} \right)}^ * }} } 2\sigma _X^2{e^{ - j\pi \left( {{k_2} - {k_1}} \right)}} \end{array}
So, the real or imaginary part of $S$ has variance $\frac{1}{2}\sigma _S^2$.