Consider two identical independently distributed sets of complex normal variables $\{X_i\}_{i=1}^N, \{Y_i\}_{i=1}^N$, with mean $\mu$ and variance $\sigma$.
Can the following expectation value $$ \mathbb E \left[\,\,\left\lvert \sum_{k=1}^N X_i^* Y_i \right\rvert \,\,\right] $$ be computed analytically?
What if the variables are real instead of complex?
The result seems to increase with $\sqrt N$, as shown by the following numerical computation:
where for each dot the average is taken over a sample of 1000 different pairs of independent complex normal $N$-dimensional vectors. The blue line is the best fit line, using as fit function $a+b\sqrt{x}$, and the parameters have been found in this case to be $a \approx -0.25$ and $b \approx 0.9$ (but of course these are only very rough estimates).
I can derive the expectation value of the square of the above absolute value, which is gives $$ \mathbb E \left[ \,\, \left\lvert \sum_{k=1}^N X_i^* Y_i \right\rvert^2 \,\, \right] = N \sigma^4, $$ for $\mu=0$. The square root of this gives the observed $\sqrt N$ dependence, but of course in general $\mathbb E[A^2] \neq \mathbb E[A]^2$, so that the argument is very handwavy.
Is there a way to compute this expectation value?
Suppose $\mu=0$ and $\sigma=1$. The products $X_iY_i$ are i.i.d with expectation 0 and variance 1. By the central limit theorem $$ Z_N=\frac{1}{\sqrt{N}} \sum_{i=1}^N X_iY_i \rightarrow Z$$ in law, with $Z \sim {\cal N}(0,1)$. We have $E(Z_N \bar{Z}_N)= \frac{1}{N}\sum_i E(X_i{Y}_i \bar{X}_j \bar{Y}_j)=1$ (use independence). Then $E\left(\;|Z_N|\; 1_{|Z_N|\geq R} \right)\ \leq \ E(|Z_N|^2/R)\leq 1/R$. This implies that $$ E\left( \left|Z_N\right| \right) \rightarrow E(|Z|)=\sqrt{\frac{2}{\pi}}=0.7978846...$$
When $\mu\neq 0$, $\sigma=1$ then $\mbox{ var} (X_iY_i)=1+2\mu^2$ and $E(X_iY_i)=\mu^2$ and $$ \frac{1}{\sqrt{N (2\mu^2+1)}} \sum_{i=1}^N (X_iY_i -N\mu^2) \sim Z$$ or $$ \sum_{i=1}^N X_iY_i \sim N\mu^2 + \sqrt{N(2\mu^2+1)}Z$$ From this $$ E\left(\left|\sum_{i=1}^N X_iY_i\right|\right) \sim N\mu^2 + o(1/N^p)$$ for any $p>0$ (the rest goes exponentially fast to zero).
When $X_i$ and $Y_i$ take complex values only the case $\mu=0$ is affected. I assume variance $E|X_i|^2=E|Y_i|^2=1$ and that the variables take complex symmetric values. The CLT then gives a complex variable $Z=U+iV$ where $U\sim{\cal N}(0,\frac{1}{2})$ and $V\sim{\cal N}(0,\frac{1}{2})$ are independent. Then $$ E(|Z|)=\frac{1}{\pi} \int\int e^{-(u^2+v^2)} |u+iv| du\; dv =\frac{1}{\pi} \int_0^{2\pi} \int_0^{\infty} e^{-r^2} r \cdot r\; dr\; d\phi$$ Thus, $$E(|Z|)=2\int_0^\infty e^{-r^2} r^2\; dr=\frac{\sqrt{\pi}}{2}=0.88622...$$ possibly fitting better with your numerical estimate above.