# The distribution of the power of the sum of inner products of two independent complex normal vectors

If I have $$\mathbf x_n=[x_0, x_1,... ,x_K]^T$$ and $$\mathbf y_n=[y_0, y_2, ..., y_K]^T$$, where $$x,y\sim\mathcal C\mathcal N(\mathbf 0,\sigma^2\mathbf I)$$.

What is the distribution of the following inner product: $$\Big|\sum_{n=0}^N \mathbf x_n^H\mathbf x_n + \sum_{n=0}^N \mathbf x_n^H\mathbf y_n\Big|^2$$ If the answer is Gamma distribution, what are the parameters of this Gamma distribution? Note that each element in both vectors is complex, random, and independent of the other elements.

$$\require{amsmath} \require{graphicx} \newcommand{\X}{\mathbf X} \newcommand{\Y}{\mathbf Y} \newcommand{\N}{\mathcal N} \newcommand{\si}{\sigma}$$ In view of my answer here, it is highly unlikely that the distribution in question can be obtained in closed form. However, it is not hard to show that this distribution is approximately normal. Indeed, the distribution in question is the distribution of the random variable (r.v.) $$U^2+V^2,$$ where $$(U,V)$$ is the sum of $$M:=(k+1)(N+1)$$ iid copies of the random pair $$(A,B):=(X_1^2+X_2^2+X_1Y_1+X_2Y_2,\;X_1Y_2-X_2Y_1),$$ where $$X_1,X_2,Y_1,Y_2$$ are iid $$\mathcal N(0,\si^2/2)$$. The mean of the pair $$(A,B)$$ is $$(\si^2,0)$$ and its covariance matrix is diagonal with $$5\si^4/2$$ and $$\si^4/2$$ on the diagonal. So, by the multivariate (here bivariate) central limit theorem, the joint distribution of $$(U,V)$$ is approximately the bivariate normal distribution $$\N(M\si^2,\;0,\;5M\si^4/2,\;M\si^4/2,\;0),$$ with means $$M\si^2$$ and $$0$$, variances $$5M\si^4/2$$ and $$M\si^4/2$$, and zero correlation.
So, if $$M$$ is large, then, by the multivariate delta method (with $$h(u,v)=u^2+v^2$$), the distribution in question, which is the distribution of $$U^2+V^2$$, is $$\approx\N(M^2\si^4,40M\si^8).$$
• @AymenKareem : How do you it works only for very large $N,K$? And what do you mean by "works"? Also, why would you prefer a gamma approximation to a normal one? In fact, I don't remember ever seeing approximations by a gamma distribution. – Iosif Pinelis Nov 22 '19 at 21:29