0
$\begingroup$

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: $$\sum_{n=0}^N \mathbf x_n^H\mathbf x_n$$ Secondly, what is the distribution of $$\sum_{n=0}^N \mathbf x_n^H\mathbf y_n$$ 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.

Thank you in advance.

$\endgroup$
0
$\begingroup$

$\require{amsmath} \require{graphicx} \newcommand{\X}{\mathbf X} \newcommand{\Y}{\mathbf Y} \newcommand{\N}{\mathcal N} \newcommand{\si}{\sigma}$ I understand the setting as follows: For $n=0,\dots,N$, let $\X_n:=(X_{n,0},\dots,X_{n,k})$ and $\Y_n:=(Y_{n,0},\dots,Y_{n,k})$, where all the $X_{n,j}$'s and $Y_{n,j}$'s are iid $\mathcal C\mathcal N(\mathbf 0,\si^2\mathbf I_k)$.

The first problem is to find the distribution of $$S_1:=\sum_{n=0}^N\X_n^H\X_n=\sum_{n=0}^N\sum_{j=0}^k|X_{n,j}|^2. $$

As for your second question, the distribution of For each pair $(n,j)$, we have $\frac2{\si^2}|X_{n,j}|^2\sim\chi^2_2$. So, $\frac2{\si^2}\,S_1\sim\chi^2_{2(k+1)(N+1)}=\text{Gamma}((k+1)(N+1),2)$. So, $$\sum_{n=0}^N\X_n^H\X_n\sim\text{Gamma}((k+1)(N+1),\si^2), $$ as you expected. If $$M:=(k+1)(N+1) $$ is large, then, by the central limit theorem, $$\text{Gamma}((k+1)(N+1),\si^2)\approx\N(M\si^2,M\si^4). $$

As for the distribution of $$S_2:=\sum_{n=0}^N\X_n^H\Y_n=\sum_{n=0}^N\sum_{j=0}^k \overline{X_{n,j}}\,Y_{n,j}, $$ it is the $(k+1)(N+1)$-fold convolution of the distribution of the complex-valued random variable $\overline X\,Y=X_1Y_1+X_2Y_2+i(X_1Y_2-X_2Y_1)$, where $X:=X_1+iX_2$, $Y:=Y_1+iY_2$, and $X_1,X_2,Y_1,Y_2$ are iid $\mathcal N(0,\si^2/2)$.

In turn, the distribution of $\overline X\,Y$ can be obtained by the transformation-of-distributions technique (i.e., change of variables in a multifold integral; see e.g. Lecture 2) and is likely unremarkable. Mathematica worked several hours on getting the distribution of $\overline X\,Y$ and came up with nothing.

However, the mean and covariance matrix of the joint distribution of $(\Re(\overline X\,Y),\Im(\overline X\,Y))=(X_1Y_1+X_2Y_2,X_1Y_2-X_2Y_1)$ are $[0,0]^T$ and $\si^4 I_2/2$. So, if, again, $M$ is large, then, by the multivariate (here bivariate) central limit theorem, the joint distribution of $(\Re S_2,\Im S_2)$ is approximately the bivariate normal distribution $$\N(0,0,M\si^4/2,M\si^4/2,0), $$ with zero means, both variances equal $M\si^4/2$, and zero correlation. That is, the real and imaginary parts of $\overline X\,Y$ are (i) zero-mean, (ii) each with variance $M\si^4/2$, (iii) jointly asymptotically normal, and (iv) asymptotically independent.

$\endgroup$
  • $\begingroup$ Thanx losif Pinelis. When you say Gamma(X,Y), does that mean X is the mean and Y is the variance of the Gamma distribution or the shape and rate parameters? $\endgroup$ – Aymen Kareem Nov 21 at 17:01
  • 1
    $\begingroup$ @AymenKareem : $\text{Gamma}(a,b)$ means the gamma distribution with shape parameter $a$ and scale parameter $b$, so that the mean is $ab$ and the variance is $ab^2$. $\endgroup$ – Iosif Pinelis Nov 21 at 22:22
  • $\begingroup$ I have added normal approximations to the distributions of the two sums. $\endgroup$ – Iosif Pinelis Nov 22 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.