# How to compute the first moment of the distribution of the convolution of Marcenko-Pastur law with a not iid matrix?

Let $$\mathbf{F}$$ denote an M × N matrix whose entries are independent zero-mean complex random variables, the limiting eigenvalue distribution is given by the Marchenko Pastur law $$MP_{\beta}$$, where $$\frac{N}{M}\rightarrow \beta$$.

It can be shown that the moments are giving by $$$$\sum_{i=1}^{k}{1 \over k} \binom{k}{i} \binom{k}{i-1}\beta^{i}=\lim_{M,N\rightarrow\infty} {1 \over M}{\rm tr}\{({\bf FF}^{H})^{k}\}$$$$

Then define $$\mathbf{S}=\frac{1}{M}\mathbf{C}^{1/2}\mathbf{FF}^H(\mathbf{C}^{1/2})^H$$, where $$\mathbf{C}$$ is a positive definite matrix, whose limiting eigenvalue distribution is denoted by $$\nu$$. It is known that the limiting eigenvalue distribution of $$\mathbf{S}$$ is given by the free multiplicative convolution of $$MP_{\beta}$$ and $$\nu$$, $$f_s=MP_\gamma\boxtimes\nu$$.

A closed form of the resulting distribution $$f_s$$ is . $$$$f_s(\lambda)=(1-\beta)^{+}\delta(\lambda)+\frac{\sqrt{(\lambda-a)^{+}(b-\lambda)^{+}}}{2\pi\lambda(1+\lambda\mu)}$$$$

where $$\mu$$ is a parameter >0

$$a=1+\beta+2\mu \beta-2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)}$$

$$b=1+\beta+2\mu \beta+2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)}$$

(see Random Matrix Theory and Wireless Communications 1)

My question is: can be found an expression of the first moment of $$f_s$$, i.e. the arithmetic mean as $$$$\int_a^b \lambda f_s(\lambda)d\lambda=\lim_{M,N\rightarrow\infty} {1 \over M}{\rm tr}\{{\bf SS}^{H}\}$$$$

• I am a bit confused by your description of the problem. On first look it seems to me that you are dealing with a general distribution for C; but your specific formula indicates that you are looking on a very specific distribution - as you did not give a specific link I had some problems to find this in Tulino-Verdu. On the other hand I wonder why you want to get the result by integration. In the limit the first moment of S is just given by the product of the first moment of C and the first moment of Marchenko-Pastur. Mar 17, 2021 at 9:30
• Thanks for your comment @RolandSpeicher. Actually, the distribution is very specific, and you can find it in Tulino-Verdù (Theorem 2.41, p.62) that, in turn, they drawn from [Mestre et al.] (ieeexplore.ieee.org/document/1203168) equations (9)-(14). So, following your approach the first moment of S should be just $\beta \cdot \sigma_1 \sigma_2$ where $\beta$ and $\sigma_1 \sigma_2$ are the first moment of F (i.e. the MP) and C, respectively. Am i right? Thanks in advance. Mar 17, 2021 at 18:23

If I correct the typo's so that the MP distribution follows when $$\mu=0$$, the definitions should be: $$f_s(\lambda)=(1-\beta)^{+}\delta(\lambda)+{{\sqrt{(\lambda-a)^+(b-\lambda)^+}}\over{2\pi \lambda(1+\mu \lambda)}},$$ $$a=1+\beta+2\mu \beta-2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)},$$ $$b=1+\beta+2\mu \beta+2\sqrt{\beta}\sqrt{(1+\mu)(1+\mu \beta)}.$$ I checked that then $$\int_a^b f_s(\lambda)\,d\lambda=\min(1,\beta)$$, so with the delta function contribution it is properly normalized to unity.
The desired integral evaluates to $$\int_a^b \lambda f_s(\lambda)\,d\lambda=\beta,$$ independent of $$\mu>0$$ for any $$\beta>0$$.
• First: Is the condition $\int_a^b f_s(\lambda)\,d\lambda=(1-\beta)^{+}=\max(0,1-\beta)$ analogous to yours? Mar 16, 2021 at 14:52
• no it is not; that integral equals 1 if $\beta>1$, not zero (you are integrating a positive function, how could the integral vanish?) Mar 16, 2021 at 15:03
• But, if $\beta>1$, is not $\int_a^b (1-\beta)^{+}\delta(\lambda)d\lambda =\int_a^b \max(0,1-\beta) \delta(\lambda) d\lambda=\int_a^b 0 \cdot \delta(\lambda) d\lambda$=0? Thanks in advance! Mar 16, 2021 at 15:25
• I'm sorry, perhaps it's other typo's, but I don't understand your reasoning; the delta function does not contribute anyway to the integral because $\lambda=0$ is not in the interval $(a,b)$. The integral from $a$ to $b$ only contains contributions from the second term in the definition of $f_s(\lambda)$; that term is positive in the interval $(a,b)$ and therefore the integral cannot become zero. Mar 16, 2021 at 15:33