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 \begin{equation} \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}\} \end{equation}

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 . \begin{equation} f_s(\lambda)=(1-\beta)^{+}\delta(\lambda)+\frac{\sqrt{(\lambda-a)^{+}(b-\lambda)^{+}}}{2\pi\lambda(1+\lambda\mu)} \end{equation}

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 \begin{equation} \int_a^b \lambda f_s(\lambda)d\lambda=\lim_{M,N\rightarrow\infty} {1 \over M}{\rm tr}\{{\bf SS}^{H}\} \end{equation}

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    $\begingroup$ 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. $\endgroup$ Mar 17, 2021 at 9:30
  • $\begingroup$ 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. $\endgroup$ Mar 17, 2021 at 18:23

1 Answer 1


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 $\beta>0$.

  • $\begingroup$ First: Is the condition $\int_a^b f_s(\lambda)\,d\lambda=(1-\beta)^{+}=\max(0,1-\beta)$ analogous to yours? $\endgroup$ Mar 16, 2021 at 14:52
  • $\begingroup$ no it is not; that integral equals 1 if $\beta>1$, not zero (you are integrating a positive function, how could the integral vanish?) $\endgroup$ Mar 16, 2021 at 15:03
  • $\begingroup$ 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! $\endgroup$ Mar 16, 2021 at 15:25
  • $\begingroup$ 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. $\endgroup$ Mar 16, 2021 at 15:33
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    $\begingroup$ the indefinite integral you need is number 2.282 in Gradshteyn & Ryzhik's table of integrals; alternatively (and more conveniently) you can enter it in Mathematica or Maple or the like. $\endgroup$ Mar 16, 2021 at 16:40

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