I need to evaluate the moment $$\mathbb{E} (AX)^n,$$ where A is an NxN Hermitian square matrix, and X is $$X=ZZ^{\ast},$$ where $Z=\mu+Y$, where $\mu$ is mean of $Z$ and $Y$ is a zero-mean complex valued matrix with independent elements, but they do not have equal variance.

Z is a NxK matrix where $K>N$.

Now, when are the moments finite as $n\to \infty$?

If it simplifies, let Y be complex Gaussian. If it simplifies, let Y be IID (equal variances).

(I will use this to find the expected inverse of $AX$)


1 Answer 1


For any $N\times N$ matrix $A$ we set $\DeclareMathOperator{\tr}{tr}$ $\newcommand{\bE}{\mathbb{E}}$

$$ |A|^2:=\sum_{i,j}|a_{ij}|^2 =\tr(A A^*). $$

There exists a constant $C=C(N)>0$ such that for any complex $N\times N$ matrices $A, B$ we have

$$ |AB|\leq C |A|\cdot |B|. $$

In particular, if $A_1,\dotsc, A_n$ are complex $N\times N$ matrices we have

$$ |A_1\cdots A_n|\leq C^{n-1}|A_1|\cdots |A_n|. $$

We have

$$ \Bigl\vert \bE(AX)^n\Bigr\vert\leq \bE\bigl(\;|(A(X)^n|\;\bigr) \leq C^{2n-1}|A|^n \bE\bigl(\;|X|^n\;\bigr) =C^{2n-1}|A|^n \bE\bigl(\;|Z|^{2n}\;\bigr) $$

$$ = C^{2n-1}|A|^n\bE\Bigl( \;\Bigl(\; \sum_{i,j}|z_{ij}|^2\;\Bigr)^n\;\Bigr). $$

Now the question reduces to the following. Suppose we are given independent, complex Gaussian variables $\zeta_1,\dotsc, \zeta_m$. Are the moments of $|\zeta_1|^2+\cdots +|\zeta_m|^2$ finite? The answer is yes, because the probablity

$$ p(R)=P(|\zeta_1|^2+\cdots +|\zeta_m|^2 > R^2) $$

goes to zero exponentially fast since the random vector $(\zeta_1,\dotsc,\zeta_m)$ is Gaussian.

Next use the fact that for any nonnegative random variable $X$ we have the equality

$$ \bE(X^k) =\int_0^\infty kx^{k-1} P(X>x) dx,\;\;\forall k\geq 1. $$

  • $\begingroup$ It is not entirely clear to me... From your answer it appears as if your conclusion is: They are always finite, no matter what $A$, the variance of $Z$, the mean $\mu$ and $n$ are...But this cannot be true. Also, the constant $C$ is important. If C>1, it is not clear at all to me that the moments are finite as $n$ grows large. I can agree that for finite $n$, the moments are finite, but the question concerns the case whether $$\lim_{n\to \infty} \mathbb{E}(AX)^n=0$$ $\endgroup$ Jan 30, 2013 at 21:24
  • $\begingroup$ The moments are finite as $n\to\infty$. They may not be bounded as $n\to\infty$. $\endgroup$ Jan 30, 2013 at 22:27
  • $\begingroup$ You need to assume something about $A$. Even the case $N=1$ you can see that if $|A|>1$ the moments go to $\infty$ like $|A|^n$. $\endgroup$ Jan 30, 2013 at 22:31
  • $\begingroup$ This is essentially the question. What is needed? I need to evaluate the expected inverse of $I+AX$, which I will do through a Neumann series. But the series is not always convergent, and in that case I will use a preconditioned in order to guarantee convergence of the series expansion. The problem is now that I don't really know how to design the preconditioned. $\endgroup$ Jan 31, 2013 at 6:34
  • $\begingroup$ The problem is tricky. Just look at the case when $A$ is diagonal and you se you need to impose some conditions on $A$. My argument shows that a condition of the type $C^2|A| <1$ will do. $\endgroup$ Jan 31, 2013 at 9:50

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