Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working on a signal processing problem where I want to model the measurement sample covariance matrix (SCM) as random matrix and hence use the results from Random Matrix Theory (RMT). Let $\mathbf{X}$ be $p\times n$ matrix with i.i.d complex gaussian random variable with zero mean and variance $1$. The sample covariance matrix is

$$\mathbf{S_n} = \frac{1}{n}\mathbf{X}\mathbf{X^H}$$

which is a Wishart Matrix.

Than Random Matrix Theory tells us the limiting eigen value distribution of $\mathbf{S_n}$ as the matrix dimension grows infinitely large i.e. as $n,p \to \infty$ and $\frac{n}{p} \to c$, the eigen value distribution converges to a deterministic distribution function $F^S(x)$.

I am having unable to understand how the results of RMT applies to the SCM which is essentially limited to finite dimension in a real world problem. Is it sufficient that the matrix dimensions $n$ and $p$ are large values? ( How large? ).

I referred to papers where people have applied RMT with array processing problems but its not clear how the infinite limiting results of RMT are applicable in finite cases.

Here is a paper I read, and in the last sentence on page 8, authors suggest that

when the number of explanatory variables is very large compared to $n$, it is natural to formulate it as a case where both $n$ and $p$ are tending to infinity.

This statements seems to contrast what I understand of RMT where both dimensions of the matrix grow infinitely large.

I would like to understand this issue from those who are working with RMT. Please help me out.

share|cite|improve this question
Search for "universality" + RMT on arxiv might help. There are various regimes of $p(n)$ for which the limits are known to hold. – zyx Apr 5 '13 at 19:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.