2
$\begingroup$

If $X$ denotes a $m \times n$ random matrix whose entries are independent identically distributed random variables with mean $\mu$ and $\sigma^2 < \infty$, let

$$Y = X X^T$$

with $X^T$ the transpose of $X$. Let $\lambda_1 , \lambda_2 ,\ldots, \lambda_m$ be the eigenvalues of $Y$ (viewed as random variables).

If $\mu = 0$, it is known that the law of $\lambda$ converges to Marchenko–Pastur distribution: https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution

My question is that in the case $\mu \neq 0$ what is the limit distribution of $\lambda$?

$\endgroup$
0
4
$\begingroup$

The addition of the same constant $\mu$ to all elements of $X$ (so that their mean becomes $\mu$) is a rank-one perturbation of the matrix, which has no effect on the distribution of the eigenvalues of $XX^T$ in the limit $n,m\rightarrow\infty$ at fixed $n/m$ --- this limiting distribution remains the Marchenko-Pastur distribution. The rank-one perturbation introduces a single outlier, but the bulk of the spectrum is unchanged. See for example The singular values and vectors of low rank perturbations of large rectangular random matrices (2011).

$\endgroup$

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.