2
$\begingroup$

Maybe this is too obvious, but I what to be sure... Let $Y$ be a $p\times p$ symmetric random matrix (i.e. you can think about $Y$ as a matrix with random entries). Define $E[Y]$, the expectation of $Y$, as the matrix with entries $(E[Y])_{ij} = E[Y_{ij}]$. I think that the next affirmation is true:

If $E[Y] = 0_{p\times p}$ then $\lambda_{\max}(Y)\geq 0$ a.s., where $\lambda_{\max}(Y)$ is the greatest eigenvalue of $Y$ (which is real since $Y$ is symmetric).

My argument is as follows. Suppose that all the eigenvalues are negative. Then $tr(Y)<0$, which implies that $E[tr(Y)]<0$ and $tr(E[Y])<0$. This is a contradiction since $E[Y] = 0_{p\times p}$. Then there exist at least one non-negative eigenvalue, one of which is $\lambda_{\max}(Y)$.

Is my argument correct? In that case, is there a generalization of this result?

$\endgroup$
5
$\begingroup$

Consider $p=1$ and $Y$ equal to the 1x1 matrix 1 with probability 1/2 or the 1x1 matrix -1 with probability 1/2.

$\endgroup$
2
  • $\begingroup$ +1, short and very clever $\endgroup$
    – gt6989b
    Jan 30 '20 at 16:08
  • $\begingroup$ Thanks. Good counterexample. $\endgroup$
    – RLC
    Jan 30 '20 at 16:44
3
$\begingroup$

To add to the existing Ian's answer, the mistake in your proof is that $tr(Y) < 0$ does not imply that $\mathbb{E}[tr(Y)] < 0$.

The reason for this is that $Y$ is some realization of the random matrix, which may be anything within the allowed range. Consider, for example, restricting $Y$ to be a $2 \times 2$ diagonal matrix with entries $u,v$. Then, $\mathbb{E}[tr(Y)] = \mathbb{E}[u+v] = 0$, but for any particular matrix, $tr(Y) = u+v$ and both $u,v$ may end up negative in any particular realization.


For another example, consider a uniform random variable $A \sim \mathcal{U}[-1,1]$. Clearly, $\mathbb{E}[A] = 0$, but if I take a sample from this distribution, generating a sequence $A_1, A_2, \ldots$, a good number of them will be negative.

$\endgroup$
7
  • $\begingroup$ Do you mean $tr(Y)<0$ does not imply $E[tr(Y)]<0$? I don't understand why not. Thanks in advance. $\endgroup$
    – RLC
    Jan 30 '20 at 16:26
  • $\begingroup$ @RLC please see the update $\endgroup$
    – gt6989b
    Jan 30 '20 at 16:33
  • 1
    $\begingroup$ Tr(Y)<0 a.s. implies E[Tr(Y)<0] but not the other way around. $\endgroup$
    – Ian
    Jan 30 '20 at 16:34
  • $\begingroup$ @Ian indeed, it is sufficient but not necessary :) $\endgroup$
    – gt6989b
    Jan 30 '20 at 16:36
  • 1
    $\begingroup$ @RLC Some confusion here, apparently. Actually, $tr(Y) < 0$ a.s. $\implies \mathbb{E}[tr(Y)]<0$, and it is true. The other way around, however, is not true. You are saying that $\mathbb{E}[tr(Y)]<0$ and are trying to imply that for almost any $Y$, you have $tr(Y) <0$, and that is false. $\endgroup$
    – gt6989b
    Jan 30 '20 at 16:55

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.