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Suppose I have a random $m \times m$ matrix $R \sim \mu$ that is possibly singular. Is it true that $ E[R] \propto I$ implies that there exists a scalar $r_{\mu, m}$ such that $E[R^+] = r_{\mu, m}I$, where ${}^+$ denotes the Moore-Penrose pseudo-inverse?

If this is false but there are certain conditions that are needed to make this true, then I'd be interested to know those conditions.

For instance it is true when $R = X^\top X$ and $X$ is $n \times m$ normally distributed (at least I have found references for this when $n\not\in [m-3, m+1]$). Not sure if this is the only non-trivial case.

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  • $\begingroup$ what does $R\sim \mu$ mean? $\endgroup$
    – Carlo Beenakker
    Commented Jan 28, 2021 at 12:12
  • $\begingroup$ The random matrix $R$ has distribution $\mu$. $\endgroup$
    – user27182
    Commented Jan 28, 2021 at 16:00
  • $\begingroup$ you write $R^+ = r_{\mu, m}I$ --- don't you want to take the expectation value of $R^+$ ? $\endgroup$
    – Carlo Beenakker
    Commented Jan 28, 2021 at 17:58
  • $\begingroup$ yes, that's a typo! thanks $\endgroup$
    – user27182
    Commented Jan 28, 2021 at 20:27

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In general no.

If $X$ and $Y$ are independent scalar random variables satisfying $\mathbb E X = \mathbb E Y = 1$ but $\mathbb E X^{-1} \neq \mathbb E Y^{-1}$, then the matrix $R = \begin{pmatrix} X & 0 \\ 0 & Y \end{pmatrix}$ is a counterexample.

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