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Suppose $C_n$ is a product of $n$ $d\times d$ matrices with IID entries coming from standard normal. The following appears to be true. Is there an elementary proof?

$$E[\|C_n\|_F^2]=d^{n+1}$$

This follows from discussion on math.SE on the moment method, but unclear how to adapt it to this, since the moment method requires fixing $n$.

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This follows from the fact that $\mathbb{E}[A^\dagger A]=d I$ (with $A^\dagger$ the conjugate transpose of $A$ and $I$ the $d\times d$ identity matrix). Hence $$\mathbb{E}[\|C_n\|_F^2]=\operatorname{tr}\mathbb{E}[A_1A_2\cdots A_nA_n^\dagger\cdots A_2^\dagger A_1^\dagger]$$ $$=\operatorname{tr}\mathbb{E}[A_2\cdots A_nA_n^\dagger\cdots A_2^\dagger]\mathbb{E}[ A_1^\dagger A_1]$$ $$=d\operatorname{tr}\mathbb{E}[A_2\cdots A_nA_n^\dagger\cdots A_2^\dagger]$$ $$=d\operatorname{tr}\mathbb{E}[A_3\cdots A_nA_n^\dagger\cdots A_3^\dagger]\mathbb{E}[ A_2^\dagger A_2]$$ $$=d^2\operatorname{tr}\mathbb{E}[A_3\cdots A_nA_n^\dagger\cdots A_3^\dagger]$$ $$=d^{n-1}\operatorname{tr}\mathbb{E}[A_n^\dagger A]=d^n\operatorname{tr} I=d^{n+1}.$$

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    $\begingroup$ Ah, that's a lot simpler than the moment method, thanks! $\endgroup$ Dec 3, 2023 at 19:42

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