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In one paper I read, the authors write $$ \mathbb{E}\left[\|\tilde{\Sigma}^{-\frac{1}{2}}\left(\hat{\Theta}-\Omega\right)\|_F^2\right]=\mathbb{E}\left[\|{\Sigma}^{-\frac{1}{2}}\left(\hat{\Theta}-\Omega\right)\|_F^2\right] $$ where $\tilde{\Sigma}$ is an unbiased estimator of $\Sigma$ and $\|\cdot\|_F$ denotes the frobenuis norm.

Why does this equation hold, and if so, does it mean that $\mathbb{E}\left[\|\hat{\Sigma}\|_F\right]=\|{\Sigma}\|_F$ is also correct?

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    $\begingroup$ The equation $$ \mathbb{E}\left[\|\tilde{\Sigma}^{-\frac{1}{2}}\left(\hat{\Theta}-\Omega\right)\|_F^2\right]=\mathbb{E}\left[\|{\Sigma}^{-\frac{1}{2}}\left(\hat{\Theta}-\Omega\right)\|_F^2\right] $$ holds due to unbiasedness of $\tilde{\Sigma}$ which allows the substitution in the expectation of the squared Frobenius norm. However, because the expectation of the Frobenius norm of an estimator does not generally equal the Frobenius norm of the expectation, the equality $\mathbb{E}\left[\|\hat{\Sigma}\|_F\right]=\|{\Sigma}\|_F$ does not hold. $\endgroup$
    – Robert Long
    Commented Nov 3 at 9:53
  • $\begingroup$ @RobertLong So $\mathbb{E}\left[\|\hat{\Sigma}\|_F^2\right]=\|\Sigma\|_F^2$ is correct, right? $\endgroup$
    – mathhahaha
    Commented Nov 3 at 11:10

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