Does transpose commute through expectation?

Working with the generalized covariance formula for a vector $x$, I have the following:

$$E[(x-\mu)(x-\mu)^T)] = E(xx^T) – \mu E(x^T)$$

But the term $E(x^T)$ doesn’t make much sense to me. Does anyone have an idea why I’m getting this term with my matrix algebra?

Answer

The short answer is “yes”, $E(x^T) = E(x)^T=\mu^T$. Your full expression will be:

$E[(x−μ)(x−μ)^T)]=E(xx^T)−μE(x^T)-E(x)\mu^T+\mu\mu^T = E(xx^T)-\mu\mu^T$

The expectation operator doesn’t care about the shape of the vector or matrix it operates on.

Attribution
Source : Link , Question Author : Swiss Army Man , Answer Author : jbowman

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