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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?

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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.

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  • $\begingroup$ Thanks, can you give me a direction for the long answer or a reference? I'm very curious about the fundamental reason. $\endgroup$
    – Swiss Army Man
    Commented Jan 10, 2012 at 18:21
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    $\begingroup$ Looking at the simplest non-trivial example can help. For instance, when $\mathbf{x}$ is a row vector of length 2, say with components $x_1$ and $x_2$, then $\mathbf{x}^t$ is a column vector of length 2 with components $x_1$ and $x_2$. The expectations are $E[x_1]$ and $E[x_2]$: these are the components of $E[\mathbf{x}]$ and $E[\mathbf{x}^t]$; the only difference is that in the first case they are arranged side-by-side in the row vector and in the second they are arranged top-to-bottom in the column vector. $\endgroup$
    – whuber ♦
    Commented Jan 10, 2012 at 18:39
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As the previous answer confirmed, the answer is yes. Actually, even complex transposing commutes through expectation, meaning: $$\mathbb{E}[x^H] = \mathbb{E}[x]^H $$

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