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I am trying to evaluate $E[f(x)f(x)^T]$ where $f(x) = Nx$ and $N \in \mathbb{R}^{N \times N}$, and also $p(x) = \mathbb{N} (x|μ, \sigma)$

So, as we know,

\begin{align*} E[f(x)f(x)^T] &= \int_{-\infty}^{\infty} f(x)f(x)^{T} p(x)dx \\ &= \int_{-\infty}^{\infty} Nx(Nx)^{T} p(x) dx \\ &= \int_{-\infty}^{\infty} Nxx^{T}N^{T} p(x) dx \end{align*}

Now, I am confused about the next step as I haven't evaluated $\mathbb{E}$ with a matrix before. Can I drag the $NN^T$ part outside as it's constant with respect to $x$? Can I break the order? What to do next?

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2 Answers 2

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Assuming the matrix doesn't depend on the random vector $x$, you can indeed take the matrix outside the expectation, since matrix multiplication is a linear operation. You just have to be careful to remember that matrix multiplication is "order dependent" i.e. not commutative:

$$ \mathbb{E}[(Nx)(Nx)^T] = \mathbb{E}[Nxx^T N^T] = N\mathbb{E}[xx^T]N^T = NC N^T $$ where $C = \mathbb{E}[xx^T]$.

Hope that helps!

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  • $\begingroup$ Thanks, it certainly helps. I want to ask what if I have $E[(Nx)^{T}(Nx)]$ instead? Then the order would be $x^{T}N^{T}Nx$. Then how would I approach? $\endgroup$ Commented Nov 10, 2020 at 21:21
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    $\begingroup$ Well, you're looking for the expected norm of the (transformed) vector $Nx$; the expected norm is the trace of the correlation matrix (see here). So, using what you learned about how the correlation matrix transforms, I think you can probably figure it out - it'll be the trace of $NCN^T$. $\endgroup$
    – icurays1
    Commented Nov 10, 2020 at 21:44
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Yes, you can pull out the $N$. Doing so leaves you with $$ N \left( \int_{-\infty}^\infty xx^T\,p(x)\,dx\right)N^T = N \,\mathbb{E}(xx^T)N^T. $$ One way to proceed is to note that $\mathbb{E}(xx^T) - \mathbb{E}(x)\mathbb E(x)^T$ is the covariance matrix of $x$.

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  • $\begingroup$ Thanks. Yeah, I know I can write, $E[xx^{T}] = Var(x) + \mu\mu^{T}$. I was basically confused about the order. Now, what happens if I have $x^{T}N^{T}Nx$. I was actually trying to understand the ordering here. Or should I ask a new question? $\endgroup$ Commented Nov 10, 2020 at 21:14
  • $\begingroup$ @AhsanulHaque You could if you want. The quick answer is that the expectation of $x^TN^TNx$ is a scalar rather than a matrix, and there is no longer an obvious way to factor out the $N$'s in the computation. However, it turns out that the expectation of $x^TN^TNx$ is simply the trace of the (matrix-valued) expectation of $Nxx^TN^T$. $\endgroup$ Commented Nov 10, 2020 at 21:44

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