Assume $x\in \mathbb{R}^N$ is a random variable vector (like a noise sequence). You now want to calculate the following term:
$E\{x^{T}Ax\}$, where $A$ is a constant matrix. How can this expression rewritten in terms of, for example, $E\{x^Tx\}$?
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1Without determining the distribution of $x$, this is impossible. – 5xum Jul 29 '15 at 07:08
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Do you know something about $A$? symmetric, PD, etc ... – Math-fun Jul 29 '15 at 07:18
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4Why don't you go back to the definition of $x^TAx$ as $\sum\limits_{ij}A_{ij}x_ix_j$? The rest follows. – Did Jul 29 '15 at 07:19
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no, A can be arbitrary – ehsank Jul 29 '15 at 07:25
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But to make it a bit simpler you may assume $A=B^TB$. – ehsank Jul 29 '15 at 07:49
2 Answers
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I do not know if this is what you are after ... there is another answer given too, but if you know the mean and the covariance of $X$ then:
using your comment ($A=B^TB$) and assuming $EX=\mu$ and $cov(X)=\Sigma$ we may write \begin{align} EX^TAX&=E(X-\mu)^TA(X-\mu)+\mu^TA\mu\\ &=E[B(X-\mu)]^TB(X-\mu)+\mu^TA\mu\\ &=\mbox{tr} B\Sigma B^T+\mu^TA\mu\\ \end{align}
Math-fun
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In general I like the convention: capital letters for random variables :-) and lower case variables for their realizations :-) – Math-fun Jul 29 '15 at 08:18
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but your result is a vector, however Ex'Ax is scalar. Note that B is a matrix. – ehsank Jul 29 '15 at 08:20
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I was not sure if "$x$" is a row or a column ... It was not clear from the question statement. – Math-fun Jul 29 '15 at 08:29
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It involves covariance between the various $x_i$.
$xx^T$ is a $n\times n$ matrix. Let $C=E\{xx^T\}$.
The answer is the trace of $AC$.
If $E\{x_ix_y\}=E\{x_i\}E\{x_j\}$, so the $x_i$ are independent of each other, then the answer is $E\{x\}^TAE\{x\}$
Empy2
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