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For standard normal distribution, the expected value of $x^2$ is $1$. A natural question is that in the multivariate case, what is the expected value of $x^t\Sigma x$ for multivariate normal distribution $x \sim N(0,\Sigma)$? I have difficulty to carry out the integral, but would guess the result is related to the norm of $\Sigma$.

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Note that $x^T \Sigma x = tr(x^T \Sigma x) = tr( x x^T \Sigma)$ by invariance of trace to cyclic permutations and trace of a scalar is itself. Also, note that trace is linear (so we can bring expectations inside the trace). Since $x$ is zero mean, $E[x x^T ] = \Sigma$.

Then we have, by the aforementioned, $E[x^T \Sigma x] = E[ tr( x x^T \Sigma) ] = tr(E[x x^T \Sigma]) = tr(E[x x^T ] \Sigma) = tr(\Sigma^2)$.

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Let $\mathbf{X} = (X_1, X_2, \ldots, X_p)$. By definition of $\mathbf{\Sigma}$ and matrix multiplication,

$$ \mathbf{X^T\Sigma X} = \sum_{i=1}^p\sum_{j=1}^p \operatorname{Cov}[X_i, X_j]X_iX_j$$

Since $E[X_i] = 0, i = 1, 2, \ldots, p$, the expected value is

$$ E[\mathbf{X^T\Sigma X}] = \sum_{i=1}^p\sum_{j=1}^p \operatorname{Cov}[X_i, X_j]^2 = \mathbf{1^T(\Sigma\circ\Sigma)1}$$

where $\mathbf{1}$ is a $p$-dimensional vector of $1$s, and $\circ$ is the Hadamard product.

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  • $\begingroup$ thx. Squared norm of $\Sigma$? My guess seems right... $\endgroup$ – ahala Jun 16 '16 at 3:17

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