# How to calculate this expectation with logarithm?

If $\mathbf{x}\sim\mathbf{N}(\mathbf{0},\mathbf{I})$, and assume that $A$ is a symmetric positive definite matrix, how can I calculate the following two expectations where there is a logarithm in it? $$\mathrm{E}_\mathbf{x}\left[\log(\mathbf{x}^\top A\mathbf{x})\right]$$ and

$$\mathrm{E}_\mathbf{x}\left[(\mathbf{x}^\top A\mathbf{x})\log(\mathbf{x}^\top A\mathbf{x})\right]$$

Using the decomposition $A=U\Lambda U^\top$ and $\mathbf{y}=U^\top\mathbf{x}$, the first expectation can be reduced to $$\mathrm{E}_\mathbf{y}\left[\log(\mathbf{y}^\top \Lambda\mathbf{y})\right]$$

$$=\mathrm{E}_\mathbf{y}\left[\log\left(\sum_i{\lambda_i\mathbf{y}_i^2}\right)\right]$$ where $y$ has the same distribution as $x$ because $U$ is a orthogonal matrix.

However, since the random variables $y_i$ are in the logarithm function, I cannot decompose the expectation further and such a expectation seems to be difficult to calculate.

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What's the distribution $\bf N(0,I)$? – Noam D. Elkies Nov 21 '11 at 17:43
@Noam: He means multivariate normal with variance matrix $\mathbf{I}.$ – Igor Rivin Nov 21 '11 at 17:49
This has nice forms for 1D, but in general maybe numerical integration might be the way to go. – Suvrit Nov 21 '11 at 19:50
@Rivin: Yes, you are right. I mean multivariate normal with identity variance. – ppyang Nov 22 '11 at 1:55

For the case $\lambda_1=\cdots=\lambda_k=1$, $\sum_i \lambda_i y_i^2$ has a chi-squared distribution with $k$ degrees of freedom. Plugging in the chi-squared density and cranking up Maple, we get $$E\Bigl(\log\bigl(\sum_{i=1}^k y_i^2\bigl)\Bigl) = \ln 2+\Psi(k/2),$$ where $\Psi(x)$ is the digamma function (the derivative of the gamma function $\Gamma(x)$). The first six values are $-\gamma-\ln 2$, $-\gamma+\ln 2$, $-\gamma+2-\ln 2$, $-\gamma+1+\ln 2$, $-\gamma+\frac83-\ln 2$, $-\gamma+\frac32+\ln 2$.

If the $\lambda_i$ are not all equal, $\sum_i \lambda_i y_i^2$ is called a weighted chi-squared distribution (and probably other names too). I don't know if there is an expression for its density that is useful here.

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The solution seems to be more complex than what I had expected. Thank you for help! – ppyang Nov 22 '11 at 2:19

Well, with Maple's help I can do the cases $n=1$ and $n=2$:

$E[\log(\lambda_1 Y_1^2)] = \log(\lambda_1/2) -\gamma$

$E[\log(\lambda_1 Y_1^2 + \lambda_2 Y_2^2)] = \log((\sqrt{\lambda_1} + \sqrt{\lambda_2})^2/2) - \gamma$

For $n=3$ I have the beginning of a series expansion: writing $\lambda_1 = \lambda_3(1 + \epsilon_1)$ and $\lambda_2 = \lambda_3 (1 + \epsilon_2)$,

\eqalign{&E[\log(\lambda_1 Y_1^2 + \lambda_2 Y_2^2 + \lambda_3 Y_3^2)] = \cr &\ln \left({\lambda_3}/2 \right) +2-\gamma+ \frac{\epsilon_1 + \epsilon_2}{3} - \frac{3 \epsilon_1^2 + 2 \epsilon_1 \epsilon_2 + 3 \epsilon_2^2}{30} + \frac{5 \epsilon_1^3 + 3 \epsilon_1^2 \epsilon_2 + 3 \epsilon_1 \epsilon_2^2 + 5 \epsilon_2^3}{105} \cr &- \frac{\epsilon_1^4}{36} - \frac{\epsilon_1^3 \epsilon_2}{63} - \frac{\epsilon_1^2 \epsilon_2^2}{70} - \frac{\epsilon_1 \epsilon_2^3}{63} - \frac{\epsilon_2^4}{36} + \ldots\cr}

Hmm: it looks like for $\epsilon_2 = 0$ this might be

$$E[\log(\lambda_1 Y_1^2 + \lambda_3 Y_2^2 + \lambda_3 Y_3^2)] = \ln(\lambda_3/2) - \gamma + \ln(1 + \epsilon_1) + 2 \arctan(\sqrt{\epsilon_1})/\sqrt{\epsilon_1}$$

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Thank you for your help! – ppyang Nov 22 '11 at 2:16