If $X$ is a random variable following a multivariate normal distribution with parameters $\mu$ and $\Sigma$, what is the expected value of $e^{-X^2}$, that is, what is $\int e^{-x^2} p(x) dx$?
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Is $X^2=X^\top X$ ? – Aug 01 '23 at 14:23
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Thanks for the clarification. Yes, X is a vector and X^2 is an inner product of this vector with itself. – Jabby Aug 01 '23 at 14:32
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1An inner product, or the standard inner product? – Aug 01 '23 at 18:26
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I mean the standard inner product – Jabby Aug 02 '23 at 11:04
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I assume you are working with $\Sigma$ (symmetric) positive definite. \begin{align}&\int_{\Bbb R^n} \frac{(2\pi)^{-n/2}}{\sqrt{\det\Sigma}} \exp(-x^\top x)\exp\left(-\frac12(x-\mu)^\top \Sigma^{-1}(x-\mu)\right)\,dx=\\&=\int_{\Bbb R^n} \frac{(2\pi)^{-n/2}}{\sqrt{\det\Sigma}} \exp\left(-\frac12\left(2x^\top x+(x-\mu)^\top \Sigma^{-1}(x-\mu)\right)\right)\,dx=\\&=\int_{\Bbb R^n} \frac{(2\pi)^{-n/2}}{\sqrt{\det\Sigma}} \exp\left(-\frac12\left(x^\top(\Sigma^{-1}+2)x^\top -2x^\top\Sigma^{-1}\mu+\mu^\top\Sigma^{-1}\mu\right)\right)\,dx= \end{align}
Now complete the square, i.e. find a linear map $T$ such that $(x-T\mu)^\top (\Sigma^{-1}+2)(x-T\mu)$ has linear term $-2x^\top \Sigma^{-1}\mu$. You obtain that $T\mu=(2+\Sigma^{-1})^{-1}\Sigma^{-1}\mu=(1+2\Sigma)^{-1}\mu$ is a valid choice.
\begin{align}&=\int_{\Bbb R^n} \frac{(2\pi)^{-n/2}}{\sqrt{\det\Sigma}} e^{-\frac12\left((x-T\mu)^\top(\Sigma^{-1}+2)(x-T\mu)+\mu^\top\Sigma^{-1}\mu-(T\mu)^\top(\Sigma^{-1}+2) T\mu\right)}\,dx=\\ &=\frac{e^{-\frac12(\mu^\top\Sigma^{-1}\mu-(T\mu)^\top(\Sigma^{-1}+2) T\mu)}}{\sqrt{(\det\Sigma)(\det(\Sigma^{-1}+2))}}\int_{\Bbb R^n} \frac{(2\pi)^{-n/2}}{\sqrt{\det(\Sigma^{-1}+2)^{-1}}} e^{-\frac12(x-T\mu)^\top(\Sigma^{-1}+2)(x-T\mu)}\,dx=\\&=\frac{e^{-\frac12(\mu^\top\Sigma^{-1}\mu-(T\mu)^\top(\Sigma^{-1}+2) T\mu)}}{\sqrt{(\det\Sigma)(\det(\Sigma^{-1}+2))}}=\\&=\sqrt{\frac{\exp(-\mu^\top\Sigma^{-1}\mu+\mu^\top\Sigma^{-1}(2+\Sigma^{-1})^{-1}(\Sigma^{-1}+2)(2+\Sigma^{-1})^{-1}\Sigma^{-1}\mu)}{\det(1+2\Sigma)}}=\\ &=\sqrt{\frac{\exp(\mu^\top((2\Sigma^2+\Sigma)^{-1}-\Sigma^{-1})\mu)}{\det(1+2\Sigma)}}\\\end{align}