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Let $\mathbf{x}\in\Bbb{R}^n$ be a multivariate normal vector with mean $\bar{\mathbf{x}}$ and covariance matrix $\Sigma$. Let also $\mathcal{H}\colon\mathbf{w}^\top\mathbf{x}+b=0$ be a hyperplane in $\Bbb{R}^n$.

We define a function $$ s\colon\Bbb{R}^n\to\Bbb{R}, \\ \mathbf{x}\mapsto\mathbf{w}^\top\mathbf{x}+b. $$ The expected value of $s(\mathbf{x})$, when $\mathbf{x}\sim\mathcal{N}(\bar{\mathbf{x}},\Sigma)$, is given obviously by $s(\bar{\mathbf{x}})=\mathbf{w}^\top\bar{\mathbf{x}}+b$. However, if I evaluate the expected value of $s(\mathbf{x})$, when $\mathbf{x}\in\Omega=\{\mathbf{x}\in\Bbb{R}^n\colon\mathbf{w}^\top\mathbf{x}+b\geq0\}$, i.e., over the "positive" halfspace defined by $\mathcal{H}$; that is, $$ s_+ = \int_\Omega s(\mathbf{x})f(\mathbf{x})\mathrm{d}\mathbf{x}, $$ where $f$ is the probability density function of $\mathbf{x}$, I get the following result $$ s_+= \frac{s(\bar{\mathbf{x}})}{2} \operatorname{erfc} \left( -\frac{s(\bar{\mathbf{x}})}{\sqrt{2\mathbf{w}^\top\Sigma\mathbf{w}}} \right) + \frac{\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}}{\sqrt{2\pi}} \exp \left( -\frac{1}{2}\left(\frac{s(\bar{\mathbf{x}})}{\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}}\right)^2 \right) $$ It seems that under a zero covariance matrix ($\Sigma\to\mathbf{0}$), this quanity tends to $s(\bar{\mathbf{x}})$, which is expected.

I would like to understand better the above result. The quantities $s(\bar{\mathbf{x}})$ and $\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}$ are in the same units; they measure distance. It also seems that the ratio $$ r = -\frac{s(\bar{\mathbf{x}})}{\sqrt{2\mathbf{w}^\top\Sigma\mathbf{w}}} $$ plays some role in how the expected value of $s(\mathbf{x})$, when $\mathbf{x}\in\Omega$ relates to the mean $s(\bar{\mathbf{x}})$. Using this ratio, the above result is rewritten in a more clear form as follows $$ s_+= \frac{s(\bar{\mathbf{x}})}{2} \left(\operatorname{erf}(r) + 1\right) + \frac{\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}}{\sqrt{2\pi}} \exp(-r^2). $$ Could anyone help me about the interpretation of the "distance" $\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}$?

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The only geometrical answer I can think of is that $\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}$ is the length of $\Sigma\mathbf{w}$ in the space $\mathbb R^n$ with inner product $\langle \mathbf a,\mathbf b\rangle= (\Sigma^{-1/2}\mathbf a)^\top (\Sigma^{-1/2} \mathbf b)$. Note $\mathbf w^\top \mathbf x=\langle \Sigma\mathbf w, \mathbf x\rangle$ so $\Sigma \mathbf w$ appears naturally in this setting. In this space $\mathbf x$ has the orthogonality you would expect of a standard normal in the sense that the covariance of $\langle \mathbf a, \mathbf x\rangle$ and $\langle \mathbf b, \mathbf x\rangle$ is $\langle \mathbf a,\mathbf b\rangle.$ Your integral is symmetric under symmetries preserving this distance.

More explicitly you can reduce to the case of a standard normal with the normal inner product:

  • $\mathbf{w}'=\Sigma^{1/2}\mathbf{w}$
  • $\mathbf{x}'=\Sigma^{-1/2}(\mathbf{x}-\overline {\mathbf{x}}')$ so $\mathbf{x}'\sim\mathcal N(0,I)$
  • $b'=s(\overline{\mathbf{x}})$

This primed version has the same integral as the original (with $\Sigma=I$). But it is then obviously rotationally invariant, the only parameters are then $|\mathbf w'|$ and $b'$. In fact it reduces to the one-dimensional case, integrating $|\mathbf w'|x+b'$ for a normal $x \sim \mathcal N(0,1)$ over $|\mathbf w'|x+b'\geq 0$.

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