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" $\sqrt{\mathbf{w}^\top\Sigma\mathbf{w}}$?

| cite | improve this question | | | | |

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 $|\mathbf w'|x+b'\geq 0$.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.