Edit: I shall try to reformulate my question in order to make it -hopefully- more clear.

Let $X$ be a random variable that follows the $n$-dimensional Gaussian distribution. The probability density of $X$ is given by the following function:

$$ f_X(\mathbf{x;\mathbf{\mu}, \Sigma}) = \frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}} exp\{ -\frac{1}{2} (\mathbf{x}-\mathbf{\mu})^T \Sigma^{-1} (\mathbf{x}-\mathbf{\mu}) \}, $$

where $\mathbf{x},\mathbf{\mu}\in\Re^n$ and $\Sigma \in S_{++}^{n}$.

In addition, let $\mathcal{H}: \mathbf{x}^T\mathbf{w}=0$ be a hyperplane in the $n$-dimensional Euclidean space $\Re^n$, where $\mathbf{w}\in\Re^n$ and $b\in\Re$. The hyperplane $\mathcal{H}$ defines two half-spaces:

$$ \Omega_{+} = \{\mathbf{x} \in \Re^n | \mathbf{x}^T\mathbf{w}+b \geq 0 \}\\ \Omega_{-} = \{\mathbf{x} \in \Re^n | \mathbf{x}^T\mathbf{w}+b < 0 \} $$

What I would like to find out is what's happening when I try to integrate the above gaussian pdf over the region $\Omega_{+}$ (or $\Omega_{-}$), as it's well-known that -by definition- integrating over the whole $\Re^n$ gives 1. This is the spirit! And this is the reason of the existence if the (normalisation) term $\frac{1}{(2\pi)^{n/2}|\Sigma|^{1/2}}$.

Please correct me if I am wrong in something above.

Next, we could observe that if $b=0$, i.e., the hyperplane crosses the origin, then - due to the symmetry of the gaussian pdf - the integral should be equal to $1/2$. But how can I compute the value of the integral when $b\neq0$? Is there some clever trick to compute how the gaussian integral changes with respect to the bias term $b$?

Thanks in advance and apologies for being kinda "newbie" or not strict enough!... Every correction or helpful comment (either on my question or my notation) would be extremely appreciated!

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    $\begingroup$ Rotate it so that you get an integral over a half-space $\{ x_n \geqslant c\}$. Then you have $n-1$ full Gaussian integrals, and one $\int_c^\infty e^{-\xi^2/2}\,d\xi$. $\endgroup$ Commented Nov 8, 2013 at 15:50
  • $\begingroup$ @DanielFischer Thank you very much for your instant answer. I will try it and I will return if I have other questions. It seems right, but I'm afraid that something is gonna go wrong! Thank you anyway! $\endgroup$ Commented Nov 8, 2013 at 15:58
  • $\begingroup$ @DanielFischer, I didn't find the time to look at your approach yet, but because of the fact that I tried something similar, I remember that I stuck at the n-dimensional rotation matrix... Could you please clarify for me what exactly you mean by ``rotate it''? Rotate around what? What is the rotation matrix in n dimensions? Apologies for being newbie... $\endgroup$ Commented Nov 10, 2013 at 17:47
  • $\begingroup$ I made some edits that I think that clarify my original question. This question could be concerned along with the following question math.stackexchange.com/questions/562332/… -- Thanks again! $\endgroup$ Commented Nov 11, 2013 at 12:37
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    $\begingroup$ What is $b$? I imagine that instead of $0$ in the definitions of $\Omega_+$ and $\Omega_-$, you may have meant to have $b$. $\endgroup$
    – robjohn
    Commented Nov 22, 2013 at 16:05

2 Answers 2


To compute the integral

$$P = \frac{1}{(2\pi)^{n/2}\sqrt{\det \Sigma}} \int_{\Omega^+} \exp \left(-\frac12 (\mathbf{x}-\mu)^T \Sigma^{-1} (\mathbf{x}-\mu)\right)\,d\mathbf{x},$$

we will need several coordinate transforms. We start with a translation to get rid of the mean, $\mathbf{y} = \mathbf{x}-\mu$. Then $\Omega^+ = \{\mathbf{x} : \mathbf{x}^T\mathbf{w}+b \geqslant 0\}$ corresponds to $\Omega_1 = \{\mathbf{y} : \mathbf{y}^T\mathbf{w} + (\mu^T\mathbf{w} + b)\geqslant 0\}$, and we have

$$P = \frac{1}{(2\pi)^{n/2}\sqrt{\det \Sigma}} \int_{\Omega_1} \exp \left(-\frac12 \mathbf{y}^T \Sigma^{-1} \mathbf{y}\right)\,d\mathbf{y}.$$

Next, since $\Sigma$ is positive definite symmetric, there is an orthogonal matrix $U$ and a diagonal matrix $D$ with positive diagonal elements such that $\Sigma = U^TDU$. Then you have $\Sigma^{-1} = (U^TDU)^{-1} = U^{-1}D^{-1}(U^T)^{-1} = U^TD^{-1}U$ since $U^T = U^{-1}$. Then let $\mathbf{z} = U\mathbf{y}$, and $\mathbf{w}_1 = U\mathbf{w}$. With $\Omega_2 = \{\mathbf{z} : \mathbf{z}^T\mathbf{w}_1 + (\mu^T\mathbf{w}+b)\geqslant 0\}$, we obtain

$$P = \frac{1}{(2\pi)^{n/2}\sqrt{\det\Sigma}} \int_{\Omega_2} \exp \left(-\frac12 \mathbf{z}^T D^{-1} \mathbf{z} \right)\,d\mathbf{z},$$

since $\lvert \det U\rvert = 1$.

Now we do a bit of rescaling. Let $\mathbf{a} = \sqrt{D^{-1}}\mathbf{z}$, or $\mathbf{z} = \sqrt{D}\mathbf{a}$, $\mathbf{w}_2 = \sqrt{D}\mathbf{w}_1$, and $\Omega_3 = \{ \mathbf{a} : \mathbf{a}^T\mathbf{w}_2 + (\mu^T\mathbf{w}+b)\geqslant 0\}$. Since $\det \sqrt{D} = \sqrt{\det\Sigma}$, that yields

$$P = \frac{1}{(2\pi)^{n/2}} \int_{\Omega_3} \exp \left(-\frac12 \mathbf{a}^T\mathbf{a}\right)\,d\mathbf{a}.$$

Now find an orthogonal matrix $B$ such that $B\mathbf{w}_2 = \lVert\mathbf{w}_2\rVert\cdot e_n$, and let $\mathbf{b} = B\mathbf{a}$, $\Omega_4 = \{\mathbf{b} : \mathbf{b}^T(\lVert \mathbf{w}_2\rVert\cdot e_n) + (\mu^T\mathbf{w}+b)\geqslant 0\} = \{\mathbf{b} : \lVert\mathbf{w}_2\rVert b_n + (\mu^T\mathbf{w}+b)\geqslant 0\}$. That yields

$$P = \frac{1}{(2\pi)^{n/2}} \int_{\Omega_4} \exp\left(-\frac12 \mathbf{b}^T\mathbf{b}\right)\,d\mathbf{b}.$$

Now $\Omega_4 = \mathbf{R}^{n-1}\times [c,\infty)$, with $c = -\dfrac{(\mu^T\mathbf{w}+b)}{\lVert \mathbf{w}_2\rVert}$, and hence

$$P = \frac{1}{\sqrt{2\pi}} \int_c^\infty \exp\left(-\frac12 x^2\right)\,dx,$$

which can be evaluated in terms of the error function.

  • $\begingroup$ What can I say?! Thanks a lot! Great answer! Could you tell me if my approach below (using the circulant matrices) is good? You get what I wanted to do with the change of variables, don't you? Thanks again! $\endgroup$ Commented Nov 25, 2013 at 21:40
  • $\begingroup$ I was afraid you'd ask. To be honest, I have no experience with circulant matrices, I can't tell you off-hand if the approach works. I have to read it very carefully to understand what's going on. Don't expect an answer to that soon. (Certainly not today, maybe tomorrow.) $\endgroup$ Commented Nov 25, 2013 at 21:47
  • $\begingroup$ Whenever you can, of course! I'd just like to know if it's gonna work! If you have something, sometime, please let me know! Thanks again! $\endgroup$ Commented Nov 25, 2013 at 21:51
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    $\begingroup$ We have $\det\sqrt{D}=\sqrt{\det\Sigma}$ since $\Sigma=U^TDU$, so $\det\Sigma=\det(U^T)\cdot\det D\cdot\det U=(\det U)^2\cdot\det D=\det D$, as $U$ is orthogonal, hence $\det U=\pm1$. Now $D=(\sqrt{D})^2$, whence $\det D=(\det\sqrt{D})^2$, and since everything is positive, $\det\sqrt{D}=\sqrt{\det D}=\sqrt{\det\Sigma}$. Then with the substitution $\mathbf{a}=(\sqrt{D})^{-1}\mathbf{z}$, the transformation formula in the short-cut version says $d\mathbf{z}=\lvert\det\sqrt{D}\rvert\cdot d\mathbf{a}$, and with the above, that cancels the $\sqrt{\det\Sigma}$. $\endgroup$ Commented Nov 26, 2013 at 20:16
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    $\begingroup$ Yes, we have $\mathbf{w}_2 = \sqrt{D}U\mathbf{w}$, but unless $\Sigma$ is a multiple of the identity, you can't unfortunately express $\lVert\mathbf{w}_2\rVert$ that easily in terms of $\mathbf{w}$. However, $\lVert\mathbf{w}_2\rVert^2 = (\sqrt{D}U\mathbf{w})^T(\sqrt{D}U\mathbf{w}) = \mathbf{w}^TU^T\sqrt{D}\sqrt{D}U\mathbf{w} = \mathbf{w}^T\Sigma\mathbf{w}$, so you can write the norm as $\sqrt{\mathbf{w}^T\Sigma\mathbf{w}}$. $\endgroup$ Commented Nov 26, 2013 at 23:04

There is something that could help... I would like some help on notation or whatever else needed, though!

Let $\mathbf{x}=R^T\mathbf{y}$. Then the hyperplane can be rewritten as follows:

$$ \mathcal{H}: \mathbf{w}^TR^T\mathbf{y}+b=0. $$

We define $R$ to be the circulant matrix generated by a vector $\mathbf{r}=(r_1,\dots,r_n)^T$. Then, I think that the following holds:

$$ \mathbf{w}^TR^T = (R\mathbf{w})^T = (W\mathbf{r})^T = \mathbf{r}^T W^T, $$

where $W$ is the circulant matrix generated by the vectors $\mathbf{w}$. Consequntly, if we demand

$$ \mathbf{r}^T W^T = \mathbf{v}^T, $$

then $\mathbf{r}^T = \mathbf{v}^T(W^T)^{-1}=\mathbf{v}^T(W^{-1})^T$. Thus,

$$ \mathbf{r} = W^{-1}\mathbf{v}. $$

If we set the vector $\mathbf{v}$ to be, for instance, $(1,0,\dots,0)^T$, then the change of variable we made will lead to a hyperplane as follows:

$$ \mathcal{H'}: y_1=c. $$

Am I right so far?

In order to find the matrix $R$, which constructs the change of variable ($\mathbf{x}=R^T\mathbf{y}$), we need to find the vector $\mathbf{r}$, which constructs the circulant matrix $R$. We can find $\mathbf{r}$ can be found from the following:

$$ \mathbf{r} = W^{-1}\mathbf{v}, $$

where $W$ is known and is generated as a circulant matrix from $\mathbf{w}$.

But $W$ is a circulant matrix, so it can be expressed as follows:

$$ W = Q\Lambda Q^{-1}, $$

where $Q$ is the matrix of the eigenvectors of $W$, while $\Lambda$ is the diagonal matrix of the corresponding eigenvalues of $W$. For a circulant matrix

$$ W= \begin{bmatrix} w_0 & w_{n-1} & \dots & w_{2} & w_{1} \\ w_{1} & w_0 & w_{n-1} & & w_{2} \\ \vdots & w_{1} & w_0 & \ddots & \vdots \\ w_{n-2} & & \ddots & \ddots & w_{n-1} \\ w_{n-1} & w_{n-2} & \dots & w_{1} & w_0 \\ \end{bmatrix}, $$

the $j$-th eigenvector is given by

$$ \mathbf{q}_j = (1,\omega_j, \omega_j^2, \dots, \omega_j^{n-1})^T, \: j=1,...,n-1 $$

where $\omega_j = exp\left(\frac{2\pi i j}{n}\right)$ are the $n$-th roots of unity, and $i$ the imaginary unit. Moreover, the corresponding $j$-th eigenvalue is given by $\lambda_j = w_0 + w_{n-1}\omega_j + \dots + c_1\omega_j^{n-1}$, $j=0,\dots,n-1$.

As a result we can find $\mathbf{r}$ as follows:

$$ \mathbf{r} = W^{-1}\mathbf{v} = Q\Lambda^{-1}Q^{-1}\mathbf{v}. $$

Thus, having the vectors $\mathbf{r}$ we can construct the circulant matrix $R$, and using the change of variable $\mathbf{x}=R^T\mathbf{y}$ we can go to a hyperplane of the form $y_n=c$, where the integration of simpler.

Am I right so far? Please give me some feedback! Thanks in advance!


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