2
$\begingroup$

Let $X$ be a random variable which follows an $n$-dimensional Gaussian distribution with mean vector $\mu\in\mathbb{R}^n$ and covariance matrix the symmetric positive definite $n\times n$ matrix $\Sigma$, i.e. $\Sigma\in\mathbb{S}_{++}^{n}$. The probability density function of $X$ is given by $f_{X}\colon\mathbb{R}^n\to\mathbb{R}^{*}_{+}$, with

$$ f_{X}(\mathbf{x};\mu,\Sigma) = \frac{1}{(2\pi)^{n/2}\lvert\Sigma\rvert^{1/2}} \operatorname{exp} \Big\{ -\frac{1}{2}(\mathbf{x}-\mu)^T\Sigma^{-1}(\mathbf{x}-\mu) \Big\}. $$

Also let $g\colon\mathbb{R}^n\to\mathbb{R}$. We would like to evaluate the following integral

$$ I = \int_{\Omega} \! \big[1-g(\mathbf{x})\big] f_{X}(\mathbf{x};\mu,\Sigma) \, \mathrm{d}\mathbf{x}, $$

or

$$ I = \frac{1}{(2\pi)^{n/2}\lvert\Sigma\rvert^{1/2}} \int_{\Omega} \! \big[1-g(\mathbf{x})\big] \operatorname{exp} \Big\{ -\frac{1}{2}(\mathbf{x}-\mu)^T\Sigma^{-1}(\mathbf{x}-\mu) \Big\} \, \mathrm{d}\mathbf{x}, $$

where $\Omega=\{\mathbf{x}\in\mathbb{R}^n \mid g(\mathbf{x})\leq 1\}$. In general, it would be very desired if for the function $g$ we know nothing but that it is a real-valued function defined on $\mathbb{R}^n$. If this is impossible, then we could assume additionally that $g$ can be written as

$$ g(\mathbf{x})=\sum_{j=1}^{N} \alpha_j k(\mathbf{x},\mathbf{x}_j), $$

where $N\in\mathbb{N}$, $\alpha_j\in\mathbb{R}$, $j=1,\dotsc,N$, and $k\colon\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}_{+}^{*}$ is symmetric.

It could be found some connection with a question I asked $8$ mounths ago.

EDIT: In fact, there has been the following "replacement": $f(\mathbf{x})=\mathbf{w}\cdot\mathbf{x}$, where $\mathbf{w}\in\mathbb{R}^n$, and $f$ belongs to a Reproducing Kernel Hilbert Space (RKHS) with associated kernel $k$ (see above). This question can be connected with this one, if someone wants to go deeper.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.