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I am attempting to marginalize a probability density function. But I got stuck on the following integral $$ \int_{-\infty}^\infty\cdots\int_{-\infty}^\infty \frac{\exp(\pmb x^T A\pmb z)} {|\exp(A\pmb z )|_1^{n+|\pmb{x}|_1}} \mathrm dz_1\cdots\mathrm dz_m $$ where $\pmb x, \pmb z \in\mathbb R^n$, $x_i\ge 0$ with large $n,m \in\mathbb N$. $|\cdot|_1$ is the sum of the components e.g. $|\pmb x|_1 = \sum_{i=1}^nx_i$.

$A\in\mathbb R^{n\times n}$ is orthogonal and the first $m$ coloums are also orthogonal to $(1,\dots,1)$.

I already stripped away some constants.

My goal is a fast evaluation of the integral on the remaining dimensions $z_{m+1},\dots,z_n$. Can anybody solve this? If there is no explicit form of the integral, a good approximation would also be very helpful!

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  • $\begingroup$ A little clarification: both $\mathbf{x}, \mathbf{z}$ have $n$ components? But you are integrating over $m$ components only where $m<n$? All the same, you will also need to integrate over the rest of the components as well? Yet, you separate the integration because of the special properties of the matrix $A$? But why? You are multiplying $z$ by $A$ anyway, so all the components come into play in the resulting scalar expression. How are $z_1,\dots,z_m$ special? I'm likely missing something here $\endgroup$ – Yuriy S Oct 16 '18 at 14:01
  • $\begingroup$ @YuriyS I think you got most of it right. Only the other components of $A$ are not special. The reason I integrate over the first $m$ components is that I need to marginalize the (density) function. Instead of having a density in an $n$ dimensional space I need it in an $n-m$ dimensional subspace where each point is the total density of the orthogonal $m$ dimensional hyperplane. $\endgroup$ – katosh Oct 16 '18 at 14:10
  • $\begingroup$ I'm pretty sure there's no exact expression for $m>2$. There's no exact expressions for much more simple integrals with sum of exponential terms in the denominator. You'd have to do the integral numerically, maybe Monte-Carlo integration? $\endgroup$ – Yuriy S Oct 16 '18 at 15:12
  • $\begingroup$ Where does the unusual denominator come from? $\endgroup$ – Matt F. Oct 19 '18 at 15:11
  • $\begingroup$ how do you define the exp of a vector ($A\bfz$) ? $\endgroup$ – G Cab Oct 20 '18 at 18:07

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