# High dimensional integral of exponentials

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!

• 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 – Yuriy S Oct 16 '18 at 14:01
• @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. – katosh Oct 16 '18 at 14:10
• 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? – Yuriy S Oct 16 '18 at 15:12
• Where does the unusual denominator come from? – Matt F. Oct 19 '18 at 15:11
• how do you define the exp of a vector ($A\bfz$) ? – G Cab Oct 20 '18 at 18:07