3
$\begingroup$

Suppose I have a p-dimensional integral:

$\int_{0}^{\infty}\int_{0}^{\infty}\dots \int_{0}^{\infty}f(x_1,x_2,\dots,x_p)dx_1dx_2\dots dx_p$

And then I make a rotation + translation transform:

$W=A^{T}(X-b)$

Question: How will the region of integration $X>0$ change in the $W$ space?

Can assume $A$ is a matrix of eigenvectors of a real symmetric positive definite matrix if this makes the answer easier.

  • 0
    @joriki - I had a feeling it would be something disgusting. Oh well, just have to try another route :(. Just would have been nice, as my function $f(.)$ "decouples" under a rotation transform into independent products. Still wouldn't mind seeing the ugly solution, if someone has the energy to work it out.2011-04-03

1 Answers 1

1

Maybe you can put $\int_{0}^{\infty}\int_{0}^{\infty}\dots \int_{0}^{\infty}f(x_1,x_2,\dots,x_p)dx_1dx_2\dots dx_p$ $ = \frac{1}{2^p} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\dots \int_{-\infty}^{\infty}f(|x_1|,|x_2|,\dots,|x_p|)dx_1dx_2\dots dx_p.$ Then, if the $f$ functions are even, you can rotate away.