3
$\begingroup$

I'm trying to compute the expected value of $ \prod_i (1-\frac{|x_i|}{L})$ on the surface of a $n$-dimensional sphere. A first step could be to integrate only on the first quadrant to take out the absolute value (since the function is symmetric).

The intuition behind it (if I'm not wrong) is that for $L>1$ that product is the probability that, if we put edge-parallel planes in every dimension separated by distance $L$, a point in the surface of the unit sphere will end up in the same hypercube as the origin.

For 2 dimensions it's easy because we can pass to polar coordinates:

$$\frac{2}{\pi}\int_{\theta=0}^{\pi/2} (1-\frac{\cos \theta}{L})(1-\frac{\sin \theta}{L}) = 1- \frac{4L-1}{\pi L^2}$$

What about higher dimensions?

$\endgroup$
4
  • $\begingroup$ Did you try looking at $\prod_i \left(1-\frac{\mid x\mid}{L} \right)$ as $\sum_i \log\left(1-\frac{\mid x\mid}{L} \right)$? $\endgroup$
    – Guilherme Thompson
    Commented Nov 28, 2016 at 17:07
  • $\begingroup$ And, on the surface of the n-hypersphere, one could say that $\mid x\mid = r$, no? $\endgroup$
    – Guilherme Thompson
    Commented Nov 28, 2016 at 17:10
  • $\begingroup$ No I didn't, but the expectation over the log cannot be directly linked to the original expectation because the log is non-linear transform, right? $\endgroup$
    – etal
    Commented Nov 28, 2016 at 17:11
  • $\begingroup$ No, it's |x_i|, not |x|. |x_i| is the absolute value of the i-th coordinate, which can be any value from 0 to 1 in the surface of the unit sphere. $\endgroup$
    – etal
    Commented Nov 28, 2016 at 18:25

0

Reset to default

You must log in to answer this question.