# Expected value

Discussion in 'Math' started by boks, Jan 10, 2009.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
Assume that two random variables (X,Y) are uniformly distributed on a circle with radius a. Then the joint probability density function is

$f(x,y) = \frac{1}{\pi a^2}, x^2 + y^2 <= a^2$,
$f(x,y) = 0, otherwise$

Find the expected value of X.

E(X) = $\int^{\infty}_{- \infty}\int^{\infty}_{- \infty}\frac{x}{\pi a^2} dxdy$

Is this correct so far? What are the limits of the integral supposed to be?

2. ### steveb Senior Member

Jul 3, 2008
2,433
469
No, it's not correct because the limits should not go to infinity, but only to the boundary of the circle. There are a couple of ways to go. You could do the integration in polar coordinates which would provide simple limits of integration (r=0 to a; and theta=0 to 2pi). Or, use rectangular coordinates and let the inner limits be functions for the circle. Outer limits would be -a and +a.

Of course, this problem is another one that can be solved instantly by noting the symmetry. Due to circular symmetry around the origin, the expected value of either x or y has to be zero.