# Expected value

#### boks

Joined Oct 10, 2008
218
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?

#### steveb

Joined Jul 3, 2008
2,431
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?
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.