Expected value and joint probability density function

[kasse]

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?

 

[Mark44]

Since your pdf is nonzero only on the circle, I think this is what you want:
\int^{a}_{-a}\int^{\sqrt{a^2 - y^2}}_{- \sqrt{a^2 - y^2}}\frac{x}{\pi a^2} dxdy

 

[rochfor1]

I think this integral would be easier in polar coordinates.

 

[HallsofIvy]

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 that is not correct because the limits of integration cannot be from -\infty to \infty. The probability outside the circle of radius a is 0, not 1/(\pi a^2).

In polar coordinates the integral would be
\frac{1}{\pi a^2}\int_r\int_\theta r cos(\theta) (r drd\theta)
and you want to cover the circle of radius a. What are the limits of integration for that?

 

[kasse]

Thanks!