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!