Joint Probability Density Function/Finding expected Value

The Joint Probability Density Function of X and Y is

f(x,y) = c(y^2 - 64x^2)e^-y, -y/8<= x <= y/8, 0<y<infinity

So the integral would look something like this?

$\displaystyle \int_{0}^{\inf} \int_{-y/8}^{y/8} (y^2-64x^2)e^{-y} dxdy $

Find c and the expected Value of X.

But I don't know how where to go from here or how to get the expected value

C is supposed to be 1 and E[X] should be 0.

Re: Joint Probability Density Function/Finding expected Value

Quote:

Originally Posted by

**bjnovak** The Joint Probability Density Function of X and Y is

f(x,y) = c(y^2 - 64x^2)e^-y, -y/8<= x <= y/8, 0<y<infinity

So the integral would look something like this?

$\displaystyle \int_{0}^{\infty} \int_{-y/8}^{y/8} (y^2-64x^2)e^{-y} dxdy $

Find c and the expected Value of X.

But I don't know how where to go from here or how to get the expected value

C is supposed to be 1 and E[X] should be 0.

If you have studied probability density functions carefully, the joint pdf should integrate to 1(by definition). so solve for "c" using the following (its simple integration), and then recall the definition of expected value to find E[X]..

$\displaystyle \int_{0}^{\infty} \int_{-y/8}^{y/8} c \cdot (y^2-64x^2) \cdot e^{-y} dxdy\;=\;1$