# Joint Probability Density Function/Finding expected Value

• Dec 5th 2011, 06:04 AM
bjnovak
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?
$\int_{0}^{\inf} \int_{-y/8}^{y/8} (y^2-64x^2)e^{-y} dxdy$

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.
• Dec 5th 2011, 01:23 PM
harish21
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?
$\int_{0}^{\infty} \int_{-y/8}^{y/8} (y^2-64x^2)e^{-y} dxdy$

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]..

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