Statistics - joint and marginal densities.

[peripatein]

Hi,

Homework Statement

I am having difficulties understand parts of the solution to the following problem in Statistics:
Let X and Y be two random, continuous variables, where X is distributed U(0,1) and Y|X=x is distributed U(x,x+1). I am asked to find the marginal density of Y.

Homework Equations

The Attempt at a Solution

For that I would first need to find the joint density. I know that the double integral of the joint density over the area defined by the lines y=x, y=x+1, x=1, x=0 (parallelogram) is 1. What I don't quite comprehend is why would the joint density then be 1/A where A is the area of the parallelogram in this case.
Next, suppose fX,Y(x,y) = 1, why do I need to separate fY(y) = ∫ (between -inf. and +inf.) dx into two integrals (one for 0 ≤ y ≤ 1 and one for 1 ≤ y ≤ 2)? Why couldn't I use one integral? And why are the integration boundaries for the second integral y-1 and 1? How could I have figured it out?

 

[lippyka]

The joint density of X and Y is 1/A because it is the probability density over the area A. The joint density is a normalized version of the probability density, so that the integral of the joint density over the given region is equal to 1. To find the marginal density of Y, you need to integrate the joint density over all possible values of X. Since X is distributed U(0,1), the integral should be taken from 0 to 1. This means that the integration boundaries for the second integral (for 1 ≤ y ≤ 2) will be y-1 and 1. Intuitively, this makes sense because when y is greater than 1, X can only take values between y-1 and 1, since X is bounded by 0 and 1.