P(|X-Y|<1): Find Probability of Difference b/w X and Y

[cse63146]

Homework Statement

Suppose X and Y are independant and uniformly distributed on the unit interval (-1,1).

Determine P(|X-Y|<1)

Homework Equations

The Attempt at a Solution

so X = Y + 1. I haven't done double integrals with absolute value, but I know for the single variable case, you need to find the roots of the equations. Is this right:

\int^0_{-1} \int^1_{Y+1} dx dy + \int^1_0\int^1_{Y+1} dx dy

 

[Dick]

The solution to |x-y|=1 is the two lines x-y=1 and -(x-y)=1. Sketch those two lines and figure out the region that intersects the square that defines your region of integration. You shouldn't even have to really integrate. You can use geometry to find the area.

 

[cse63146]

My prof wants to do this through integration.

so the lines are X - Y = 1 and -X + Y = 1 and re - writing in terms of X, I get X = 1 + Y and X = Y - 1

so \int^0_{-1} \int^1_{Y+1} dx dy + \int^1_0\int^1_{Y-1} dx dy ?

 

[Dick]

You CAN solve it by integration. But you still need to sketch a picture of the region you are integrating over. Why do you think the upper limit in x is always 1? My picture says otherwise.

 

[cse63146]

http://img34.imageshack.us/img34/1728/75248295.jpg

so I need to integrate what's under the area between the 2 equations?

 

[Dick]

You need to integrate what's between the two equations and inside of the square -1<=x<=1 and -1<=y<=1, yes.

 

[cse63146]

Let me give this another try

\int^0_{-1} \int^{o}_{Y+1} dx dy + \int^1_0\int^1_{Y-1} dx dy

There's another part of the question where it wnats me to find E[|X-Y|]. Would I just multiply the integral in the bottom right quadrant by y - x and the integral of the top left quadrant by x - y?

 

[Dick]

You still haven't got it. Take your picture and split it into triangles. The total area of the square is four. The area of the region between the curves is 3=3/2+3/2. That's the answer you are looking for. Now figure out why your integrals don't give you this. For E[|x-y|] you don't 'multiply' integrals by anything. You integrate |x-y| instead of 1. |x-y|=x-y if x>y and -(x-y) if y<x. You'll have to subdivide your region some more.

 

[cse63146]

3 was the correct answer? Than the following integrals give me that answer

\int^0_{-1} \int^{Y+1}_{-1} dx dy + \int^1_0\int^1_{Y-1} dx dy

cause when you integrate both double integrals, you get 3/2(each), like you said (unless I made a mistake somewhere)

 

[Dick]

That's right. You may have to solve these problems by showing an integration but that doesn't mean you can't use geometry to find the area and check your answer. Of course, you still have to figure out the probability from the area.

 

[cse63146]

Thanks for all your help.