1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determine P(|X-Y|<1)

  1. Jul 25, 2009 #1
    1. The problem statement, all variables and given/known data

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

    Determine P(|X-Y|<1)

    2. Relevant equations



    3. 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:

    [tex]\int^0_{-1} \int^1_{Y+1} dx dy + \int^1_0\int^1_{Y+1} dx dy[/tex]
     
    Last edited: Jul 25, 2009
  2. jcsd
  3. Jul 25, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
    Last edited: Jul 25, 2009
  4. Jul 29, 2009 #3
    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 [tex]\int^0_{-1} \int^1_{Y+1} dx dy + \int^1_0\int^1_{Y-1} dx dy[/tex] ?
     
  5. Jul 29, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  6. Jul 29, 2009 #5
    http://img34.imageshack.us/img34/1728/75248295.jpg [Broken]

    so I need to integrate what's under the area between the 2 equations?
     
    Last edited by a moderator: May 4, 2017
  7. Jul 29, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You need to integrate what's between the two equations and inside of the square -1<=x<=1 and -1<=y<=1, yes.
     
  8. Jul 29, 2009 #7
    Let me give this another try

    [tex]\int^0_{-1} \int^{o}_{Y+1} dx dy + \int^1_0\int^1_{Y-1} dx dy[/tex]

    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?
     
    Last edited: Jul 30, 2009
  9. Jul 30, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  10. Jul 30, 2009 #9
    3 was the correct answer? Than the following integrals give me that answer

    [tex]\int^0_{-1} \int^{Y+1}_{-1} dx dy + \int^1_0\int^1_{Y-1} dx dy[/tex]

    cause when you integrate both double integrals, you get 3/2(each), like you said (unless I made a mistake somewhere)
     
  11. Jul 30, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  12. Jul 30, 2009 #11
    Thanks for all your help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Determine P(|X-Y|<1)
Loading...