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!

Find P(X+Y<1) in a different way

  1. Oct 17, 2012 #1
    1. Let the pdf of X,Y be [tex]f(x,y) = x^2 + \frac{xy}{3}, 0<x<1, 0<y<2[/tex]
    Find P(X+Y<1) two ways:

    a) P(X+Y<1) = P(X<1-Y)
    b) Let U = X + Y, V=X, and finding the joint distribution of (U,V), then the marginal distribution of U.

    3. The attempt at a solution
    a) P(X<1-Y) = ?
    [tex]P(x<1-y) = \int_0^1 \int_0^{1-y} (x^2 + \frac{xy}{3})dx = 7/72[/tex]
    b) I have no idea how to start.
     
    Last edited: Oct 17, 2012
  2. jcsd
  3. Oct 17, 2012 #2
    Resolved.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find P(X+Y<1) in a different way
  1. Determine P(|X-Y|<1) (Replies: 10)

Loading...