1. Not finding help here? Sign up for a free 30min 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!

Conditional pmf

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data

    let [itex]f_{X,Y}(x,y)=2e^{-(x+y)} [/itex] for [itex] 0 \le x \le y[/itex] and [itex] y \ge 0 \\

    [/itex]find [itex]P(Y<1 | X < 1)
    [/itex]

    2. Relevant equations
    [itex]
    f(X=x | y=y) = \dfrac{f_{X,Y}(x,y)}{f_y(y)}
    [/itex]

    3. The attempt at a solution
    [itex]P(Y<1 | X<1) = \int_0^1 \dfrac{f_{X,Y}(x,y)}{\int_0^1 f_X(x) dx} dy [/itex]

    before I integrate, I want to make sure I understand the concept, which I don't think I do.
     
  2. jcsd
  3. Oct 26, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Use the formula
    [tex] P(A|B) = \frac{P(A\, \& \, B)}{P(B)}. [/tex]
    with appropriately-defined A and B.

    RGV
     
  4. Oct 26, 2012 #3
    I'm thinking that would be something like this:

    P(A) = P(Y<1)
    P(B) = P(X<1)=[itex]\int_0^1 f_X(x) dx [/itex]
    so then [itex]P(A \cap B) =P(Y<1 \cap X<1) = \int_0^1 \int_0^y f_{X,Y}(x,y) dxdy [/itex]

    and then [itex] \dfrac{\int_0^1 \int_0^y f_{X,Y}(x,y) dxdy}{\int_0^1 f_X(x) dx}[/itex]

    or am I still not understanding?
     
  5. Oct 27, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    This is OK. Now all you need do is find fX, and do the integrations in the numerator and the denominator.

    RGV
     
  6. Oct 27, 2012 #5
    Great. Thanks for the help again.
     
  7. Oct 29, 2012 #6
    I thought I understood it, but something is not correct. just working on the denominator, I get [itex]f_X(x)=-2(e^{-(x+y)}-e^{-x}) [/itex] and then if I try to integrate that I get [itex]\int_0^1 f_X(x) dx = 2e^{-y-1}+2e^{-y}+e^{-1}+1[/itex] What am I missing here?
     
  8. Oct 29, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You are missing the fact that f_X(x) cannot have y in it. You need to start again.

    RGV
     
  9. Oct 29, 2012 #8
    You're right. I see my mistake: [itex]f_X(x)=2e^{-x} [/itex] makes much more sense.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Conditional pmf
  1. Statistics pmf (Replies: 0)

  2. Finding a marginal pmf (Replies: 13)

  3. Joint PMF (Replies: 6)

  4. PMF Probability (Replies: 6)

Loading...