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Find the constant that makes f(x,y) a PDF

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the value of k that makes this a probability density function.
    The question does not specify whether X and Y are independent or dependent. That is actually another part this question.

    2. Relevant equations
    Let X and Y have a joint density function given by
    f(x,y) =e^(-kX), 0<=X<inf, 0<=Y<=1,
    k is a constant, and 0 elsewhere

    3. The attempt at a solution
    I need quite a bit of direction with this question, I have no idea where to start.
     
  2. jcsd
  3. Oct 5, 2011 #2
    should I be posting this question in the statistics section of pf?
     
  4. Oct 5, 2011 #3

    Mark44

    Staff: Mentor

    This is the right place for this question.

    How can you tell whether a function is a "probability density function"?
     
  5. Oct 5, 2011 #4
    The integral of a joint PDF = 1:
    f(x,y) dxdy = 1
    Sorry I don't have a better response.
     
  6. Oct 5, 2011 #5

    Mark44

    Staff: Mentor

    That's the direction I was looking for.

    For a PDF, the integral should be 1, right?

    What does k need to be so that this integral is 1?
    [tex]\int_{y = 0}^1 \int_{x = 0}^{\infty} e^{-kx} dx~dy = 1 [/tex]
     
  7. Oct 5, 2011 #6
    I end up with k=1, still not sure if I'm right...
     
  8. Oct 6, 2011 #7
    got it, thanks.
     
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