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Probability density function

  1. Dec 18, 2007 #1
    Find a constant c such that f(x,y)=cx2 + e-y, -1<x<1, y>0, is a proper probability density function.

    My idea:
    f(y)
    1
    =∫ f(x,y) dx
    -1

    So I have found f(y), now I set the following integral equal to 1 in order to solve for c:


    ∫ f(y) dy = 1
    0

    Integrating, I get something like (c)(∞)+...=1

    If this is the case, how can I solve for c? (can't divide something by infinity) Is this question even possible?

    Thanks!
     
    Last edited: Dec 18, 2007
  2. jcsd
  3. Dec 18, 2007 #2

    Dale

    Staff: Mentor

    Try c=0. It is the only number that could possibly have a finite limit when multiplied by infinity.
     
  4. Dec 18, 2007 #3

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    How? Can you describe?
     
  5. Dec 19, 2007 #4
    Performing the integartion, I get
    f(y)=
    1
    ∫ cx2 + e-y dx
    -1
    =2c/3 + 2e-y

    Setting

    ∫ f(y)dy=1
    0
    I get 2c/3 (∞) + 2 =1
     
  6. Dec 19, 2007 #5

    Dale

    Staff: Mentor

    Look at your 2c/3 term, it is a constant wrt y. Any constant, integrated from 0 to infinity, is infinite. It must be zero in order for the integral to converge.
     
  7. Dec 19, 2007 #6
    For 2c/3 (∞) + 2 =1
    Put c=0
    => 2=1 (if the first term is zero)
    So c=0 doesn't give a proper probability density function...
     
  8. Dec 19, 2007 #7

    EnumaElish

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    Science Advisor
    Homework Helper

    No such real c exists.
     
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