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Expected Value of pmf

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  • Expected Value of pmf

    I am a college undergrad just starting my preparations for my P/1 test. The following question was in a book I am using to study:
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    Consider a random variable X whose probability mass function is given by
    p(x) = 0.2 , x = −1
    0.3 , x = 0
    0.1 , x = 0.2
    0.1 , x = 0.5
    0.3 , x = 4
    0 otherwise

    Find E(p(x)).

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    My method was to multipy all of the probabilities by the possible outcomes, .2*-1 + .3*0 + .1*.2 + .1*.5 + .3*4 and i got 1.07. The solution given was .24 and I also had arrived at a different answer in a similar question involving expected values of cdfs. I'm pretty sure there is something fundamental I am missing here. Someone please help! THANKS!

  • #2
    That's because you found E(X), E(p(X)) = sum(p(x)*p(x)) over all x. So in this case it's .2^2+.3^2+.1^2+.1^2+.3^2 = .24. Remember what expected value formula is.
    NoMoreExams
    Actuary.com - Posting Master
    Last edited by NoMoreExams; June 26 2009, 06:17 PM. Reason: Good point about capitalization

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    • #3
      Are we sure that the original problem is properly phrased? E(p(x))=? Please tell me what kind of random variable p(x) is. What is the sample space? How is it partitioned so that we could end up with a random variable p(x)? Thanks.

      Edit: Never mind. If the x in p(x) is capitalized to make p(X), then E(p(X)) would indeed make sense. E(p(x)) should just be a typo.
      winvicta
      Actuary.com - Level II Poster
      Last edited by winvicta; June 26 2009, 04:42 PM.
      P(7/09) FM(8/09)

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      • #4
        Thanks! I knew i was missing something...

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