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Finding the PMF of a function of a discrete random variable

  1. Sep 17, 2014 #1
    The discrete random variable K has the following PMF:

    p(k) = { 1/6 if k=0
    2/6 if k=1
    3/6 if k=2
    0 otherwise
    }

    Let Y = 1/(1+K), find the PMF of Y


    My attempt:
    So, I am really confused about what this is asking.

    I took all of my possible K values {0, 1, 2} and plugged them in to the formula for Y to get:

    Y = {1,1/2,1/3}
    Then the only Y value that is also a K value is 1 so:

    p(y): {1 if y=1
    0 ptherwise
    }

    This does not look right to me.

    Am I approaching this the right way?
     
  2. jcsd
  3. Sep 17, 2014 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Your question is confusing (I don't know what PMF is supposed to be).

    The density function for y:
    p(y=1) = 1/6, p(y=1/2) = 1/3, p(y=1/3) = 1/2
     
  4. Sep 17, 2014 #3

    statdad

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    Homework Helper

    PMF = probability mass function, since this is a discrete random variable the term "density" isn't typically used.
    You don't want to base your probabilities for Y only on the ones for K. Think this way.
    k = 0 if and only if Y = 1/(1+0) = 1, so p(y=1) = (fill in the blank)
    K = 1 if and only if Y = 1/(1 + 1) = 1/2, so p(y = 1/2) = (again, fill in the blank)

    Keep going this way.
     
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