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PMF of a random varible

  1. Nov 5, 2011 #1
    for a random variable X with parameter λ, Y = m if m < X < m + 1
    what is pmf of Y?

    it's basically asking for P[m< X < m+1]
    i know how to solve this for P[m < X < m + 1] ... it would be e-λm - e-λ(m+1) because
    P[a < X < b] = Fx(b) - Fx(a) and i know the PDF of an exponential random variable. however, in this problem the inequality does not match that so i'm wondering how to solve. i am unsure how to decompose P[m < X < m + 1]. i don't know of any property of the CDF that matches this
     
  2. jcsd
  3. Nov 6, 2011 #2

    Ray Vickson

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    It doesn't matter. For any random variable X that is (absolutely) continuous--that is, which has a density--- the probability of a single point = 0, so if a < b we have P{a <= X <= b} = P{a < X <= b} = P{a <= X < b} = P{a < X < b} = F(b) - F(a).

    RGV
     
  4. Nov 7, 2011 #3
    ok. thanks.
     
  5. Nov 7, 2011 #4
    so it would be e-λm - e-λ(m+1) which can be written as
    e-λm(1-e) .. would this be binomial or geometric?
     
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