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PMF of a random varible

  • Thread starter magnifik
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  • #1
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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
 

Answers and Replies

  • #2
Ray Vickson
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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
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
 
  • #3
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ok. thanks.
 
  • #4
360
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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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