PMF of Y for Exponential Random Variable

[magnifik]

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] = F_x(b) - F_x(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

 

[Ray Vickson]

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] = F_x(b) - F_x(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

 

[magnifik]

ok. thanks.

 

[magnifik]

so it would be e^-λm - e^-λ(m+1) which can be written as
e^-λm(1-e^-λ) .. would this be binomial or geometric?