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Math Help - Expectation of a geometric random varable

  1. #1
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    Expectation of a geometric random varable

    Let X be a geometric random variable with parameter p. Find E[min(X, 3)]

    My work:

    min(X, 3) = X if X < 3 and 3 if X ≥ 3 so

    E[min(X, 3)] = 0* P(X = 0) + 1* P(X = 1) + 2*P(X = 2) + 3* P(X ≥ 3)

    = 1*p + 2*p*(1-p) + 3* (1-p)^2

    as
    P(X ≥ 3) = P(X > 2) = (1-p)^2

    Is my answer of
    = 1*p + 2*p*(1-p) + 3* (1-p)^2 correct?

    Last edited by Capital; April 19th 2014 at 12:42 PM.
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  2. #2
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    Re: Expectation of a geometric random varable

    Quote Originally Posted by Capital View Post
    Let X be a geometric random variable with parameter p. Find E[min(X, 3)]
    min(X, 3) = X if X < 3 and 3 if X ≥ 3 so

    E[min(X, 3)] = 0* P(X = 0) + 1* P(X = 1) + 2*P(X = 2) + 3* P(X ≥ 3)

    = 1*p + 2*p*(1-p) + 3* (1-p)^2

    as
    P(X ≥ 3) = P(X > 2) = (1-p)^2

    Is my answer of
    = 1*p + 2*p*(1-p) + 3* (1-p)^2 correct?
    That is correct.
    Thanks from Capital
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