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Thread: mean and variance from mgf

  1. August 23rd 2009, 11:32 AM #1
    bigdoggy
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    mean and variance from mgf

    The mgf is given by M_x(t) = {pe^t \over 1 - (1-p)e^t}

    I have the mean as M_x'(0) = {1 \over p }\mbox{ and the variance as } M_x''(0) - (M_x'(0))^2 \mbox{ and the variance comes to }{1-p \over p^2}
    is this correct??
    Last edited by bigdoggy; August 24th 2009 at 02:47 AM.
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  2. August 23rd 2009, 12:48 PM #2
    Moo
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    Hello,

    the best way to spot your mistake is that you show your working
    then we'll see what's wrong

    for your information, this is the mgf of the geometric distribution
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  3. August 24th 2009, 03:05 AM #3
    bigdoggy
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    For the mean : M_X'(t)= {(1-(1-p)e^t)pe^t + pe^t(1-p)e^t \over (1-(1-p)e^t)^2}
    and so M_X'(0)= {1 \over p}
    and V[X]=M_x''(0)-{[M_x'(0)]}^2={1-p \over p^2}<br /> does that look correct?
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  4. August 24th 2009, 03:10 AM #4
    Moo
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    okay, you changed your result... now it is correct.

    and it's positive, because 1-p \geq 0 (p is a probability, so necessarily \leq 1)
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  5. August 24th 2009, 03:18 AM #5
    bigdoggy
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    Quote Originally Posted by Moo View Post
    okay, you changed your result... now it is correct.

    and it's positive, because 1-p \geq 0 (p is a probability, so necessarily \leq 1)
    Yeah, I spotted my mistake, but needed to check what I had is correct because I don't have the answer
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  6. August 24th 2009, 03:20 AM #6
    Moo
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    Quote Originally Posted by bigdoggy View Post
    Yeah, I spotted my mistake, but needed to check what I had is correct because I don't have the answer
    i told you it was the mgf of a geometric distribution
    the article in wikipedia gives the variance, you could've checked
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