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Math Help - Moment-generating function of a continuous variable

  1. #1
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    Moment-generating function of a continuous variable

    I am having a tough time with these. For instance, take the following distribution: f(x) = 1, 0<x<1, and f(x) = 0 elsewhere. Now I was under the impression that I had to integrate (e^(tx) f(x) dx) over all values of x, which yields e^t / t. But the back of the book says that the moment generating function is (2e^t / (3 - e^t)). I have no clue how they got it. Is it simply a matter of integrating x from negative to positive infinity, instead of from 0 to 1? Or is there more to it than that?
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  2. #2
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    Re: Moment-generating function of a continuous variable

    Hey phys251.

    The MGF(t) = E[e^(tX)]Integral (over all X) e^(tx)*f(x)dx for any continuous random variable.

    This means that MGF(t) = Integral [0,1] e^(tx)*1*dx = (e^t)/t - 1/t = (e^t - 1)/t. Wikipedia confirms this:

    http://en.wikipedia.org/wiki/Uniform...ating_function
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  3. #3
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    Re: Moment-generating function of a continuous variable

    Then why the heck does the back of the book say it's (2e^t)/(3 - e^t)??
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  4. #4
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    Re: Moment-generating function of a continuous variable

    I don't know but I as well as Wikipedia got a different answer: if your book is wrong then its wrong.

    Try doing the calculation yourself and see what you get, and I recommend you plug the integral into Wolfram Alpha to get another confirmation.

    www,wolframalpha.com
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