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Math Help - Exponentially Distributed RV - MGF

  1. #1
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    Exponentially Distributed RV - MGF

    I need help. The problem is this:

    X is an exponentially distributed random variable with p.d.f.

    f(x) = \lambda{e}^{-\lambda*x}
    0<x<infinity

    Show that the m.g.f. of X is {(1 - t/\lambda)} for {t<\lambda}, and use this to find E(X) and Var(X).


    I have no idea what to do with this problem or where to start. Please help point me somewhere? The only thing I could think of to potentially do is to undo the p.d.f. to get the original random variable, but...yeah. I'm really, really lost.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by notgod View Post
    I need help. The problem is this:

    X is an exponentially distributed random variable with p.d.f.

    f(x) = \lambda{e}^{-\lambda*x}
    0<x<infinity

    Show that the m.g.f. of X is {(1 - t/\lambda)} for {t<\lambda}, and use this to find E(X) and Var(X).


    I have no idea what to do with this problem or where to start. Please help point me somewhere? The only thing I could think of to potentially do is to undo the p.d.f. to get the original random variable, but...yeah. I'm really, really lost.

    just combine the exponential terms.

    \lambda\int_0^{\infty}e^{-x(\lambda-t)}dx

    You can now integrate BUT it's better to notice another exponential density

    ={\lambda\over \lambda-t} (\lambda-t) \int_0^{\infty}e^{-x(\lambda-t)}dx

    ={\lambda\over \lambda-t}

    since that is another valid density, just let \lambda^* = \lambda-t
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  3. #3
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    problem from my book

    Let X and Y be two Statistically independent Exponentially distributive random variables with parameters \lambda1 and \lambda2 respectively.

    a) find the CDF (commulative density function) of Z = X/X+Y , (use property of exponential pdf to evaluate the integrals in this question).

    b) find the expected value of Z when \lambda1=1, and \lambda2= 2


    So please can any one solve this, with all the steps of integration , Thanks a lot.
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  4. #4
    MHF Contributor matheagle's Avatar
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    I think Laurent solved this a couple of days ago here.
    Z is a beta.
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