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Math Help - [SOLVED] Expectation - geometric series?

  1. #1
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    [SOLVED] Expectation - geometric series?

    hi everyone, can anyone help me on what looks like a simple problem?
    i dont need you guys to do the whole problem, only hint or tell me how to begin : )

    im working with E(e^t*N) = (e^-lambda)*[(sum of n = 0:infinity)*((lambda*e^t)^n))/n!]

    this is equal to exp(lambda(e^t - 1))

    how do you equal this to exp(lambda(e^t - 1)) ? geometric progression?
    i am trying geometric progression formula = a/1-r
    but getting nowhere : (

    thank you all. sorry i dont have word or latex atm so i couldnt type the whole formula in maths form
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  2. #2
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    Use the series for e and let x={\lambda}e^{t}

    e=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}

    Therefore, e^{{\lambda}e^{t}}\cdot e^{-{\lambda}}=e^{{\lambda}(e^{t}-1)}
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  3. #3
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    Quote Originally Posted by galactus View Post
    Use the series for e and let x={\lambda}e^{t}

    e=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}

    Therefore, e^{{\lambda}e^{t}}\cdot e^{-{\lambda}}=e^{{\lambda}(e^{t}-1)}


    i follow you on this, almost what i dont understand is..
    where does the 'n!' go in ((lambda*e^t)^n)/n!? and about about the 'n' in (lambda*e^t)^n ? does it vanish?

    thank you : )

    sorry it's hard to follow and messy because i cant write up formulas here

    i thought you can use the forumla a/1-r for geometric progression having infinity











    Nvm. i got it
    !
    thanks all of you who read this thread
    Last edited by matlabnoob; June 2nd 2010 at 02:41 AM.
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