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Math Help - Poisson Distributiom

  1. #1
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    Poisson Distributiom

    i just started stats uni this year so stats is completely new to me. can someone please help me with this question which i'm sure must be really easy, but i dont understand a thing!


    Suppose that X has a poisson distribution with mean lamda.
    a) find by summation the mean of X
    b) find also the variance of X



    If W is a poisson random variable with mean 2, what is P(W>3 given W>1) ?
    [for this question we have to use the tables right? and not the equation?)




    If x1 = 3 ; x2 = 2 ; x3 = 4 ; x4 = 2 ; x5 = 5, and all are equally likely values for X, what is E[X(X-1)] ?


    thanks so much...
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  2. #2
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    Quote Originally Posted by star_Asia View Post
    i just started stats uni this year so stats is completely new to me. can someone please help me with this question which i'm sure must be really easy, but i dont understand a thing!


    Suppose that X has a poisson distribution with mean lamda.
    a) find by summation the mean of X
    b) find also the variance of X
    If X is Poission it means it takes values for n=0,1,2,... with probability P(X=n) = \frac{\lambda^n e^{-\lambda}}{n!}.

    To find the mean you need to compute: E[X] = \sum_{n=0}^{\infty} n\frac{\lambda^n e^{-\lambda}}{n!} = \lambda \sum_{n=1}^{\infty}\frac{\lambda^{n-1} e^{-\lambda}}{(n-1)!}=\lambda \sum_{n=0}^{\infty}\frac{\lambda ^n e^{-\lambda}}{n!} = \lambda e^{\lambda} e^{-\lambda} = \lambda

    To find the variance you need to compute: \text{Var}(X) = E[X^2] - E[X]^2

    It is helpful to consider: E[X(X-1)] = \sum_{n=0}^{\infty}n(n-1) \frac{\lambda ^ne^{-\lambda}}{n!} = \lambda^2 \sum_{n=2}^{\infty} \frac{\lambda^{n-2}e^{-\lambda}}{(n-2)!} = \lambda^2 \sum_{n=0}^{\infty} \frac{\lambda^n e^{-\lambda}}{n!} = \lambda^2.

    This means that: \lambda^2 = E[X(X-1)] = E[X^2] - E[X] \implies E[X^2] = \lambda^2 + E[X] = \lambda^2 + \lambda.

    Thus: E[X^2] - E[X]^2 = \lambda^2 + \lambda - \lambda^2 = \lambda.
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  3. #3
    Senior Member tukeywilliams's Avatar
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    Or use moment generating function:  \phi(t) = \text{exp} \{\lambda(e^{t}-1) \} .

    Then  \phi '(t) = \lambda e^{t} \text{exp} \{\lambda(e^{t}-1) \} and  \phi ''(t) = (\lambda e^{t})^{2} \text{exp} \{\lambda(e^{t}-1) \} + \lambda e^{t} \text{exp} \{\lambda(e^{t}-1) \} .

    So  E[X] = \phi '(0) = \lambda and  E[X^2] = \phi ''(0) = \lambda^{2} + \lambda .

    And  \text{Var}(X) = E[X^2] - (E[X])^{2} = \lambda .

    You get the mgf as follows:  E[e^{tX}] = \sum_{n=0}^{\infty} \frac{e^{tn} e^{-\lambda} \lambda^{n}}{n!} which can be considered as summation.
    Last edited by tukeywilliams; January 1st 2008 at 09:43 PM.
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