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Math Help - Deriving method of moments

  1. #1
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    Deriving method of moments

    Let's hypothesize that I have n iid observations from a Poisson distribution with parameter lambda, how do I derive the method of moments estimator of lambda?

    X_1,...,X_n \sim Poisson(\lambda)

    \mbox{Derive the method of moments estimator of } \lambda

    Here's what I have.

    Pr\{X_i = k\} = \frac{\lambda^k}{k!}e^{-\lambda}

    \mu_1 = E[X] = \sum_{i=1}^{k} k \frac{ \lambda^k}{k!}e^{- \lambda} = e^{- \lambda} \sum_{i=1}^{k}k \frac{ \lambda^k}{k!}

    How do I simply this more?
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  2. #2
    MHF Contributor harish21's Avatar
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    For a Poisson Distribution:

    \displaystyle{E[X] = \sum_{x=0}^{\infty} x \frac{e^{-\lambda}\lambda^{x}}{x!}}

    \displaystyle{= \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^{x}}{(x-1)!}}

    \displaystyle{=\lambda  \sum_{x=1}^{\infty} \frac{e^{-\lambda}\lambda^{(x-1)}}{(x-1)!}}

    substitute y = x-1

    \displaystyle{=\lambda \sum_{y=0}^{\infty} \frac{e^{-\lambda}\lambda^{y}}{y!}}

    \displaystyle{= \lambda}


    Note that \displaystyle{\frac{e^{-\lambda}\lambda^{y}}{y!}}} is the pmf of a Poisson distribution, and the pmfs of all discrete distributions sum up to 1
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by financial View Post
    Let's hypothesize that I have n iid observations from a Poisson distribution with parameter lambda, how do I derive the method of moments estimator of lambda?

    X_1,...,X_n \sim Poisson(\lambda)

    \mbox{Derive the method of moments estimator of } \lambda

    Here's what I have.

    Pr\{X_i = k\} = \frac{\lambda^k}{k!}e^{-\lambda}

    \mu_1 = E[X] = \sum_{i=1}^{k} k \frac{ \lambda^k}{k!}e^{- \lambda} = e^{- \lambda} \sum_{i=1}^{k}k \frac{ \lambda^k}{k!}

    How do I simply this more?
    First show that $$ \lambda is the mean of a Poisson RV with parameter $$ \lambda


    Since the $$ Xs are iid each with mean $$ \lambda the sample mean of the $$ Xs has expectation equal to $$ \lambda

    CB
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