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Thread: 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?

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

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

    Here's what I have.

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

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

    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:

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

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

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

    substitute y = x-1

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

    $\displaystyle \displaystyle{= \lambda}$


    Note that $\displaystyle \displaystyle{\frac{e^{-\lambda}\lambda^{y}}{y!}}}$ 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?

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

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

    Here's what I have.

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

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

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


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

    CB
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