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Math Help - Poission estimators problem

  1. #1
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    Poission estimators problem

    Hi all, I need to determine the variance of the estimator for Poission mean parameter \lambda

    Let  \theta = { 1 \over n} {\sum_{i=1}^{n} } X_i

    From our knowledge of Poission rv,
     E[X_i] = \lambda

     E[{\sum_{i=1}^{n} }X_i] = n{\lambda}

    similarly,
     E[{X_i}^2] = \lambda + \lambda^2

     E[ {\sum_{i=1}^{n}} {X_i}^2 ] = n{\lambda} + n{\lambda^2}

    To find the variance, we make use of the above results:

     V[ \theta] = {1 \over n^2} V[{\sum_{i=1}^{n}} {X_i}] = {1 \over n^2} (E[ {\sum_{i=1}^{n}} {X_i}^2] - E[{\sum_{i=1}^{n} }X_i]^2)
     = { 1 \over n}{\lambda} + {1 \over n}{\lambda^2} - {\lambda^2}


    but its supposed to equal  {1 \over n} \lambda which makes it an unbiased estimator. But the above equations don't arrive at this conclusion. Where am I wrong?
    Last edited by chopet; August 28th 2008 at 07:38 PM.
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  2. #2
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    Hi !
    Quote Originally Posted by chopet View Post
    Hi all, I need to determine the variance of the estimator for Poission mean parameter \lambda

    Let  \theta = { 1 \over n} {\sum_{i=1}^{n} } X_i

    From our knowledge of Poission rv,
     E[X_i] = \lambda

     E[{\sum_{i=1}^{n} }X_i] = n{\lambda}

    similarly,
     E[{X_i}^2] = \lambda + \lambda^2

     E[ {\sum_{i=1}^{n}} {X_i}^2 ] = n{\lambda} + n{\lambda^2}

    To find the variance, we make use of the above results:

     V[ \theta] = {1 \over n^2} V[{\sum_{i=1}^{n}} {X_i}] = {1 \over n^2} (E[ {\sum_{i=1}^{n}} {X_i}^2] - E[{\sum_{i=1}^{n} }X_i]^2)
     = { 1 \over n}{\lambda} + {1 \over n}{\lambda^2} - {\lambda^2}


    but its supposed to equal  {1 \over n} \lambda which makes it an unbiased estimator. But the above equations don't arrive at this conclusion. Where am I wrong?
    It think I've figured out your mistake.

    Imagine that Y=\sum_{i=1}^n X_i

    So V[\theta]=V \left[\frac 1n Y\right]=\frac 1{n^2} V [Y]=\frac 1{n^2} \left(E[Y^2]-(E[Y])^2\right)

    \left(E[Y^2]-(E[Y])^2\right)=\left(E \left[{\color{red}\left(\sum_{i=1}^n X_i\right)^2}\right]-\left(E \left[\sum_{i=1}^n X_i\right]\right)^2 \right)


    Do you see how the red part makes all the difference... ?
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  3. #3
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    Quote Originally Posted by chopet View Post
    Hi all, I need to determine the variance of the estimator for Poission mean parameter \lambda

    Let  \theta = { 1 \over n} {\sum_{i=1}^{n} } X_i

    From our knowledge of Poission rv,
     E[X_i] = \lambda

     E[{\sum_{i=1}^{n} }X_i] = n{\lambda}

    similarly,
     E[{X_i}^2] = \lambda + \lambda^2

     E[ {\sum_{i=1}^{n}} {X_i}^2 ] = n{\lambda} + n{\lambda^2}

    To find the variance, we make use of the above results:

     V[ \theta] = {1 \over n^2} V[{\sum_{i=1}^{n}} {X_i}] = {1 \over n^2} (E[ {\sum_{i=1}^{n}} {X_i}^2] - E[{\sum_{i=1}^{n} }X_i]^2)
     = { 1 \over n}{\lambda} + {1 \over n}{\lambda^2} - {\lambda^2}


    but its supposed to equal  {1 \over n} \lambda which makes it an unbiased estimator. But the above equations don't arrive at this conclusion. Where am I wrong?
    Why make such heavy weather of things ......

    If \theta = { 1 \over n} {\sum_{i=1}^{n} } X_i and the X_i are i.i.d. random variables with Var(X_i) = \sigma^2 then it's simple to prove that Var(\theta) = \frac{1}{n^2} \left( n \sigma^2 \right) = \frac{\sigma^2}{n}.

    If X_i is a Poisson random variable then you know that  Var(X_i) = \lambda.

    Therefore Var(\theta) = \frac{\lambda}{n}.
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