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Thread: 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 $\displaystyle \lambda$

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

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

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

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

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

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

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


    but its supposed to equal $\displaystyle {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; Aug 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 $\displaystyle \lambda$

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

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

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

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

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

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

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


    but its supposed to equal $\displaystyle {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 $\displaystyle Y=\sum_{i=1}^n X_i$

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

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


    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 $\displaystyle \lambda$

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

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

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

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

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

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

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


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

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

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