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Let $X \sim Poiss(\lambda)$.

As, $\displaystyle \sum_{i=1}^{N} X_i $ is sufficient statistic for both mean (and variance) of $Y$, so we can define the unbiased estimate for mean as , $ s=\frac{1}{N} \sum_{i=1}^{N} X_i$, where all samples are i.i.d.

I am wondering how to find the estimator if parameter of interest is second moment i-e $E[X^2]=\lambda + \lambda^2?$

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  • $\begingroup$ Are $X_1,X_2,\ldots,X_n$ i.i.d? $\endgroup$
    – Janak
    Commented Jan 30, 2015 at 13:54
  • $\begingroup$ oh yes, thanks for correction. $\endgroup$
    – kaka
    Commented Jan 30, 2015 at 13:56
  • $\begingroup$ variance of $s$ is constant. $\endgroup$
    – kaka
    Commented Jan 30, 2015 at 14:16
  • $\begingroup$ Yes you right, I am still trying to solve it. $\endgroup$
    – Janak
    Commented Jan 30, 2015 at 14:18
  • $\begingroup$ please refer the question back, I have change the parameter of interest, I think that ll be more straight forward to think. $\endgroup$
    – kaka
    Commented Jan 30, 2015 at 14:23

1 Answer 1

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Let $(X_1, \dots, X_n)$ denote a random sample of size $n$ drawn from a population random variable $X$. By the 'fundamental expectation result', for any distribution whose moments exist, the $r^{th}$ sample raw moment $\acute{m}_r=\frac{1}{n} \sum _{i=1}^n X_i^r$ is an unbiased estimator of the $r^{th}$ population raw moment $E[X^r]$.

Thus, $\acute{m}_2 = \frac{1}{n} \sum _{i=1}^n X_i^2$ is an unbiased estimator of $E[X^2]$.

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    $\begingroup$ The 'fundamental expectation result' [ Stuart and Ord (1994), Kendall's Advanced Theory of Statistics, Section (12.5)] is a beautifully general way of finding unbiased estimators of complicated population moments. This is just a very simple case of its application. Of course, we can also show directly that $E[\acute{m}_r] = \frac{1}{n} \sum E[X^r] = E[X^r]$, but it's useful to have the more general approach for more difficult problems. $\endgroup$
    – wolfies
    Commented Jan 30, 2015 at 16:58

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