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Let's say I have a variable $X_i$ distributed as $Poisson(\lambda)$ for $i=0,1,2,...,m$ Now, we assume that we can write $\lambda = n_ip$ such that $X_i$ is distributed as $Poisson(n_ip)$. We assume that the probability $p$ is constant for all $m$ years. Also $n_i$ is large and $p$ is small.

Now If I'm to find the maximum likelihood estimator of p, my question is if the following is correct.

$ \hat{\lambda} = n_ip = \frac{\sum_{i=1}^mX_i}{m} $

$ \hat{p} = \frac{\sum_{i=1}^mX_i}{n_im} $

And if this is correct, how do I find the expectation and variance of the estimator of p?

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I think that $\lambda$ should be replaced here by $\lambda_{i}$. (If not then $p$ by $p_{i}$, but you are talking about a constant $p$). Secondly it is wrong to use $i$ as index of a sum and loose at the same time as you did for instance in expression $\frac{\sum_{i=1}^{m}X_{i}}{n_{i}m}$ –  drhab Oct 24 '13 at 8:46
    
I thought there was something weird about that myself, but I don't have any precedent or similiar questions to draw on :( –  L1meta Oct 24 '13 at 10:18

1 Answer 1

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Assume that $X_1,\ldots,X_m$ are independent with $X_i\sim\mathrm{po}(pn_i)$ for some fixed positive constants $n_1,\ldots,n_m$ and an unknown $p>0$. Suppose $x_1,\ldots,x_m$ are realizations of $X_1,\ldots,X_m$. Then the likelihood function is $$ L(p)=\prod_{i=1}^m \frac{(p n_i)^{x_i}}{x_i!}\exp(-pn_i), $$ and hence the log-likehood function is $$ l(p)=\log L(p)=\sum_{i=1}^m\log(n_i^{x_i}/x_i!)+\sum_{i=1}^m x_i\log(p)-\sum_{i=1}^mpn_i. $$ The derivative with respect to $p$ is $$ l'(p)=\frac1p\sum_{i=1}^mx_i-\sum_{i=1}^m n_i $$ and hence $$ \hat{p}=\frac{\sum\limits_{i=1}^m x_i}{\sum\limits_{i=1}^m n_i}. $$ It should be pretty straightforward to find its expectation and variance.

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