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Assume that $X_i \sim \text{Poisson}(\lambda^i)$, then we want to find the maximum likelihood estimate (MLE) of $\lambda$ and its asymptotics. I did in the following way, but got stuck here.

Since $\mathbb{P}(X_i=x_i)=e^{\lambda^i}\frac{\lambda^{i x_i}}{x_i!}$. Then the likelihood is $\mathcal L(\lambda;X)=\prod_i^n e^{\lambda^i}\frac{\lambda^{i x_i}}{x_i!}$. And the loglikelihood is $\ell(\lambda;X)=C-\sum_i^n \lambda^i+(\log\lambda)\sum_{i=1}^n ix_i$. By taking derivative w.r.t. $ \lambda$, I got $-\sum_i^n i\lambda^{i-1}+\frac{\sum_{i=1}^n ix_i}{\lambda}=0$, i.e. $\sum_i^n i \lambda^i=\sum_{i=1}^n ix_i$. But I don't know how to proceed to find the MLE of $\lambda$, then.

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Did you try summation of geometric series on $\sum_i \lambda^i$? –  Dilip Sarwate Dec 13 '11 at 2:11
    
Note that in taking the derivative of $\sum_{i=1}^n \lambda^i$, the index $i$ has no way of finding its way outside of the sum. :) –  cardinal Dec 13 '11 at 2:21
    
It's not $i \sum_i \lambda^i$, it's $\sum_i i \lambda^i$. –  Robert Israel Dec 13 '11 at 2:21
    
I've tried to make edits to the math including correcting two equations. Please make sure I haven't inadvertently added more typos. –  cardinal Dec 13 '11 at 2:24
    
Sorry for the typo. I tried to use the geometric series. But it still ends up with a polynomial with high order. –  Julie Dec 13 '11 at 2:29
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1 Answer

The left hand side of your equation, $$\sum_{i=1}^n i \lambda^i=\sum_{i=1}^n ix_i$$, is an arithmetico-geometric sequence, which provides the identity

$$\sum_{i=1}^n i \lambda^i = \frac{\lambda + n\lambda^n}{1-\lambda} + \frac{\lambda-\lambda^n}{(1-\lambda)^2} - \lambda$$.

This may be helpful for your problem.

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Thank you! I think in this case, we may not be able to get an formula for the MLE of $\lambda$. So I simply assume it will be the solution of the normal equation and then derive the asymptotic properties. I am surprised to get an answer for question asked one year ago! –  Julie Feb 6 '13 at 4:25
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