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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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    Sorry for the typo. I tried to use the geometric series. But it still ends up with a polynomial with high order.2011-12-13

2 Answers 2

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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!2013-02-06
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Notice that $\sum_{i} \lambda^i = (1+\lambda)^n - 1$. Use this and then calculate the derivative in respect to $\lambda$.

you will have $\lambda(1+\lambda)^{n-1} = \sum_i i x_i$, then you need to use some iterative method to solve the equation(Newton method and it converges quadratically).