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April 10th, 2017, 01:32 PM   #1
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From: london

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Maximum Likelihood estimation

Trying to find MLE for $\displaystyle \lambda $ of the truncated Poisson distribution tPoisson(λ, k) when k=1


$\displaystyle p_x = \frac{\lambda^x e^{-\lambda}}{q_kx!}$

where:

$\displaystyle q_k = 1- \sum_{i=0}^{k} \frac{\lambda^ie^{-\lambda}}{i!} $

$\displaystyle x = k+1, k+2, k+3......$
$\displaystyle \lambda > 0$
$\displaystyle i >= 2$

Log likelihood function:

$\displaystyle ln(L(\lambda) = ln [\prod_{}^{\infty}\frac{\lambda^x e^{-\lambda}}{q_kx!} ] $

$\displaystyle ln(L(\lambda) = ln (\lambda) \sum_{}^{\infty} x - n\lambda - nln(q_k) - \sum_{}^{\infty} x! $

when k=1, $\displaystyle q_k = - \lambda e^{-\lambda} $

Normally I would differentiate $\displaystyle ln(L(\lambda) $ with respect to $\displaystyle \lambda$ and equate to 0, to find solution for $\displaystyle \lambda=$, however $\displaystyle ln(q_k)$ results in an error as you cant take natural log of negative numbers, so im not sure how to continue
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April 12th, 2017, 09:37 AM   #2
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Please ignore this question, I went wrong further up the question stack, so dont need to do this.
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