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April 10th, 2017, 01:32 PM  #1 
Senior Member Joined: Feb 2015 From: london Posts: 116 Thanks: 0  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 
April 12th, 2017, 09:37 AM  #2 
Senior Member Joined: Feb 2015 From: london Posts: 116 Thanks: 0 
Please ignore this question, I went wrong further up the question stack, so dont need to do this.


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estimation, likelihood, maximum 
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