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May 14th, 2017, 02:40 AM  #1 
Senior Member Joined: Feb 2015 From: london Posts: 121 Thanks: 0  Likelihood function
Example in book  For the zerotruncated Poisson distribution, the probability mass function is given by: $\displaystyle P(X = x) = p_x = \frac{λ^x}{ (e^λ − 1)x!} $ x = 1,2, ... Suppose that we have data in which we have observed x, $\displaystyle n_x$ times, x ≥ 1. Thus we have $\displaystyle \sum_{x=1}^{\infty} n_x$ observations in total. The log likelihood can be written as: $\displaystyle L(λ; n) = (\frac{λ^1}{(e^λ1)1!})^{n_1} (\frac{λ^2}{(e^λ1)2!})^{n_2} = \prod_{x=1}^{n} (\frac{λ^{x_i}}{ (e^λ − 1)x_i!})^{n_x}$ My Question:  If I look at other resources, it looks like the likelihood function is normally the product of the probability mass function, i.e: $\displaystyle L(λ; x) = \prod_{x=1}^{n} \frac{λ^{x_i}}{ (e^λ − 1)x_i!} $ for example: https://onlinecourses.science.psu.edu/stat504/node/31 Can anyone please explain why in the first example, they have added the power of $\displaystyle n_x$ in the formula 
May 14th, 2017, 09:59 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,237 Thanks: 637 
This is a mess. a) do you mean you've observed $X=k$ occur $n_k$ times (It's standard to use i,j,k,l,m,n for discrete values, and x, y, u,v,y,z etc. for continuous ones) If this is this case this problem is virtually identical to the previous one. b) your first formulation of $L(\lambda, x)$ has the variable $i$ in it. That can't be correct. I'm pretty sure this problem is just like the last one. 

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