My Math Forum Likelihood function

 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 zero-truncated 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: $\displaystyle L(λ; n) = (\frac{λ^1}{(e^λ-1)1!})^{n_1} (\frac{λ^2}{(e^λ-1)2!})^{n_2} = \prod_{x=1}^{n} (\frac{λ^{x_i}}{ (e^λ − 1)x_i!})^{n_x}$ 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: Southern California, USA Posts: 1,403 Thanks: 713 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. Thanks from 123qwerty

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