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My question concerns using Maximum likelihood to estimate unknown parameters used by several (poisson) distributions.
The parameters are the pairs $(a_1,b_1),\dots,(a_N,b_N)$, and for each pair $(i,j), i\neq j$ we'll have an observation $k_{ij}$ of poissoin distribution with parameter $\lambda=a_i/b_j$.
So, I want to estimate the parameters, is it correct if I use the likelihood function? $$L=\prod_{(i,j),i\neq j} \frac{(a_i/b_j)^{k_{ij}} e^{-a_i/b_j}}{k_{ij}!}$$

If yes, and if I define $\sum 1/b_i = 1$ (since the all the parameters are redundant modulo multiplication) then I get $$a_i \sum_{j\neq i}k_{ij} = 1-1/b_i$$ $$b_j \sum_{i\neq j}k_{ij} = \sum_{i\neq j} a_i$$

Thank you in advance.

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  • $\begingroup$ I would be explicit about the fact that $L$ is a function of $((a_1,b_1),\ldots,(a_N,b_N))$ for $i=1,\ldots,N$ with the values of $k_{i,j}$ fixed. If the same expression is views as a function of $k_{i,j}$, $i,j=1,\ldots,N$ with the values of $((a_1,b_1),\ldots,(a_N,b_N))$ fixed, then it's a probability mass function is is a quite different function. $\endgroup$
    – Michael Hardy
    Commented Mar 8, 2015 at 0:42
  • $\begingroup$ This is correct. $\endgroup$
    – Michael Hardy
    Commented Mar 8, 2015 at 0:50
  • $\begingroup$ There is any explicit solution for the system of equations that I got? $\endgroup$
    – jpceia
    Commented Mar 8, 2015 at 2:47

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