1
$\begingroup$

I consider data {$y_{ij}$} which is a rectangular table of counts, and that follows a Poisson distribution, $y_{ij}\sim Poi(\mu_{ij})$. The mean value, $\mu_{ij}$ depends on a variable $x_{ij}$, which for each pair $(i,j)$ is known. As a model I have the following: $$\mu_{ij}=\alpha_i\beta_jx_{ij}^\gamma,$$ and I want to derive the likelihood equation system. I'm not at all sure, but wonder if the log-likelihood is given by $$\sum_i\sum_j(y_{ij}\log \mu_{ij}-\mu_{ij}).$$ I more or less got this from the fact that I know that for a Poisson distributed random variable $y_i\sim Poi(\mu_i)$, the log-likelihood is given by $$\sum_i (y_{i}\log \mu_{i}-\mu_{i}).$$ Am I way off here, or am I thinking somewhat right?

$\endgroup$
1
  • $\begingroup$ I think you are right. $\endgroup$
    – user158565
    May 18 '17 at 14:01
1
$\begingroup$

If $X \sim Pois(\theta)$ then $f(x ; \theta) = \frac{\theta^x e^{-\theta}}{x!}$ so if you have $Y_{ij}$ independent then $$ \log f(\vec y ; \vec \mu) = \sum_{ij} \log f(y_{ij} ; \mu_{ij}) = \sum_{ij} (y_{ij}\log \mu_{ij} - \mu_{ij} - \log y_{ij}!). $$

If you are only interested in relative values of the log likelihood (such as when you optimize it) you can ignore the constant $-\sum_{ij} \log y_{ij}!$, but that is part of the log likelihood.

You don't say anywhere that you are making an independence assumption although your result suggests it; if you're not willing to do that, then this model is not correct.

Now if we want the likelihood in terms of our model $\mu_{ij} = \alpha_i \beta_j x_{ij}^\gamma$ we can just substitute that in.

$\endgroup$
1
  • $\begingroup$ Great, thanks. Are their any need for restrictions to achieve uniqueness of the parameterization? $\endgroup$
    – StatStud
    May 25 '17 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.