I understood the Log-likelihood using the link “log” for poisson, λ=exp(α+βx). But I can’t get the Log-likelihood in the case of “identity”, λ=α+βx. How do I get it?. The example is the following data. I’d like to get the l(θ) of identity about the data. Please give me some advice.
Log-likelihood using link=log for poisson, λ=exp(α+βx).
x=1,2,10
y=10,15,20
l(θ)=45α+240β−∑log(yi!)−exp(α+β)−exp(α+2β)−exp(α+10β)
With the Poisson parameter $\lambda_i=\alpha+\beta x_i$, $i=1, \dotsc,n$, for $Y_i \sim \mathcal{Pois}(\lambda_i)$, independently. The Poisson pmf is $e^{-\lambda} \frac{\lambda^y}{y!}$, to get the likelihood just substitute in there for $\lambda$ the expression $\alpha+\beta x_i$, and multiply, giving $$ L(\alpha,\beta) = \prod_{i=1}^n e^{-\alpha -\beta x_i} \frac{(\alpha+\beta x_i)^{y_i}}{y_i!} $$ Taking the logarithm gives you the loglikelihood.
