# Why do we use quasi-ML methods when estimating a quasi-Poisson regression?

To make clear what I want to ask, I want to begin with the negative binomial distribution. The first two moments are $E(y)=\mu$ with variance $Var(y)=\mu+\mu^2/k$. When we solve the scorefunction of the log-likelihood with an iterationalgorithm, we get an estimator for $\widehat{\mu}$ (because of the estimator of $\widehat{\beta}$) and an estimator for the shape parameter $\widehat{k}$. So there are two parameters to be estimated.

When I look at the quasi-Poisson approach we have mean $E(y)=\mu$ and variance $Var(y)=\phi\cdot\mu$. This approach is not solvable since we do not know the probability function of $y$. Thus we use quasi-ML methods.
My question is: Why don't we create a distribution where we have the two moments of the quasi-Poisson regression and use a classic ML-estimation? It should be solvable with classic ML-methods since an estimation with negbin distribution (where we have also two estimators for $\mu$ and $k$ compared to $\mu$ and $\phi$).

• Yes this is clear. But what if there exists a distribution with a shape Parameter $\phi$ that has mean $E(y)=\mu$ and $Var(y)=\phi\cdot\mu$. In this case we should get rid of the QMLE estimation, no? – MarkDollar Jul 9 '11 at 11:33
• @Mark While there is actually a distribution in the exponential family with the variance property of the quasi-Poisson (a scaled Poisson) - it's not a distribution directly suitable for modelling count data (unless $\phi=1$ it's not defined where the counts are) and it's also not suitable for continuous data -- eliminating the two main things people use GLMs for. So people use quasi versions of count-distributions on count data not because they believe that the distribution is an exact description, but because the mean-variance behaviour is a reasonable description of what they're dealing with – Glen_b Sep 7 '16 at 5:20