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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 $\phi$).

I'm curious about your answers!

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You can't just "create a distribution" with a given form for the mean-variance relationship. In this case and others where quasi-likelihood is used, there is no probability distribution with the properties you desire.

That's why it's called quasi-likelihood:

"a quasi-likelihood function is not the log-likelihood corresponding to any actual probability distribution".

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  • $\begingroup$ 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? $\endgroup$ – MarkDollar Jul 9 '11 at 11:33
  • $\begingroup$ @Mark Is there a particular reason for using QMLE? $\endgroup$ – suncoolsu Jul 9 '11 at 11:55
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    $\begingroup$ @MarkDollar Yes, if such a distribution does exist you'd probably prefer full MLE these days. In the past, quasilikelihood may have had some computational advantages, but that's less likely to be an issue now. $\endgroup$ – onestop Jul 9 '11 at 12:03
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    $\begingroup$ @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 $\endgroup$ – Glen_b Sep 7 '16 at 5:20
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There are two ways here. The first way is to model the scaled Poisson data by the mixture of Normal distributions with parameters α1N(μ1,ϕ⋅μ1)+α2N(μ2,ϕ⋅μ2)+... You can easily write ML estimation for α, μ and ϕ. This approach works for relatively large mean values μ>10. The μ is small, you can use "scaled Poisson" distribution.

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