5
$\begingroup$

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!

$\endgroup$
0

3 Answers 3

5
$\begingroup$

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".

$\endgroup$
4
  • $\begingroup$ 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, 2011 at 11:33
  • $\begingroup$ @Mark Is there a particular reason for using QMLE? $\endgroup$
    – suncoolsu
    Jul 9, 2011 at 11:55
  • 1
    $\begingroup$ @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, 2011 at 12:03
  • 3
    $\begingroup$ @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, 2016 at 5:20
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

This is a great question. The issue with creating a distribution is that two moments are not sufficient to define a distribution. In fact, an infinite number of moments is required to perfectly define a single distribution. There is short discussion on this topic here on mathoverflow.

Simply put, there are an infinite number of distributions which have the desired first two moments. Therefore, which of the infinite number are appropriate to define for ML? There is no way to answer this question. Hence why Quasi-Likelihood methods are used instead.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.