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I am learning about quasi-Poisson and i'm stuck at the concept of quasi-likelihood function.

In wikipedia, it is said that:

The term quasi-likelihood function was introduced by Robert Wedderburn in 1974 to describe a function that has similar properties to the log-likelihood function but is not the log-likelihood corresponding to any actual probability distribution

My question is that: if we estimate a variable $Y$ which depends on the coefficients $\beta$ by maximizing the quasi-likelihood function (for example, quasi-Poisson), is it valid ? I mean $Y$ must follow "some unknown" distribution, but the quasi-likelihood, instead, does not correspond to "any" distribution. So why using a function that correspond to "nothing" to estimate a random variable with a valid probability distribution ?

Thank you very much!

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Quasilikelihood (like pseudolikelihood, M-estimation, unbiased estimating functions, and composite likelihood) came from the realisation that many of the desirable properties of maximum likelihood estimators could be obtained without using the actual likelihood.

If you have an objective function $Q(\theta)$ that has its maximum at the right place (so $\theta_0$ maximises $E_{\theta_0}[Q(\theta)]$), plus some conditions on smoothness and variance, $\hat\theta$ will be consistent and asymptotically Normal by essentially the same arguments as for a maximum likelihood estimator. So, it's like an MLE except that you don't need to use the likelihood.

There isn't any requirement that the objective function you use isn't a likelihood -- it's just that when it is a likelihood you might not call it quasilikelihood, you might use some other name. Nowadays, I think the distinction between quasilikelihoods and other objective functions that aren't the likelihood is becoming less important

So why not just use the likelihood? Well, you might not know it. If you have, say, count data that are not a good fit to a Poisson distribution, you might want to have a straightforward analysis that doesn't require you finding a distribution that does have a good fit. A quasilikeliood estimator that captures the dependence of the mean $\mu$ on $X$ and the dependence of the variance on $\mu$ will give reasonable estimates and not require inventing a new parametric distribution and then programming it (this was the 1970s, remember)

Quasilikelihood estimators will typically not be fully efficient -- if you did know the likelihood you could do better -- but they may be simple enough or well-behaved enough that you are prepared to live with the efficiency loss. They also don't necessarily satisfy the information equality, so log quasilikelihood ratios don't necessarily have an asymptotic $\chi^2$ distribution (though they do have a known asymptotic distribution)

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  • $\begingroup$ Thank you so much for your answer. It is amazing. There is one point I want to make clear. So, given the dependent variables $Y$ and the independents variables $X_1, .., X_n$, what we want is to know the dependency of $Y$ on the $X$ (i.e. calculate the coefficients $\beta$) and we don't really care about what is the true distribution of $Y$, right ? I mean, we use the log likelihood when we have already the distribution of $Y$, information of the mean $\mu$ and relationship of mean and variance. But for the case of quasi-likelihood, we don't know any thing of the distribution of $Y$ $\endgroup$ Jul 20, 2021 at 5:07
  • $\begingroup$ We typically know something about the distribution of $Y$, otherwise we wouldn't know which quasilikelihoods make sense, but we either aren't sure about the exact distribution of $Y$ or we do know it but it's too complicated to evaluate $\endgroup$ Jul 20, 2021 at 5:57
  • $\begingroup$ Thank you very much for your help. I am considering, in the case we don't know exactly the distribution of $Y$, ok, then we use the quasi-likelihood. But if, we know, say, exactly the distribution of $Y$, can we use the quasi-likelihood instead of the usual log likelihood ? or in that case, it should be better to use the usual log likelihood ? Thank you! $\endgroup$ Jul 20, 2021 at 6:17

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