What Quasi-Likelihood Does Not Specify

Moments Are Not a Full Probability Model

Quasi-likelihood is useful when a mean model and a variance function are credible but a complete distribution for the observations is not. It uses those specified features to construct estimating equations and, with appropriate conditions, to support large-sample inference. It does not fill in all the probabilities that a full probability model would require.

The distinction matters even for simple count data. Two distributions can have exactly the same mean and variance while assigning very different probabilities to extreme observations. As a result, quasi-likelihood calculations that depend only on the specified moments can agree for both distributions, even though exact tail probabilities and likelihoods do not.

For one observation with response \(y\), mean \(\mu\), and positive variance function \(V(\mu)\), a quasi-likelihood \(Q(y;\mu)\) is defined, up to a term not depending on \(\mu\), by

Thus, for a fixed reference value \(\mu_0\), its difference is

provided \(V(t)>0\) and the integrand is integrable between the two values. This construction specifies a function used for estimation; it does not turn that function into a normalized probability distribution. In particular, there is no general rule that the exponential of a quasi-likelihood, after normalization, must equal the probability of observing \(y\).

Worked Example: Two Count Distributions With the Same Moments

What the Quasi-Score Does Determine

Suppose \(Y_1,\ldots,Y_n\) are independent, \(E[Y_i]=\mu_i(\beta)\), and \(\operatorname{Var}(Y_i)=\phi V(\mu_i(\beta))\). Here \(\beta\) is a vector of parameters, and \(D_i=\partial\mu_i(\beta)/\partial\beta\) is a column vector. The quasi-score is

This expression involves the observations through their residuals and uses their assumed means and variances. Under the stated moment and independence conditions, its expectation and covariance can be calculated without specifying the full distribution.

To prove the result, write \(V_i=V(\mu_i(\beta))\). Linearity of expectation and the mean assumption give

For the covariance, independence makes all cross-covariances between distinct observations zero. The covariance of the \(i\)th score contribution is therefore

Adding these contributions gives the stated covariance. No step in this calculation uses probabilities for individual outcomes, such as \(P(Y_i=0)\) or \(P(Y_i\geq 5)\). The result therefore applies equally to different distributions that satisfy the same stated mean and variance assumptions. \(\square\)

When a quasi-score equation is solved for an estimator, its large-sample behavior can often be studied using this score covariance and regularity conditions for the estimating equation. Such inference is not a claim that the observations have a particular distribution. It is an approximation justified by the model for the moments and by the sampling conditions. For instance, independence matters: marginal means and variances alone do not determine the covariance of a sum when observations are dependent.

Worked Example: A Moment-Based Estimate and Its Standard Error

What Requires a Full Distribution

A full probability model supplies information beyond the first two moments. With such a model, one can calculate the likelihood \(L(\theta)\) for observed data and derive exact probabilities for specified events, assuming the model is correct. A quasi-likelihood specification alone does not identify those quantities. It cannot, for example, determine an exact finite-sample p-value for a test statistic whose null distribution depends on the tails of the observations.

The difference also applies to familiar likelihood calculations. If \(V(\mu)=\mu\), integrating the quasi-score definition gives \(Q(y;\mu)=y\log\mu-\mu\), up to a term independent of \(\mu\). This is the parameter-dependent part of a Poisson log-likelihood. That algebraic match does not prove that the observations are Poisson: a quasi-likelihood uses this expression because of the chosen mean-variance relationship. A different distribution can share a mean and variance at a specified mean without having the Poisson probabilities.

Common Mistake: Treating a Quasi-Likelihood as a Likelihood

A tempting but incorrect argument is: “The quasi-likelihood has the same algebraic form as a Poisson log-likelihood, so the data must be Poisson and Poisson tail probabilities apply.” The formula for \(Q\) follows from the quasi-score and variance function; it does not establish the probability of any observed outcome. As the examples show, different distributions can give the same moment-based estimating calculations but assign very different probabilities to data and tail events.

The correction is to match the inference to what has actually been specified. Use quasi-score methods for conclusions justified by the mean and variance assumptions and the relevant large-sample conditions. Use exact likelihoods, exact finite-sample tests, and model-specific tail probabilities only when a full probability model is supplied and its assumptions are defensible. A quasi-likelihood may be a useful working tool, but it does not silently add the missing distributional information.