What Quasi-Likelihood Cannot Provide

What a Quasi-Likelihood Specifies

Quasi-likelihood methods can support useful estimation and inference without specifying a complete probability distribution for the response. That flexibility has a boundary: knowing a response’s mean and variance does not, by itself, tell us the probabilities or densities of its possible values. Recognizing this boundary is essential when deciding which calculations are justified by a quasi-likelihood analysis.

For a response $Y$, suppose its mean is $\mu=E[Y]$ and its variance is $\operatorname{Var}(Y)=\phi V(\mu)$, where $\phi>0$ is a dispersion parameter and $V$ is a specified positive variance function on the relevant mean domain. These are moment conditions. They do not state that $Y$ follows a particular distribution. In a regression model, the mean $\mu_i$ can depend on predictors and parameters, but the same distinction remains: specifying $\mu_i$ and $\phi V(\mu_i)$ is not the same as specifying the distribution of $Y_i$.

Definition: A quasi-likelihood function for a response value $y$ and mean $\mu$ is a function $Q(y;\mu)$ whose derivative with respect to $\mu$ is $$ \frac{\partial Q(y;\mu)}{\partial\mu} = \frac{y-\mu}{\phi V(\mu)}. $$ Here $\phi>0$, and $V(\mu)>0$ on an interval of allowed means. The derivative condition determines $Q$ only up to an additive term that may depend on $y$, but not on $\mu$. A quasi-likelihood function is not necessarily the log of a probability density or mass function.

The derivative in this definition gives the quasi-score for the mean. In a regression model, applying the chain rule to the mean model gives the corresponding quasi-score equations for the regression parameters. Those equations can support estimation, and the sandwich covariance approach can support robust uncertainty calculations under appropriate conditions. Neither operation supplies the missing distribution of the response.

Same Moments, Different Distributions

A direct construction shows why the first two moments do not identify a distribution. Consider two probability mass functions on $\{0,1,2,3\}$. The first assigns probability $1/4$ to each value. The second assigns probabilities $7/24, 3/24, 9/24,$ and $5/24$, respectively.

Worked Example: Matching the Mean and Variance

For the uniform distribution on $\{0,1,2,3\}$, the probabilities sum to $1$, and $$ E[Y]=\frac{0+1+2+3}{4}=\frac{3}{2}, \qquad E[Y^2]=\frac{0^2+1^2+2^2+3^2}{4}=\frac{14}{4}=\frac{7}{2}. $$ Thus $\operatorname{Var}(Y)=\frac{7}{2}-(\frac{3}{2})^2=\frac{5}{4}$.

For the second distribution, the probabilities sum to $(7+3+9+5)/24=1$, and $$ E[Y]=\frac{0(7)+1(3)+2(9)+3(5)}{24} =\frac{36}{24}=\frac{3}{2}, $$ $$ E[Y^2]=\frac{0^2(7)+1^2(3)+2^2(9)+3^2(5)}{24} =\frac{84}{24}=\frac{7}{2}. $$ Its variance is also $\frac{7}{2}-(\frac{3}{2})^2=\frac{5}{4}$.

But the probability of $Y\geq 2$ differs. Under the first distribution it is $1/4+1/4=1/2$; under the second it is $(9+5)/24=7/12$. The two distributions have identical means and variances, yet assign different probabilities to the same event.

This example proves a general point about what moment information can establish. A mean and variance function may be fully specified across the range of means, but they still do not generally select one probability law at any particular mean. To specify a law, one must add a distributional assumption or other information beyond those moments.

Result: A specified mean and variance do not, in general, determine a unique response distribution. Consequently, they do not determine all event probabilities or the full predictive distribution. The example above proves nonuniqueness even when the response has finite support.

A Quasi-Likelihood Is Not Automatically a Log-Likelihood

The derivative defining quasi-likelihood can be integrated, but integration does not turn the result into a probability model. For example, suppose the variance function is $V(\mu)=\mu$, the dispersion is $\phi=1$, and $\mu>0$. The derivative condition becomes $$ \frac{\partial Q(y;\mu)}{\partial\mu}=\frac{y}{\mu}-1. $$ One convenient choice of integration constant gives $Q(y;\mu)=y\log\mu-\mu+1$, with the convention $0\log\mu=0$ when $y=0$.

Worked Example: Comparing a Poisson Log-Likelihood and a Quasi-Likelihood

If a separate assumption states that $Y\sim\operatorname{Poisson}(\mu)$, its mass function is $P(Y=y)=e^{-\mu}\mu^y/y!$ for nonnegative integers $y$. Its log-likelihood contribution is $$ \ell(\mu)=y\log\mu-\mu-\log(y!). $$ For $V(\mu)=\mu$ and $\phi=1$, the quasi-likelihood choice above is $Q(y;\mu)=y\log\mu-\mu+1$. Subtracting gives $$ \ell(\mu)-Q(y;\mu) = [y\log\mu-\mu-\log(y!)]-[y\log\mu-\mu+1] =-1-\log(y!). $$ The difference is independent of $\mu$, but its sign is important: the Poisson log-likelihood minus this $Q$ is $-1-\log(y!)$. Equivalently, $Q$ minus the Poisson log-likelihood is $1+\log(y!)$.

This equality holds because a Poisson distribution was explicitly assumed for the comparison. The derivative condition defining $Q$ alone does not make that assumption. In particular, another distribution can have the same mean and variance as a Poisson response without having its mass function or log-likelihood.

More generally, if $V$ is positive and continuous on an interval of allowable means, a primitive can be written as $$ Q(y;\mu)=\int_{\mu_0}^{\mu}\frac{y-t}{\phi V(t)}\,dt+c(y), $$ where $\mu_0$ is a reference mean in that interval and $c(y)$ does not depend on $\mu$. The fundamental theorem of calculus gives $$ \frac{\partial Q(y;\mu)}{\partial\mu}=\frac{y-\mu}{\phi V(\mu)}. $$ If $Q_1$ and $Q_2$ both satisfy this derivative condition on the interval, then the derivative of $Q_1-Q_2$ is zero there. Hence their difference is constant as a function of $\mu$, though it may depend on $y$. This proves the stated freedom in the integration constant; it does not show that either primitive is the logarithm of a normalized probability model. $\square$

Exact Tails and Predictive Distributions Need More

An exact tail probability such as $P(Y\geq c)$ depends on how probability is distributed across the response values, not just on the first two moments. The same issue applies to prediction: a fitted mean and variance do not determine the probability that a future response falls in an interval, exceeds a threshold, or takes a particular value. Such calculations require a specified predictive distribution.

Worked Example: A Different Tail Probability With Poisson Moments

Let $Y\sim\operatorname{Poisson}(1)$. For an integer-valued response, modify its probabilities at $0,1,2,3$ by adding $0.01(1,-3,3,-1)$, respectively, and leave all probabilities at $k\geq4$ unchanged. The original probabilities at these four values are $e^{-1}, e^{-1}, e^{-1}/2,$ and $e^{-1}/6$. After modification they are approximately $0.3779, 0.3379, 0.2139,$ and $0.0513$, all nonnegative; all other probabilities remain nonnegative.

The changes sum to $0.01(1-3+3-1)=0$, so the modified probabilities still sum to $1$. The change in the mean is $0.01[0(1)-1(3)+2(3)-3(1)]=0$. The change in the second moment is $0.01[0^2(1)-1^2(3)+2^2(3)-3^2(1)]=0$. Thus the modified distribution still has mean $1$ and second moment $2$, hence variance $2-1^2=1$, the same as $\operatorname{Poisson}(1)$.

Nevertheless, $P(Y\geq3)$ changes. For the Poisson distribution, $$ P(Y\geq3)=1-P(Y=0)-P(Y=1)-P(Y=2) =1-\frac{1+1+\tfrac12}{e} \approx 0.0803. $$ The modification subtracts $0.01$ from the probability at $3$ and makes no changes at $k\geq4$, so the modified distribution has $P(Y\geq3)\approx0.0703$. Matching the Poisson mean and variance does not determine this tail probability.

The example also clarifies the scope of prediction from quasi-likelihood. A fitted mean and variance can provide moment-based summaries, and large-sample methods can sometimes justify approximate inference. But the quasi-likelihood specification alone cannot produce an exact individual-response prediction distribution or exact prediction probabilities. A normal approximation, for example, is an additional approximation with its own conditions; it is not implied as an exact law by knowing two moments.

What Remains Available—and What Does Not

The absence of a full probability model does not make quasi-likelihood analysis uninformative. Given an appropriate mean model and variance function, quasi-score equations can estimate regression parameters. With the required regularity conditions, sandwich covariance estimates and the associated large-sample methods can quantify uncertainty in those estimators and in functions of them. These procedures address estimating the mean structure; they do not reconstruct the response distribution.

The practical distinction is between inference about model parameters and calculations about the complete sampling law. A quasi-likelihood analysis may support an estimate of a regression coefficient, a confidence interval for a contrast, or an asymptotic test based on estimating equations. But the moment specification alone does not provide an ordinary likelihood function for the data. Without a specified likelihood, likelihood-based quantities and procedures cannot be calculated from the quasi-likelihood assumptions alone. Likewise, exact tail probabilities, quantiles, and predictive distributions are not identified.

Key takeaway: A quasi-likelihood encodes a mean–variance relationship through an estimating function. It does not, by itself, specify a response distribution. Before reporting an exact likelihood-based or predictive-distribution calculation, verify that a full probability model has been supplied.

Common Mistake

Wrong reasoning: “The variance function matches that of a familiar distribution, so the response must have that distribution. I can therefore use its likelihood and tail probabilities.”

Correction: Matching a mean and variance function does not establish the corresponding probability mass function or density. The worked examples give distinct distributions with the same first two moments but different event probabilities. Likelihoods and exact distributional calculations require an additional distributional assumption; they do not follow from the quasi-likelihood derivative alone.

Check Your Understanding

Answer each question using the distinction between moment information and a complete distribution.

What does the derivative defining a quasi-likelihood specify, and what freedom remains in the integrated function?
In the finite-support example, verify that both distributions have variance $5/4$. Why does their different probability of $Y\geq2$ matter?
For the Poisson comparison, what is the Poisson log-likelihood minus $Q(y;\mu)=y\log\mu-\mu+1$?
Can the mean and variance alone determine the exact probability $P(Y\geq c)$? Explain.
Name one kind of inference that quasi-likelihood can support and one distributional calculation that needs an additional model assumption.