Why Quasi-Poisson Is Not A Probability Likelihood

Why a Variance Adjustment Is Not a Count Distribution

For count data, a mean model alone does not say how likely each possible count is. Quasi-Poisson adds a variance relationship to the mean model, allowing variability beyond the Poisson variance. That can be useful for estimating mean-model coefficients and adjusting their uncertainty. But a mean and variance do not determine the probabilities of zero, one, two, or more events. This distinction matters especially in a generalized linear mixed model (GLMM), where a full likelihood would need a conditional response distribution as well as a distribution for the random effects.

The usual count support is the set of nonnegative integers, $0,1,2,\ldots$. Quasi-Poisson is used for responses on that support, but the support and two moments do not specify a probability mass function. Its name signals a mean-variance specification and associated estimating approach, not a uniquely defined probability distribution.

Definition: In a quasi-Poisson mean model, the response $Y_i$ has conditional mean $\mu_i=E(Y_i\mid\text{model information})$ and conditional variance $$ \operatorname{Var}(Y_i\mid\text{model information})=\phi\mu_i, $$ where $\phi>0$ is a dispersion parameter. A common log-link mean model is $\log(\mu_i)=\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}$. These statements specify moments and a mean structure, not a complete probability law for $Y_i$.

For a GLMM with random effect $b_j$, the conditional mean might instead be $\mu_{ij}(b_j)=\exp(\mathbf{x}_{ij}^\mathsf{T}\boldsymbol{\beta}+b_j)$, with conditional variance $\phi\mu_{ij}(b_j)$. One may specify a distribution for $b_j$, such as a normal distribution, but that still does not specify the conditional probability distribution of each count given $b_j$. A full likelihood needs both pieces.

What the Quasi-Score Uses

The estimating equation for the mean parameters uses residuals scaled by the mean-variance relationship. For a log-link model, its quasi-score has the form below (with a common dispersion parameter).

U(\boldsymbol{\beta}) =\sum_i \frac{(Y_i-\mu_i)}{\phi\mu_i} \frac{\partial\mu_i}{\partial\boldsymbol{\beta}} =\frac{1}{\phi}\sum_i\mathbf{x}_i(Y_i-\mu_i).

The second equality follows because $\partial\mu_i/\partial\boldsymbol{\beta}=\mu_i\mathbf{x}_i$ under the log link. The equation uses the observed residuals and the specified variance function; it does not require assigning a probability to every possible count. The factor $1/\phi$ does not change the root of this equation when $\phi$ is common to all observations, though estimating $\phi$ affects standard errors and tests.

Justification: A quasi-likelihood function $Q_i$ can be defined through $\partial Q_i/\partial\mu_i=(Y_i-\mu_i)/V(\mu_i)$, where $V(\mu_i)$ is the variance function before multiplying by $\phi$. For $V(\mu_i)=\mu_i$, integrating with respect to $\mu_i$ gives $Q_i=Y_i\log(\mu_i)-\mu_i$, up to terms independent of $\mu_i$. This resembles the parameter-dependent part of the Poisson log likelihood. Resemblance does not supply the omitted probability normalization or establish a Poisson response law; for general quasi-Poisson use, the specified moments do not produce a unique normalized mass function.

In a GLMM, a likelihood-based analysis ordinarily starts with a conditional mass function $f(y_{ij}\mid b_j)$ and a random-effect density $g(b_j)$. For a cluster of observations, its marginal likelihood contribution is obtained by integrating the conditional joint response probability over $b_j$. Quasi-Poisson moments alone do not provide $f$, so they cannot determine that integral. A quasi-likelihood or estimating-equation procedure can still be used, but its inferential basis is not a fully specified response likelihood.

Worked Example: Same Mean and Variance, Different Probabilities

Suppose a count has mean $2$ and variance $2$, the moments specified by a Poisson mean of $2$ with $\phi=1$. Show that these moments do not determine the probability of zero.

One distribution with these moments is Poisson with mean $2$. Its probability of zero is $P(Y=0)=e^{-2}\approx0.1353$.

A different distribution assigns probabilities $0.25, 0.50, 0.25$ to counts $0,2,4$, respectively. Its mean is \[ E(Y)=0(0.25)+2(0.50)+4(0.25)=2. \] Its second moment is $E(Y^2)=0+4(0.50)+16(0.25)=6$, so its variance is $6-2^2=2$. Yet its probability of zero is $0.25$, not $0.1353$.

Both distributions satisfy the same mean and variance. Therefore, the quasi-Poisson moment specification cannot select between them or give a unique probability for observing zero. A full probability model is needed for such a calculation.

Worked Calculations and the Mixed-Model Distinction

Worked Example: Estimating a Dispersion Multiplier

For illustration, suppose four hypothetical counts have fitted means $1,2,2,3$, and the observed counts are $0,2,3,5$. A mean model has $p=2$ estimated regression coefficients. Calculate the Pearson estimate of dispersion.

The Pearson statistic is the sum of squared residuals divided by their fitted means: \[ X_P^2=\sum_i\frac{(y_i-\hat{\mu}_i)^2}{\hat{\mu}_i}. \] Here the four contributions are $1,0,0.5,$ and $4/3$, so $X_P^2=2.833$. Using $n-p=4-2=2$ residual degrees of freedom gives \[ \hat{\phi}=\frac{X_P^2}{n-p}=\frac{2.833}{2}=1.417. \]

This suggests more residual variation than the Poisson variance would predict, relative to this fitted model. It is an imprecise estimate: with only two residual degrees of freedom, it should not be treated as a reliable description of a population distribution. In a mixed model, the effective estimation and uncertainty calculations also depend on the fitted random-effects structure and method; this small-sample formula is an illustration, not a universal GLMM rule.

Worked Example: Random Effects Do Not Fill In the Missing Response Law

Consider a hypothetical conditional mean $\mu(b)=2\exp(b)$, where $b\sim N(0,0.25)$, and suppose the conditional variance is specified as $\phi\mu(b)$ with $\phi=1.4$. Find the marginal mean and variance from these moment assumptions.

For a normal random effect with variance $0.25$, $E\{\exp(b)\}=\exp(0.25/2)=\exp(0.125)\approx1.133$. Also, $\operatorname{Var}\{\exp(b)\}=\exp(0.25)\{\exp(0.25)-1\}\approx0.3647$. Thus $E\{\mu(b)\}=2(1.133148)=2.2663$, and $\operatorname{Var}\{\mu(b)\}=4(0.3647)\approx1.4588$.

By the law of total variance, the marginal variance is $E\{\operatorname{Var}(Y\mid b)\}+\operatorname{Var}\{E(Y\mid b)\}$. Consequently, \[ \operatorname{Var}(Y)=1.4(2.2663)+1.4588\approx4.632. \]

These moments follow from the stated assumptions, but they do not give $P(Y=0)$, $P(Y=1)$, or any other count probability. Specifying the normal random-effect distribution supplies a law for $b$, not a conditional mass function for $Y\mid b$. A response likelihood still requires an additional full distributional assumption.

What Quasi-Poisson Can and Cannot Support

The dispersion multiplier adjusts the modeled variance relative to the mean. Under a log-link mean model, quasi-score estimates of the regression coefficients can be useful even when Poisson variability is too small. In common quasi-likelihood implementations, uncertainty is adjusted using an estimated dispersion; the precise procedure and finite-sample properties depend on the fitting method and design. These results concern estimation and inference for the mean structure, not the probability of an individual count.

Approach	What it specifies	What it can support
Poisson likelihood	A Poisson response mass function and a mean model	Likelihood-based estimation and probability calculations, subject to model adequacy
Negative-binomial likelihood	A specified count mass function with an additional dispersion structure	Likelihood-based inference and count probabilities under that distributional assumption
Quasi-Poisson	A mean model and variance proportional to the mean	Quasi-score estimation and dispersion-adjusted uncertainty under the method's assumptions; not a unique count probability law

Because quasi-Poisson does not provide a full likelihood, ordinary likelihood-based comparisons such as likelihood-ratio tests and AIC do not follow from it. Nor can a fitted quasi-Poisson model, by itself, produce a fully specified predictive distribution, a probability of a particular count, or a likelihood contribution for integrating over random effects. A prediction of the mean is possible; a prediction interval or tail probability requires more distributional information or a separately justified procedure.

Diagnostics, Alternatives, and Limitations

A dispersion estimate greater than one can flag a mismatch between observed residual variability and a Poisson variance assumption. It does not diagnose the cause. A single multiplier rescales the variance but does not address every way a count model can fail. Examine residual patterns against fitted means and predictors, check whether the mean structure is plausible, and look for systematic discrepancies such as too many zeros or an overly heavy upper tail. In a GLMM, also examine whether the random-effects structure and conditional dependence assumptions are defensible. Moment checks cannot substitute for checking a full probability model when probability predictions are required.

Quasi-Poisson is most useful when inference about the mean is the primary goal and a variance adjustment is appropriate for the design and estimating method. If probability calculations, likelihood-based comparison, or a full predictive distribution are central, a specified count likelihood is needed. Poisson and negative-binomial likelihoods are alternatives when their distributional assumptions are credible; neither should be selected solely because a dispersion statistic crosses a threshold.

A common mistake is to say, “The variance is $\phi\mu$, so the model gives the probability of every count.” The correction is that the variance and mean constrain probabilities but do not uniquely determine them. Another mistake is to use a quasi-Poisson fit as though it supplied a likelihood for AIC or random-effect integration. It supplies no such likelihood unless an additional response distribution is explicitly imposed. That added assumption changes the model and should be stated and assessed, rather than silently inferred from the quasi-Poisson variance formula.

Key takeaway: Quasi-Poisson describes a mean and a mean-linked variance, and supports quasi-likelihood estimation under suitable conditions. It is not a full probability distribution for counts and does not, on its own, provide likelihood-based probabilities or a GLMM response likelihood.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

If two count distributions have the same mean and variance, must they have the same zero probability? Why?
For a log-link quasi-Poisson mean model, what role does the dispersion multiplier play in the quasi-score?
Does specifying normally distributed random effects complete a quasi-Poisson GLMM likelihood? Explain.
A quasi-Poisson analysis estimates $\phi=1.6$. What does this indicate, and what does it not establish?
Why is ordinary AIC not supplied by a quasi-Poisson specification alone?

Solutions. (1) No. Mean and variance do not determine a unique count distribution; the worked example gives zero probabilities $0.1353$ and $0.25$ with the same moments. (2) A common factor $1/\phi$ scales the quasi-score and hence does not change its root, but dispersion affects uncertainty calculations. (3) No. The random-effects law specifies $b$, but a conditional response mass function for $Y\mid b$ is still needed. (4) It indicates residual variance above the Poisson mean-variance level relative to the fitted model and estimation method; it does not identify the cause or determine a count probability distribution. (5) AIC requires a specified likelihood, while quasi-Poisson moments alone do not define a normalized response likelihood.