Dispersion in Quasi-Likelihood Models

Why Dispersion Has a Separate Role

A quasi-likelihood model specifies how the response mean changes with predictors and how the response variance depends on that mean. A scalar dispersion parameter controls the overall size of the variance relative to the mean–variance function. It therefore matters for uncertainty and information, but under a common scalar dispersion it does not change the quasi-score equations that determine the regression coefficients.

This distinction is useful in overdispersed count and proportion data. A Poisson or binomial mean model can still describe how the expected response depends on predictors even when the observed variation is larger than the reference variance. The quasi-likelihood formulation represents this by multiplying the variance function by \(\phi\), rather than by changing the coefficient equation itself.

Definition: For independent responses \(Y_1,\ldots,Y_n\), a scalar-dispersion quasi-likelihood model specifies \(E[Y_i]=\mu_i\) and \(\operatorname{Var}(Y_i)=\phi V(\mu_i)\), where \(\phi>0\) is common to all observations and \(V\) is a positive variance function on the relevant mean domain. The mean model is \(\eta_i=x_i^\mathsf{T}\beta=g(\mu_i)\), with differentiable link \(g\). The variance and mean specifications do not, by themselves, specify a full probability distribution.

Write \(D_i=d\mu_i/d\eta_i\), evaluated at the model mean. In regression, the quasi-score for \(\beta\) is

U_\beta(\beta;\phi) =\sum_{i=1}^n \frac{Y_i-\mu_i}{\phi V(\mu_i)}D_i x_i =\frac{1}{\phi}U_{\beta,0}(\beta),

where \(U_{\beta,0}\) denotes the same equation with the factor \(1/\phi\) removed. Thus the dispersion parameter appears as a common positive multiplier on the entire coefficient score. The implications for the coefficient estimates and their uncertainty follow from this structure.

Quasi-Likelihood and the Coefficient Equation

For a scalar response with a mean in an interval where \(V\) is positive, define its scaled quasi-likelihood contribution by integrating the quasi-score with respect to the mean:

q_\phi(y,\mu) =\frac{1}{\phi}\int_y^\mu \frac{y-t}{V(t)}\,dt.

The integral is a quasi-likelihood construction, not a log probability. Differentiating with respect to its upper endpoint gives \(\partial q_\phi(y,\mu)/\partial\mu=(y-\mu)/\{\phi V(\mu)\}\). Applying the chain rule with \(\mu_i=\mu_i(\beta)\) yields the regression quasi-score above. In particular, for a differentiable log link \(D_i=\mu_i\), while for a logit link \(D_i=\mu_i(1-\mu_i)\).

Result: If \(\phi\) is a common positive scalar, the solutions of \(U_\beta(\beta;\phi)=0\) are exactly the solutions of \(U_{\beta,0}(\beta)=0\). Therefore, whenever the coefficient equation has a unique solution, that coefficient estimate does not depend on the value assigned to \(\phi\).

Proof. Since \(\phi>0\), multiplication or division by \(\phi\) does not change whether a vector is zero. Hence \(U_\beta(\beta;\phi)=0\) if and only if \(U_{\beta,0}(\beta)=0\), at every \(\beta\). The two equations have identical solution sets; if that set contains a unique value, the estimate is the same for every positive \(\phi\). \(\square\)

This result concerns a common scalar dispersion multiplying every observation’s variance. It does not say that all changes to a variance model leave coefficient estimates unchanged. If dispersion varies by observation, it changes the relative weights in the score and can change its root. Nor does the result ensure a unique finite solution: existence and uniqueness depend on the mean model and data.

Worked Example: An Intercept-Only Quasi-Poisson Model

Suppose three independent counts are \(1,2,6\), with a common mean \(\mu=e^\beta\) and variance \(\phi\mu\). Thus \(V(\mu)=\mu\), \(D_i=\mu\), and the intercept-only quasi-score is

U_\beta(\beta;\phi) =\sum_{i=1}^3\frac{Y_i-\mu}{\phi\mu}\mu =\frac{1}{\phi}\{(1-\mu)+(2-\mu)+(6-\mu)\} =\frac{9-3\mu}{\phi}.

Setting this equal to zero gives \(9-3\mu=0\), so \(\hat\mu=3\) and \(\hat\beta=\log 3\). The fitted coefficient is the same for every \(\phi>0\).

For this intercept, the quasi-information derived below is \(3\mu/\phi\). Evaluated at \(\hat\mu=3\), it is \(9/\phi\), so its inverse, the model-based local variance for \(\hat\beta\), is \(\phi/9\). The coefficient equation fixes the fitted mean; the dispersion controls the scale of its uncertainty.

How Dispersion Scales Information

For independent observations, define the sensitivity of the quasi-score as the negative expected derivative of the score with respect to \(\beta\), and define its covariance as the variance of the score. These two matrices describe the local response of the estimating equation and the score’s variability. Under the stated variance model they are equal, a fact that can be shown directly without assuming a particular response distribution.

Theorem: Suppose the responses are independent, \(E[Y_i]=\mu_i(\beta)\), and \(\operatorname{Var}(Y_i)=\phi V(\mu_i)\), with the mean model differentiable and \(V(\mu_i)>0\). Assume differentiation and expectation can be interchanged for the score. Then

A(\beta) =-E\!\left[\frac{\partial U_\beta}{\partial\beta^\mathsf{T}}\right] =\frac{1}{\phi}\sum_{i=1}^n \frac{D_i^2}{V(\mu_i)}x_ix_i^\mathsf{T}, \qquad B(\beta)=\operatorname{Var}(U_\beta)=A(\beta).

Proof. The \(i\)th score contribution is \(u_i=(Y_i-\mu_i)D_i x_i/\{\phi V(\mu_i)\}\). Because \(E[Y_i-\mu_i]=0\), \(E[u_i]=0\). Independence therefore gives \(\operatorname{Var}(U_\beta)=\sum_i\operatorname{Var}(u_i)\). Using the specified response variance,

\operatorname{Var}(u_i) =\frac{D_i^2x_ix_i^\mathsf{T}}{\phi^2V(\mu_i)^2} \operatorname{Var}(Y_i) =\frac{D_i^2}{\phi V(\mu_i)}x_ix_i^\mathsf{T}.

Summing gives the displayed expression for \(B(\beta)\). For the sensitivity, differentiate \(u_i\) with respect to \(\beta^\mathsf{T}\). The derivative includes terms from differentiating \(\mu_i\), \(D_i\), and \(V(\mu_i)\). Every term other than the derivative of \(Y_i-\mu_i\) is multiplied by \(Y_i-\mu_i\), so its expectation is zero. Since \(\partial\mu_i/\partial\beta^\mathsf{T}=D_i x_i^\mathsf{T}\), the remaining expected derivative is

E\!\left[\frac{\partial u_i}{\partial\beta^\mathsf{T}}\right] =-\frac{D_i^2}{\phi V(\mu_i)}x_ix_i^\mathsf{T}.

Summing and changing sign gives the stated \(A(\beta)\), equal to \(B(\beta)\). \(\square\)

The matrix is scaled by \(1/\phi\): larger dispersion means less quasi-score information. If the design has full column rank and the weights \(D_i^2/V(\mu_i)\) are positive, \(A\) is nonsingular. In that case its inverse is the usual local, model-based covariance approximation for a coefficient estimator, and is proportional to \(\phi\). This is a quasi-score information statement; it does not turn the quasi-likelihood into a probability likelihood or establish a full distribution for the estimator.

Worked Example: Quasi-Binomial Intercept With Different Dispersion

Two independent groups have 10 and 20 trials, with 6 and 10 successes, respectively. Model each group’s success proportion with a common probability \(p\), and use the quasi-binomial variance \(\phi p(1-p)/m_i\) for the group proportion, where \(m_i\) is its number of trials. Equivalently, for the success count the variance is \(\phi m_i p(1-p)\).

Using the count formulation and the logit intercept \(\beta=\log\{p/(1-p)\}\), the score is

U_\beta =\frac{1}{\phi}\sum_{i=1}^2 \frac{Y_i-m_ip}{m_ip(1-p)}\,m_ip(1-p) =\frac{1}{\phi}\{(6-10p)+(10-20p)\} =\frac{16-30p}{\phi}.

Setting the score to zero gives \(\hat p=16/30=8/15\), and hence \(\hat\beta=\log\{(8/15)/(7/15)\}=\log(8/7)\), regardless of \(\phi\). For the intercept, \(D=p(1-p)\); the information is

A(\hat\beta) =\frac{1}{\phi}\sum_{i=1}^2m_i\hat p(1-\hat p) =\frac{30}{\phi}\frac{8}{15}\frac{7}{15} =\frac{112}{15\phi}.

Thus the inverse-information variance is \(15\phi/112\). Changing dispersion leaves the fitted proportion and log-odds unchanged, but changes the model-based uncertainty.

Worked Example: A Two-Group Quasi-Poisson Regression

Consider four independent counts, with two observations at \(x=0\), having values 1 and 3, and two at \(x=1\), having values 2 and 6. Let \(\log(\mu_i)=\beta_0+\beta_1x_i\) and \(\operatorname{Var}(Y_i)=\phi\mu_i\). For the log link, \(D_i=\mu_i\); therefore the score equations are proportional to the groupwise residual sums. At \(x=0\), the fitted mean must be \((1+3)/2=2\), and at \(x=1\), it must be \((2+6)/2=4\). Thus

\hat\beta_0=\log 2,\qquad \hat\beta_1=\log 4-\log 2=\log 2.

For each observation, the information weight is \(D_i^2/\{\phi V(\mu_i)\}=\mu_i/\phi\). The two observations in the first group contribute total weight \(4/\phi\); the two in the second contribute \(8/\phi\). With design rows \((1,0)\) and \((1,1)\), the information and its inverse are

A(\hat\beta) =\frac{1}{\phi} \begin{pmatrix} 12&8\\ 8&8 \end{pmatrix}, \qquad A(\hat\beta)^{-1} =\phi\begin{pmatrix} 1/4&-1/4\\ -1/4&3/8 \end{pmatrix}.

For example, the local model-based variance for the slope is \(3\phi/8\). The fitted group means, and therefore the coefficients, are unchanged by \(\phi\); the covariance scale is not.

Coefficient Estimation Is Not Dispersion Estimation

The coefficient estimate solves a mean-model equation. When the dispersion is common, it can be found without first choosing \(\phi\), because the common factor cancels from the equation. Dispersion estimation asks a different question: how large is the response variability relative to \(V(\mu)\)? The coefficient equation alone does not answer that question. Estimating dispersion requires information about observed variation as well as fitted means, and its procedures require additional choices and conditions.

The scaled quasi-likelihood \(q_\phi\) should not be mistaken for a full log-likelihood. A probability likelihood contains distribution-specific terms, including any normalization that depends on its parameters. A variance function and a mean do not supply those terms. Consequently, one cannot generally estimate \(\phi\) by treating the integrated quasi-likelihood as though it were a complete probability likelihood. The coefficient-score cancellation is a valid property of the estimating equation; it is not a general recipe for estimating dispersion.

Common Mistake

Wrong reasoning: “If \(\phi\) is unknown, the regression coefficients must also change, because \(\phi\) occurs in their score.”

Correction: The score is \(U_\beta=U_{\beta,0}/\phi\). For a common \(\phi>0\), setting it to zero is equivalent to setting \(U_{\beta,0}\) to zero, so the coefficient roots are unchanged. The information, however, is proportional to \(1/\phi\), and its inverse is proportional to \(\phi\). Thus dispersion affects uncertainty even when it does not affect the coefficient estimate.

Key takeaway: In a scalar-dispersion quasi-likelihood model, \(\phi\) multiplies the response variance, divides the coefficient quasi-score and information, and multiplies the model-based coefficient covariance. A common positive \(\phi\) leaves the coefficient-score roots unchanged; estimating \(\phi\) is a separate task.

Check Your Understanding

Use the definitions and results above to answer these questions.

Why does multiplying every coefficient score by the same positive dispersion parameter leave its root set unchanged?
In a quasi-Poisson model with a log link, what is \(D_i=d\mu_i/d\eta_i\), and how does \(\phi\) enter the score?
Under the theorem’s assumptions, how do the score covariance and sensitivity depend on \(\phi\)?
Why does an unchanged coefficient estimate not imply unchanged standard errors?
Give one reason that an integrated quasi-likelihood should not be treated automatically as a full probability likelihood for estimating dispersion.