Extended Quasi-Likelihood for Mean and Dispersion

Why Extend Quasi-Likelihood?

Quasi-likelihood uses a specified mean and variance relationship to construct an estimating equation for the mean parameters. In many applications the variance is written as $\operatorname{Var}(Y_i)=\phi V(\mu_i)$, where $V$ is a known variance function and $\phi>0$ is an unknown dispersion parameter. Extended quasi-likelihood (EQL) adds a term involving $\phi$ to the quasi-likelihood criterion, so that mean and dispersion can be considered together.

The extension is useful, but its interpretation needs care. Its mean equation is the familiar quasi-score equation. Its dispersion equation instead uses the quasi-deviance; in general, the resulting scale estimate is not guaranteed to estimate the variance multiplier consistently under every distribution having the stated mean and variance.

Definition: For an observation $y$, a mean $\mu$, and a variance function $V$ that is positive on the relevant range, define the unit quasi-likelihood contribution $$ Q(y;\mu)=\int_y^\mu \frac{y-t}{V(t)}\,dt. $$ For dispersion $\phi>0$, the extended quasi-likelihood contribution is $$ Q^+(y;\mu,\phi)=\frac{Q(y;\mu)}{\phi}-\frac{1}{2}\log\{2\pi\phi V(y)\}. $$ For independent observations, the extended quasi-likelihood criterion is the sum of these contributions.

This formula is used when $V(y)>0$, so the logarithm is defined. EQL is not necessarily the log-density of a specified probability distribution. For discrete or boundary-valued data, it is often a criterion built from the mean-variance specification, rather than an exact likelihood. In particular, the normalization term involving $V(y)$ should not be taken as proof that a full sampling distribution has been specified.

Estimating Equations

Suppose the mean depends on a parameter vector $\beta$ through $\mu_i=\mu_i(\beta)$, and let $\phi$ be common to the independent observations. Write $Q_i=Q(y_i;\mu_i)$. The key derivative follows directly from the fundamental theorem of calculus: the integrand in the definition of $Q$, evaluated at its upper endpoint, is $(y_i-\mu_i)/V(\mu_i)$.

\frac{\partial Q_i}{\partial \mu_i} =\frac{y_i-\mu_i}{V(\mu_i)}.

Because $V(y_i)$ does not depend on the fitted parameters, differentiating the sum of EQL contributions with respect to $\beta$ gives the mean estimating equation below. Here $\partial\mu_i/\partial\beta$ is a column vector.

U_\beta(\beta,\phi) =\sum_{i=1}^n \frac{y_i-\mu_i}{\phi V(\mu_i)} \frac{\partial\mu_i}{\partial\beta} =0.

Since $\phi$ is a common positive scalar, it does not change the roots of this equation. Thus estimating $\beta$ by EQL gives the same mean equation as quasi-likelihood. Differentiating with respect to $\phi$ gives a separate equation:

\frac{\partial}{\partial\phi}\sum_{i=1}^n Q^+(y_i;\mu_i,\phi) =-\frac{\sum_{i=1}^n Q_i}{\phi^2}-\frac{n}{2\phi}=0.

Solving this equation gives $\hat\phi=-2\sum_i Q_i/n$, evaluated at the fitted means. The unit quasi-deviance is $d_i=-2Q_i$, so the estimate is the average fitted unit quasi-deviance. This profile calculation requires $\sum_i Q_i<0$; if the sum is zero, there is no positive interior maximizer in $\phi$.

Result: If the fitted means are fixed while profiling over $\phi>0$, and $\sum_i Q_i<0$, the EQL criterion has a unique maximum in $\phi$, at $$ \hat\phi=-\frac{2}{n}\sum_{i=1}^n Q(y_i;\mu_i) =\frac{1}{n}\sum_{i=1}^n d_i. $$ The same formula applies after estimating the means, with each $\mu_i$ replaced by its fitted value.

To verify the maximum, let $S=\sum_i Q_i<0$. The terms depending on $\phi$ are $S/\phi-(n/2)\log\phi$. Their derivative vanishes at $\phi=-2S/n>0$, and their second derivative there is $-n/(2\phi^2)<0$. The derivative is positive for smaller positive $\phi$ and negative for larger $\phi$, so this is the unique maximum. The remaining terms do not involve $\phi$.

Worked Examples

Worked Example: Gaussian Data and the Residual Mean Square

Suppose $Y_i$ has mean $\mu_i$ and variance $\phi$, so $V(\mu)=1$. Consider three observations $4,5,7$ and fit an intercept, making each fitted mean the sample mean.

For $V(t)=1$, $$ Q(y;\mu)=\int_y^\mu(y-t)\,dt=-\frac{(y-\mu)^2}{2}. $$ The fitted mean is $\hat\mu=(4+5+7)/3=16/3$. The residual sum of squares is $$ (4-16/3)^2+(5-16/3)^2+(7-16/3)^2 =\frac{16+1+25}{9}=\frac{14}{3}. $$ Therefore $\sum_i Q_i=-(1/2)(14/3)=-7/3$, and $$ \hat\phi=-\frac{2}{3}\left(-\frac{7}{3}\right)=\frac{14}{9}. $$ This is the residual sum of squares divided by $n$. In the Gaussian case the EQL criterion is the exact normal log-likelihood, apart from the usual parameter-independent terms.

Worked Example: A Poisson Variance Function

Take $V(\mu)=\mu$, as for a Poisson mean-variance relationship, and suppose the three positive counts are $2,4,5$, with specified means $3,4,4$. For positive $y$ and $\mu$, $$ Q(y;\mu)=\int_y^\mu \frac{y-t}{t}\,dt =y\log(\mu/y)-(\mu-y). $$

For $y=2,\mu=3$, the contribution is $2\log(3/2)-1$, approximately $-0.18907$. For $y=4,\mu=4$, it is $0$. For $y=5,\mu=4$, it is $5\log(4/5)+1$, approximately $-0.11572$. Thus $$ \sum_i Q_i\approx-0.30479,\qquad \hat\phi=-\frac{2}{3}(-0.30479)\approx0.20319. $$ The fitted unit quasi-deviance total is approximately $0.60958$. This calculation illustrates the EQL dispersion equation; it does not establish that $0.20319$ estimates a true Poisson variance multiplier under an arbitrary count distribution. The formula also has a boundary limitation: at a zero count, $V(y)=0$, so the displayed EQL normalization term is not defined.

Worked Example: A Gamma-Type Variance Function

Let $V(\mu)=\mu^2$, and consider observations $y_1=2,y_2=3$ with means $\mu_1=1,\mu_2=2$. Integrating gives $$ Q(y;\mu)=\int_y^\mu\left(\frac{y}{t^2}-\frac{1}{t}\right)dt =\log(y/\mu)+1-y/\mu. $$ For the first observation, $Q(2;1)=\log 2-1\approx-0.30685$. For the second, $Q(3;2)=\log(3/2)-1/2\approx-0.09453$. Therefore $\sum_i Q_i\approx-0.40138$, and with $n=2$, $$ \hat\phi=-\frac{2}{2}(-0.40138)\approx0.40138. $$ The calculation shows how the chosen variance function changes the quasi-deviance and hence the EQL scale estimate. It does not require assuming a Gamma sampling density.

Large-Sample Interpretation

The mean equation has a direct estimating-equation interpretation. If $E[Y_i]=\mu_i(\beta_0)$, the mean model is correctly specified, and the relevant expectations exist, then at $\beta_0$, $$ E\left[ \frac{Y_i-\mu_i(\beta_0)}{\phi V(\mu_i(\beta_0))} \frac{\partial\mu_i(\beta_0)}{\partial\beta} \right]=0. $$ This unbiasedness is a key basis for large-sample inference from quasi-likelihood. It does not by itself guarantee consistency or asymptotic normality: those conclusions also require suitable identification, regularity, and information conditions.

The dispersion equation has a different target. For clarity, suppose there is a sequence of independent observations, the fitted means satisfy $\hat\beta\xrightarrow{p}\beta_0$, and the average quasi-likelihood contribution obeys a uniform law of large numbers near $\beta_0$. Suppose also that $$ \frac{1}{n}\sum_{i=1}^n E[Q(Y_i;\mu_i(\beta_0))]\longrightarrow q_0<0. $$ Continuity and the uniform convergence assumption then imply $$ \frac{1}{n}\sum_{i=1}^n Q(Y_i;\mu_i(\hat\beta))\xrightarrow{p}q_0. $$

Large-sample result: Under the conditions just stated, the EQL dispersion estimate satisfies $$ \hat\phi=-\frac{2}{n}\sum_{i=1}^n Q(Y_i;\mu_i(\hat\beta)) \xrightarrow{p}-2q_0. $$ Thus its probability limit is determined by the expected quasi-deviance per observation.

The conclusion follows by substituting the convergent average into the exact profile formula; multiplication by $-2$ preserves convergence in probability. The limit equals the true variance multiplier only when the expected quasi-deviance per observation has the corresponding value, namely $-2q_0=\phi$. This equality holds in the correctly specified Gaussian model with fixed means, since $E[(Y_i-\mu_i)^2]=\phi$ implies $E[Q_i]=-\phi/2$. It need not hold exactly for a general non-Gaussian distribution specified only through its first two moments.

Why the Distinction Matters

EQL offers a coherent way to use the same criterion for mean fitting and a deviance-based scale estimate. But the criterion's dispersion equation is not interchangeable with every other scale estimator. Pearson estimation, for example, uses squared residuals divided by the fitted variance function; EQL uses the quasi-deviance contributions. Their finite-sample values and large-sample targets can differ.

A common mistake is to reason that because $\operatorname{Var}(Y_i)=\phi V(\mu_i)$, any estimate obtained by differentiating a criterion containing $\phi$ must consistently estimate $\phi$. That conclusion does not follow. The EQL derivative yields $\hat\phi=n^{-1}\sum_i d_i$, so consistency depends on the expected quasi-deviance, not just on the variance equation. Treat EQL as a genuine likelihood only when a probability model justifying that likelihood is available; otherwise, state the estimating criterion and its assumptions explicitly.

Key takeaway: Extended quasi-likelihood leaves the mean quasi-score unchanged and gives a dispersion estimate equal to fitted quasi-deviance divided by $n$. Its large-sample scale target is determined by expected quasi-deviance and need not equal the variance multiplier without additional conditions.

Check Your Understanding

Try these questions before moving on.

Why does the EQL mean estimating equation have the same roots as the quasi-likelihood mean equation when $\phi$ is a common positive scalar?
Starting from $Q^+(y;\mu,\phi)$, derive the estimating equation for $\phi$ and state the condition for a positive interior solution.
For $V(\mu)=1$, show that the unit quasi-deviance is the squared residual.
Why does a Poisson variance function make the displayed EQL normalization problematic at a zero observation?
What quantity determines the large-sample limit of the EQL dispersion estimate, and why need not it equal the true variance multiplier?