Constructing Quasi-Likelihood From a Variance Function

From a Variance Function to a Quasi-Likelihood

A quasi-likelihood uses a specified relationship between a response's mean and variance without requiring a complete probability distribution for that response. Its construction is driven by a simple requirement: as a function of the mean, its derivative should equal the quasi-score contribution. Integrating that derivative produces a criterion that can be used to study and estimate the mean.

Let $Y$ be a response with $E[Y]=\mu$ and $\operatorname{Var}(Y)=\phi V(\mu)$, where $\phi>0$ is a fixed scale and $V$ is the variance function. The mean parameter $\mu$ ranges over an open interval $I$, and $V$ is continuous and strictly positive on $I$. These conditions make the reciprocal variance function well-defined and integrable over every compact subinterval of $I$.

Definition: Fix a reference value $\mu_*\in I$. A quasi-likelihood contribution $Q(y;\mu)$ for an observed response $y$ is a function of $\mu$ satisfying $$ \frac{\partial Q(y;\mu)}{\partial \mu} = \frac{y-\mu}{\phi V(\mu)}. $$ One convenient construction, up to an additive term that may depend on $y$ but not on $\mu$, is $$ Q(y;\mu) = \int_{\mu_*}^{\mu}\frac{y-t}{\phi V(t)}\,dt. $$

The reference point only sets the level of the contribution; it does not affect its derivative. In particular, if two differentiable functions of $\mu$ have the required derivative on the connected interval $I$, their difference is constant in $\mu$. The construction is therefore determined up to an observation-dependent additive constant. Such constants do not change the estimating equation.

For independent responses $Y_1,\ldots,Y_n$, with means $\mu_i$ and variance functions $\phi V(\mu_i)$, the total quasi-likelihood is the sum of the individual contributions. If the means depend on a parameter vector $\beta$, the chain rule gives $$ \frac{\partial}{\partial\beta} \sum_{i=1}^n Q(y_i;\mu_i(\beta)) = \sum_{i=1}^n \frac{y_i-\mu_i}{\phi V(\mu_i)} \frac{\partial\mu_i}{\partial\beta}. $$ Thus integrating the reciprocal variance function recovers the quasi-score estimating equation.

Worked Constructions

The examples below integrate the derivative with respect to $\mu$. Additive terms that do not depend on $\mu$ may be chosen for convenience; the score is unchanged.

Worked Example: Normal Mean-Variance Relationship

Suppose $Y$ has mean $\mu$ and variance $\phi$, so $V(\mu)=1$. Take $\phi=2$, an observed value $y=5$, and evaluate the contribution at $\mu=3$.

The required derivative is $$ \frac{\partial Q(y;\mu)}{\partial\mu}=\frac{y-\mu}{\phi}. $$ Integrating with respect to $\mu$ gives $$ Q(y;\mu)=\frac{y\mu}{\phi}-\frac{\mu^2}{2\phi}+c(y), $$ where $c(y)$ does not depend on $\mu$. Choosing $c(y)=-y^2/(2\phi)$ gives the familiar form $$ Q(y;\mu)=-\frac{(y-\mu)^2}{2\phi}. $$ At the specified values, $Q(5;3)=-(5-3)^2/4=-1$. Differentiating gives $(5-3)/2=1$, which agrees with $(y-\mu)/\phi$.

This contribution is proportional to the negative squared error. That resemblance to a normal log-likelihood follows from the variance function, but the construction itself has not required a normal probability model.

Worked Example: Poisson Mean-Variance Relationship

For a Poisson mean-variance relationship, $V(\mu)=\mu$ for $\mu>0$. Allow a fixed scale $\phi>0$, with the ordinary Poisson case obtained by setting $\phi=1$. For an observed count $y$, integrate the derivative $$ \frac{\partial Q(y;\mu)}{\partial\mu} = \frac{y-\mu}{\phi\mu} = \frac{1}{\phi}\left(\frac{y}{\mu}-1\right). $$

An antiderivative is $$ Q(y;\mu)=\frac{1}{\phi}\bigl(y\log\mu-\mu\bigr)+c(y). $$ When $y>0$, choosing $c(y)=-(y\log y-y)/\phi$ yields $$ Q(y;\mu)=\frac{1}{\phi}\left\{y\log\left(\frac{\mu}{y}\right)-(\mu-y)\right\}. $$ For $y=0$, the expression $y\log y$ is interpreted by its limit $0\log 0=0$; equivalently, use $Q(0;\mu)=(-\mu+c)/\phi$ for any constant $c$.

For example, with $\phi=1$, $y=4$, and $\mu=2$, $$ Q(4;2)=4\log(2/4)-(2-4)=2-4\log 2\approx -0.773. $$ Its derivative at $\mu=2$ is $4/2-1=1$, exactly the required score $(4-2)/2=1$.

Worked Example: Grouped-Binomial Mean-Variance Relationship

Let $Y$ be a binomial count from $m$ trials, with mean $\mu=mp$. Then $0<\mu<m$, and $$ \operatorname{Var}(Y)=mp(1-p)=\mu\left(1-\frac{\mu}{m}\right). $$ Thus $V(\mu)=\mu(1-\mu/m)$, with $\phi=1$ for the usual binomial variance.

The derivative to integrate is $$ \frac{y-\mu}{\mu(1-\mu/m)} = \frac{m(y-\mu)}{\mu(m-\mu)} = \frac{y}{\mu}+\frac{y-m}{m-\mu}. $$ The last equality follows by combining the two fractions: their numerator is $y(m-\mu)+(y-m)\mu=m(y-\mu)$. Integrating gives $$ Q(y;\mu)=y\log\mu+(m-y)\log(m-\mu)+c(y). $$ For $0<y<m$, choose $c(y)=-y\log y-(m-y)\log(m-y)$. Then $$ Q(y;\mu)=y\log\left(\frac{\mu}{y}\right) +(m-y)\log\left(\frac{m-\mu}{m-y}\right). $$

For $m=10$, $y=7$, and $\mu=5$, this gives $$ Q(7;5)=7\log(5/7)+3\log(5/3)\approx -0.823. $$ Differentiating the unnormalized expression gives $7/5-3/5=0.8$. The required score is also $(7-5)/[5(1-5/10)]=2/2.5=0.8$. At $y=0$ or $y=m$, the same formulas are interpreted by continuity, using $0\log 0=0$.

Two Properties of the Construction

The derivative condition is not merely a recipe for producing a formula. It identifies exactly which aspect of the contribution is fixed by the variance function, and it gives the expected contribution a useful shape around the true mean.

Theorem: Suppose $V$ is continuous and positive on a connected open interval $I$. Any two differentiable functions of $\mu$ satisfying $\partial Q(y;\mu)/\partial\mu=(y-\mu)/[\phi V(\mu)]$ differ by a term that is constant in $\mu$.

Proof. Let $Q_1$ and $Q_2$ satisfy the derivative condition. Their difference $D(\mu)=Q_1(y;\mu)-Q_2(y;\mu)$ has derivative $$ D'(\mu)=\frac{y-\mu}{\phi V(\mu)}-\frac{y-\mu}{\phi V(\mu)}=0 $$ for every $\mu\in I$. By the mean value theorem, for any two points $\mu_a,\mu_b\in I$, there is a point between them at which $$ D(\mu_b)-D(\mu_a)=D'(\xi)(\mu_b-\mu_a)=0. $$ Hence $D(\mu_b)=D(\mu_a)$ for all points in $I$, so the difference is constant in $\mu$. $\square$

Theorem: Let $Y$ have finite mean $E[Y]=\mu_0\in I$. For a fixed $\phi>0$ and positive continuous $V$, the expected quasi-likelihood contribution is strictly increasing for $\mu<\mu_0$ and strictly decreasing for $\mu>\mu_0$, wherever it is defined. Consequently, its unique maximum over $I$ occurs at $\mu_0$.

Proof. Use the reference-point construction $$ Q(Y;\mu)=\int_{\mu_*}^{\mu}\frac{Y-t}{\phi V(t)}\,dt. $$ On any compact subinterval of $I$, the functions $1/V(t)$ and $t/V(t)$ are bounded. The integral can be written as a term linear in $Y$ plus a term independent of $Y$, so its expectation is finite whenever $E[|Y|]<\infty$. Taking expectations and using linearity gives $$ E[Q(Y;\mu)] = \int_{\mu_*}^{\mu}\frac{\mu_0-t}{\phi V(t)}\,dt. $$ The integrand is continuous, so the fundamental theorem of calculus yields $$ \frac{d}{d\mu}E[Q(Y;\mu)] = \frac{\mu_0-\mu}{\phi V(\mu)}. $$ Since $\phi V(\mu)>0$, this derivative is positive when $\mu<\mu_0$ and negative when $\mu>\mu_0$. Therefore the expected contribution rises up to $\mu_0$ and falls after it, establishing the unique maximum. $\square$

This result concerns the expected contribution as a function of the mean. It does not assert that a quasi-likelihood is a probability density or mass function. Its value is that the variance function determines a criterion whose expected shape is centered on the true mean.

Using the Contribution in a Mean Model

For observations with modelled means $\mu_i(\beta)$, the total contribution is $Q(\beta)=\sum_i Q(y_i;\mu_i(\beta))$. Differentiating with respect to a component of $\beta$ multiplies each mean-scale score by the corresponding derivative of the mean. In vector notation, $$ \frac{\partial Q(\beta)}{\partial\beta} = \sum_{i=1}^n \frac{y_i-\mu_i(\beta)}{\phi V(\mu_i(\beta))} \frac{\partial\mu_i(\beta)}{\partial\beta}. $$ This is why the integration is useful: it packages the quasi-score into a scalar criterion while preserving the specified mean-variance relationship. It also makes clear that changing the variance function changes the relative weight assigned to residuals of different means.

Common Mistakes and Interpretation

Reversing the sign. Integrating $(\mu-y)/[\phi V(\mu)]$ gives the negative of the desired score. Check the derivative against $(y-\mu)/[\phi V(\mu)]$, rather than relying on how the resulting expression looks.
Dropping the scale inconsistently. If the target derivative includes $1/\phi$, the integral must include the same factor. Setting $\phi=1$ is valid only when that is the intended variance relationship.
Treating the result as a likelihood automatically. A quasi-likelihood contribution is constructed from a mean and variance specification. It need not be the log of any normalized probability model, so likelihood interpretations do not follow just from the integration.
Letting the integration constant change the score. A term $c(y)$ that depends only on the observation disappears on differentiation with respect to $\mu$. A term that also depends on $\mu$ is not an arbitrary integration constant and can alter the estimating equation.

The central check is always differentiation: after integrating, differentiate the proposed expression and confirm both the residual $y-\mu$ and the denominator $\phi V(\mu)$. That check catches sign errors, missing scale factors, and incorrect algebra.

Check Your Understanding

Use the derivative definition to answer the following.

If $V(\mu)=1$, what is a quasi-likelihood contribution, up to a term depending only on $y$?
For $V(\mu)=\mu$, differentiate $y\log\mu-\mu$ and identify the resulting score.
Why can an additive term depending only on $y$ be omitted when deriving the quasi-score?
For the grouped-binomial variance function, what is the allowed interval for $\mu$, and why must $V(\mu)$ be positive there?
State the sign of the derivative of the expected quasi-likelihood when $\mu$ is below the true mean $\mu_0$.