Integrating the Quasi-Score

From a Quasi-Score to a Function of the Mean

A quasi-score specifies how an estimating equation changes when the fitted mean changes. Integrating that score gives a function whose differences compare candidate means for the same observed response. This construction is useful because it summarizes the mean–variance specification without requiring a probability mass function or density for the response.

The integration variable is the mean, not the regression coefficient. For one observation $y$, suppose its variance is specified by $\operatorname{Var}(Y)=\phi V(\mu)$, where $\mu=E[Y]$, $V$ is a known variance function, and $\phi>0$ is a scale factor. The mean-scale quasi-score, with the factor $1/\phi$ omitted, is $(y-\mu)/V(\mu)$. We seek a function of $\mu$ whose derivative is exactly this expression.

Definition: Let the possible means belong to an interval $I$, and let $V:I\to(0,\infty)$ be continuous. For an observed response $y\in I$, define the mean-scale quasi-likelihood, normalized to be zero at $\mu=y$, by $$ Q(y;\mu)=\int_y^\mu \frac{y-t}{V(t)}\,dt,\qquad \mu\in I. $$ The integral is taken with its usual orientation: reversing its limits changes its sign. If an endpoint is not in $I$, the definition can be extended there only when the corresponding improper integral converges.

The word “likelihood” in this name does not mean that $Q$ is the logarithm of a response probability. It is a function constructed from the specified mean–variance relationship. The reference value $Q(y;y)=0$ fixes an otherwise arbitrary additive constant for each observed $y$.

Basic Properties and the Regression Score

The first result confirms that integration recovers the desired score and identifies the best-fitting mean for a single observation. The conditions in the definition ensure that the integrand is continuous between $y$ and $\mu$, so the Fundamental Theorem of Calculus applies.

Theorem: If $V$ is positive and continuous on $I$, then $Q(y;\mu)$ is differentiable for interior $\mu\in I$, with the corresponding one-sided derivative at any included endpoint given by $$ \frac{\partial Q(y;\mu)}{\partial \mu}=\frac{y-\mu}{V(\mu)}. $$ Furthermore, $Q(y;\mu)\leq 0$, with equality if and only if $\mu=y$.

Proof. By the Fundamental Theorem of Calculus, differentiating the integral with respect to its upper limit gives $\partial Q(y;\mu)/\partial\mu=(y-\mu)/V(\mu)$. Since $V(\mu)>0$, this derivative is positive when $\mu<y$ and negative when $\mu>y$. Thus $Q$ increases up to $\mu=y$ and decreases after $\mu=y$. As $Q(y;y)=0$, its unique maximum is zero at $y$, and it is strictly negative at every other mean. $\square$

In a regression model, write $\eta_i=x_i^{\mathsf T}\beta$ and let $\mu_i=h(\eta_i)$ be the mean, where $h$ is differentiable. For observations $y_1,\ldots,y_n$, define the sum of the individual mean-scale functions as $Q(\beta)=\sum_{i=1}^n Q(y_i;\mu_i)$. Differentiation connects this function directly to the regression quasi-score.

Result: If each $\mu_i$ is differentiable in $\beta$, then $$ \frac{\partial Q(\beta)}{\partial\beta} =\sum_{i=1}^n \frac{y_i-\mu_i}{V(\mu_i)} \frac{d\mu_i}{d\eta_i}x_i. $$ If the variance is $\phi V(\mu_i)$ for a common positive $\phi$, then $Q(\beta)/\phi$ has derivative equal to this expression divided by $\phi$.

Derivation. For each observation, apply the chain rule to $Q(y_i;\mu_i)$. The derivative with respect to $\mu_i$ is $(y_i-\mu_i)/V(\mu_i)$, and the derivative of $\mu_i=h(x_i^{\mathsf T}\beta)$ with respect to the vector $\beta$ is $(d\mu_i/d\eta_i)x_i$. Multiplying these factors and summing over $i$ gives the displayed result. Multiplying the function by $1/\phi$ multiplies its derivative by the same factor. $\square$

This calculation explains the term “integrating the quasi-score”: the regression score follows by differentiating the summed function with respect to the regression parameters. It does not require independence or a joint distribution for the observations; it uses the specified mean and variance functions and the differentiability stated above.

Worked Forms for Common Variance Functions

The integral can be evaluated for familiar variance functions. These forms give relative values for candidate means when the observation $y$ is fixed. They depend on $V$, not on assigning a probability to $y$.

Worked Example: Poisson-Type Variance Function

Take $V(t)=t$, with $t>0$, and observed response $y=4$. Compare candidate means $\mu=6$ and $\mu=4$.

For $y>0$, integrating the definition gives

Q(y;\mu)=\int_y^\mu \frac{y-t}{t}\,dt =y\log\!\left(\frac{\mu}{y}\right)-(\mu-y).

At $\mu=6$, substitute $y=4$:

Q(4;6)=4\log(6/4)-(6-4) =4\log(1.5)-2 \approx 4(0.405465)-2 =1.621860-2=-0.378140.

At $\mu=4$, the defining integral has identical limits, so $Q(4;4)=0$. Thus the candidate mean 4 has the larger relative value for this observation. The variance function $V(\mu)=\mu$ is often called Poisson-type; using it here does not assert a Poisson response distribution.

Worked Example: Binomial-Type Variance Function

Let $V(t)=t(1-t)$ for $0<t<1$, and take an observed response proportion $y=0.4$ and candidate mean $\mu=0.6$. Partial fractions give

\frac{y-t}{t(1-t)} =\frac{y}{t}+\frac{y-1}{1-t}.

Integrating from $y$ to $\mu$ yields

Q(y;\mu) =y\log\!\left(\frac{\mu}{y}\right) +(1-y)\log\!\left(\frac{1-\mu}{1-y}\right).

For the stated values,

\begin{aligned} Q(0.4;0.6) &=0.4\log(0.6/0.4)+0.6\log(0.4/0.6)\\ &=0.4\log(1.5)+0.6\log(2/3)\\ &\approx 0.162186-0.243279\\ &=-0.081093. \end{aligned}

The negative value agrees with the theorem because $0.6\ne0.4$. The binomial-type variance function is defined on the open interval $(0,1)$. Values at $y=0$ or $y=1$, when needed, are obtained as limits if the relevant improper integral converges.

Worked Example: Constant Variance Function

For $V(t)=1$, the integral is

Q(y;\mu)=\int_y^\mu(y-t)\,dt =y(\mu-y)-\frac{\mu^2-y^2}{2} =-\frac{1}{2}(\mu-y)^2.

If $y=2$ and $\mu=3$, then $Q(2;3)=-\tfrac12(3-2)^2=-0.5$. At the observed mean $\mu=2$, $Q(2;2)=0$. Here the relative value decreases quadratically with the distance between the observed response and the candidate mean.

Worked Example: Gamma-Type Variance Function

For positive means, set $V(t)=t^2$, and take $y=2$, $\mu=3$. Then

\begin{aligned} Q(y;\mu) &=\int_y^\mu \left(\frac{y}{t^2}-\frac{1}{t}\right)\,dt\\ &=\left[-\frac{y}{t}-\log t\right]_y^\mu\\ &=\log(y/\mu)+1-y/\mu. \end{aligned}

Consequently, $$ Q(2;3)=\log(2/3)+1-2/3 \approx -0.405465+0.333333=-0.072132. $$ The result is negative, as it must be when $\mu\ne y$. This calculation uses the variance function $V(\mu)=\mu^2$; it does not specify a response density.

Why These Values Are Relative, Not Probabilities

For a fixed observation and a fixed variance function, differences in $Q(y;\mu)$ describe how the integrated mean-scale score changes between candidate means. For multiple observations, the summed function gives a corresponding criterion for comparing regression mean structures. The normalization $Q(y;y)=0$ is convenient, but it does not make the function a probability or ensure that exponentiating it gives a density or mass function.

A probability model must assign probabilities or densities that are nonnegative and normalized over the response space. The integral defining $Q$ imposes no such normalization. Distinct response models can share the same mean and variance function and therefore the same mean-scale quasi-likelihood, while assigning different probabilities to particular outcomes. The quasi-likelihood records the information supplied by the mean and variance specification, not a full response distribution.

A common extension is to include a dispersion factor by using $Q(y;\mu)/\phi$. This changes the scale of the function but not the basic integration: its derivative with respect to $\mu$ is $(y-\mu)/\{\phi V(\mu)\}$. The mean-scale function presented here keeps $\phi$ separate so that the role of the variance function is visible.

Common Mistake: Reversing the Integral or Treating It as a Log Probability

A frequent sign error is to integrate $(t-y)/V(t)$ from $y$ to $\mu$ while claiming to recover the quasi-score $(y-\mu)/V(\mu)$. Differentiating that incorrectly signed integral gives $(\mu-y)/V(\mu)$, the negative of the intended score. The correction is to use the integrand $(y-t)/V(t)$ with limits from $y$ to $\mu$, as in the definition.

Another mistake is to read $Q(y;\mu)$ as a log probability because it is called a quasi-likelihood. For example, the Poisson-type calculation $Q(4;6)=-0.378140$ is a relative function value, not the logarithm of the probability of observing 4. No probability for that response was assigned in the calculation. The score and integral are justified by the chosen variance function alone.

Key takeaway: Integrating the mean-scale quasi-score gives $Q(y;\mu)=\int_y^\mu (y-t)/V(t)\,dt$. For positive $V$, this function has its unique maximum at $\mu=y$. Its relative values encode the specified mean–variance relationship, not a response probability model.

Check Your Understanding

Use the definition and conditions above to answer these questions.

Differentiate $Q(y;\mu)$ with respect to $\mu$, and state which regularity condition permits this step.
For $V(t)=1$, derive the expression for $Q(y;\mu)$ and explain why it is nonpositive.
For $V(t)=t$, calculate $Q(2;4)$ and check its sign.
Explain why a quasi-likelihood value does not, by itself, give the probability of the observed response.
In a regression model with $\mu_i=h(x_i^{\mathsf T}\beta)$, identify the chain-rule factors that produce the regression quasi-score from the summed quasi-likelihood.

What You'll Learn