Quasi-Likelihood Under Variance-Function Misspecification

Why Variance Misspecification Need Not Move the Target

A quasi-score uses a variance function to weight residuals. If that function is wrong, observations may receive inefficient weights, and uncertainty calculations based on the working variance may be wrong. Yet the estimator can still target the correct mean parameter: at the true parameter, each conditional residual has expectation zero, whatever positive weights the working variance supplies. This conclusion requires a correctly specified conditional mean and suitable conditions ensuring that the estimating equation has a unique, stable root.

The distinction is important. Correctness of the mean model concerns the equation’s target; correctness of the variance model concerns, among other things, the equation’s sensitivity and the variability of its random part. The latter two determine the asymptotic covariance.

Quasi-Score and the Two Variance Functions

Let $Y_1,\ldots,Y_n$ be independent responses, with fixed covariates $x_i$, and suppose their conditional means are $\mu_i(\beta)=\mu(x_i;\beta)$, where $\beta\in\mathbb{R}^p$. Write $D_i(\beta)=\partial\mu_i(\beta)/\partial\beta$, a column vector. Assume the true conditional mean is $\mu_i(\beta_0)$. The working variance function $V_w(\mu)>0$ need not equal the true conditional variance. We use the quasi-score with any common dispersion multiplier omitted, since a common positive multiplier does not change its root.

Definition: The quasi-score contribution and estimating equation are $$ \psi_i(\beta)=\frac{D_i(\beta)\{Y_i-\mu_i(\beta)\}}{V_w(\mu_i(\beta))}, \qquad U_n(\beta)=\sum_{i=1}^n\psi_i(\beta). $$ A quasi-score estimator $\hat\beta$ is a solution of $U_n(\hat\beta)=0$. The working variance is the function used in this equation; the true conditional variance is $\sigma_i^2=\operatorname{Var}(Y_i\mid x_i)$.

At $\beta_0$, correct conditional mean specification gives $E[\psi_i(\beta_0)\mid x_i]=0$. This is the key reason the working variance can be wrong without shifting the target. It does not by itself guarantee consistency: the population estimating equation must have no other relevant root, and the sample equation must converge to it uniformly enough.

Probability Limit and Asymptotic Covariance

Here is a standard fixed-design version of the result. Assume $\beta_0$ is an interior point of the parameter space; $\mu_i(\beta)$ and $V_w$ are sufficiently smooth near $\beta_0$, with $V_w$ bounded away from zero there. Assume a uniform law of large numbers for $n^{-1}U_n(\beta)$ on a compact neighborhood, with continuous limit $g(\beta)$, where $g(\beta_0)=0$ and $\beta_0$ is its unique root in that neighborhood. Assume the estimator lies in that neighborhood with probability tending to one and the uniform convergence and root-identification conditions imply consistency. For asymptotic normality, also assume the derivative averages converge uniformly near $\beta_0$, the score contributions satisfy a central limit theorem, and the limiting matrices below exist and are nonsingular where required.

Define the sensitivity matrix $A$ and variability matrix $B$ by

A=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n \frac{D_i(\beta_0)D_i(\beta_0)^T}{V_w(\mu_i(\beta_0))}, \qquad B=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n \frac{\sigma_i^2D_i(\beta_0)D_i(\beta_0)^T} {V_w(\mu_i(\beta_0))^2}.

Suppose $A$ is positive definite, $B$ is finite, and $n^{-1/2}U_n(\beta_0)\xrightarrow{d}N(0,B)$. Then

Theorem: Under the stated consistency and regularity conditions, $$ \hat\beta\xrightarrow{p}\beta_0, \qquad \sqrt n(\hat\beta-\beta_0)\xrightarrow{d} N\!\left(0,A^{-1}BA^{-T}\right). $$ The matrix $A$ is determined by the working weights and the mean model; $B$ also contains the true conditional variances. Thus the covariance generally is not the covariance obtained by treating the working variance as true.

To see the form of $A$, differentiate $\psi_i(\beta)$. The derivative includes terms from differentiating $D_i(\beta)$ and the working weight, each multiplied by $Y_i-\mu_i(\beta)$, as well as the derivative of that residual. At $\beta_0$, conditional expectation makes the first terms vanish. Since $\partial\mu_i/\partial\beta=D_i$, the remaining conditional expected derivative is

E\!\left[\left.\frac{\partial\psi_i(\beta_0)}{\partial\beta^T}\right|x_i\right] =-\frac{D_i(\beta_0)D_i(\beta_0)^T}{V_w(\mu_i(\beta_0))}.

Independence and the zero conditional means also give $\operatorname{Var}\{\psi_i(\beta_0)\mid x_i\} =\sigma_i^2D_i(\beta_0)D_i(\beta_0)^T/V_w(\mu_i(\beta_0))^2$, which explains $B$.

Proof of the Limit

First, the uniform law of large numbers gives $n^{-1}U_n(\beta)\xrightarrow{p}g(\beta)$ uniformly on the specified neighborhood. The root-identification and consistency conditions ensure that a root of the sample equation converges in probability to the unique root $\beta_0$; hence $\hat\beta\xrightarrow{p}\beta_0$.

For the distributional result, use the integral form of Taylor’s theorem, which is valid for a vector estimating equation. It avoids assuming that one intermediate point represents the change in every component of a vector-valued function. Define

C_n=-\int_0^1\frac{1}{n}\sum_{i=1}^n \frac{\partial\psi_i\{\beta_0+t(\hat\beta-\beta_0)\}}{\partial\beta^T}\,dt.

The fundamental theorem of calculus, applied along the line segment from $\beta_0$ to $\hat\beta$, gives the exact identity

0=\frac{1}{\sqrt n}U_n(\beta_0)-C_n\sqrt n(\hat\beta-\beta_0).

Consistency and the assumed uniform derivative convergence imply $C_n\xrightarrow{p}A$. Since $A$ is nonsingular, $C_n$ is invertible with probability tending to one, and then $\sqrt n(\hat\beta-\beta_0)=C_n^{-1}n^{-1/2}U_n(\beta_0)$. The central limit theorem gives $n^{-1/2}U_n(\beta_0)\xrightarrow{d}N(0,B)$. Slutsky’s theorem therefore yields the stated limit with covariance $A^{-1}BA^{-T}$. $\square$

Worked Examples

Worked Example: A Common Mean With Two Variances

Suppose half the independent observations come from a group with mean $\mu_0=10$ and variance $4$, and half from a group with the same mean and variance $16$. Estimate the common mean using working variances $1$ and $4$, respectively. Here $D_i=1$, so the equation is the sum of weighted residuals.

Let $\bar Y_1,\bar Y_2$ be the sample means in the two groups, each of size $n/2$. The equation is $(n/2)(\bar Y_1-\hat\mu)+(n/8)(\bar Y_2-\hat\mu)=0$. Solving, $\hat\mu=(4\bar Y_1+\bar Y_2)/5$. As both group means converge to 10, $\hat\mu\xrightarrow{p}10$, despite the wrong working variances.

For the asymptotic variance, the per-observation sensitivity and variability limits are $A=\frac12(1)+\frac12(1/4)=5/8$ and $B=\frac12(4/1^2)+\frac12(16/4^2)=5/2$. Thus $\operatorname{Var}(\hat\mu)$ is asymptotically $A^{-1}BA^{-1}/n=(2.5)/(0.625^2n)=6.4/n$. For comparison, treating the working variances as actual variances would give $A^{-1}/n=1.6/n$. The target is correct, but that model-based variance is not.

Worked Example: A Linear Mean With Unequal Group Variances

Let $x_i$ be 0 or 1, with half the observations at each value, and suppose $E[Y_i\mid x_i]=\beta_0+\beta_1x_i$. The true conditional variances are 1 at $x=0$ and 9 at $x=1$, but use the constant working variance $V_w=1$. At the true parameter, $D_i=(1,x_i)^T$.

The sensitivity is $A=\frac12\begin{pmatrix}1&0\\0&0\end{pmatrix} +\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix} =\begin{pmatrix}1&1/2\\1/2&1/2\end{pmatrix}$. The variability is $B=\frac12\begin{pmatrix}1&0\\0&0\end{pmatrix} +\frac92\begin{pmatrix}1&1\\1&1\end{pmatrix} =\begin{pmatrix}5&9/2\\9/2&9/2\end{pmatrix}$.

The determinant of $A$ is $1/4$, and $A^{-1}=\begin{pmatrix}2&-2\\-2&4\end{pmatrix}$. Multiplication gives $A^{-1}BA^{-1}=\begin{pmatrix}2&-2\\-2&20\end{pmatrix}$. Therefore the asymptotic covariance of $\hat\beta$ is $n^{-1}\begin{pmatrix}2&-2\\-2&20\end{pmatrix}$. This agrees with the group-mean calculation: the intercept estimates the $x=0$ mean, while the slope is the difference of the two group means.

Worked Example: Poisson Log-Mean With a Constant Working Variance

Suppose $Y_i\mid x_i\sim\operatorname{Poisson}(\mu_i)$, with $\log\mu_i=\beta_0+\beta_1x_i$. Half the covariates are 0 and half are 1; at the true parameter the corresponding means are 1 and 2. Use the constant working variance $V_w=1$, although the true variances are 1 and 2. The derivative is $D_i=\mu_i(1,x_i)^T$.

The correct mean implies $E[\psi_i(\beta_0)\mid x_i]=0$, so the unique population root is $\beta_0$. At $x=0$, $D_i=(1,0)^T$; at $x=1$, $D_i=(2,2)^T$. Consequently, $A=\frac12\begin{pmatrix}1&0\\0&0\end{pmatrix} +\frac12\begin{pmatrix}4&4\\4&4\end{pmatrix} =\begin{pmatrix}5/2&2\\2&2\end{pmatrix}$.

For $B$, multiply each $D_iD_i^T$ by the true variance. Thus $B=\frac12\begin{pmatrix}1&0\\0&0\end{pmatrix} +\frac12(2)\begin{pmatrix}4&4\\4&4\end{pmatrix} =\begin{pmatrix}9/2&4\\4&4\end{pmatrix}$. Since $A^{-1}=\begin{pmatrix}2&-2\\-2&5/2\end{pmatrix}$, direct multiplication gives $A^{-1}BA^{-1}=\begin{pmatrix}2&-2\\-2&3\end{pmatrix}$. The estimator targets the correct log-mean coefficients, while the variance is determined by both the Poisson variability and the constant working weights.

Interpretation and a Common Pitfall

The working variance affects the estimator through its weights, and therefore affects both $A$ and $B$. In a correctly specified mean model, the expected score at the true parameter remains zero, but that fact alone does not establish a unique root, consistency, or efficiency. In particular, a misspecified variance can yield a valid target and an invalid conventional standard error at the same time.

Wrong reasoning: “The conditional mean is correct, so the working variance can be used as the true variance when computing standard errors.”
Correction: The mean condition establishes zero expected residuals, not the true variability of the score. Compute $B$ using $\operatorname{Var}(Y_i\mid x_i)$, and combine it with $A$ to obtain $A^{-1}BA^{-T}$.

Another important limitation is that this result concerns variance-function misspecification with the conditional mean correctly specified. If the mean model is wrong, the population estimating equation may have a different root; zero conditional residual means can no longer be invoked at the intended parameter.

Check Your Understanding

Check that you can distinguish the target of the estimating equation from its asymptotic variability.

Why does a positive working weight preserve a zero expected score at the true parameter when the conditional mean is correct?
In the two-group common-mean example, identify which quantities enter $A$ and which enter $B$.
Why is an integral Taylor expansion appropriate for a vector-valued estimating equation?
If the working variance is multiplied by a common positive constant, what happens to the estimating-equation root? Explain.
What additional condition, beyond a zero expected score at $\beta_0$, is needed to support consistency?