Dispersion Parameters and Scale Invariance

Why Dispersion Can Change Uncertainty Without Changing the Fit

A mean model describes how the expected response varies with covariates. A variance model describes how much observations vary around that mean. In quasi-likelihood methods, these are linked but distinct pieces: a multiplicative dispersion parameter can enlarge or shrink the variance while leaving the mean model unchanged.

When the same positive dispersion parameter applies to every observation, it divides every component of the quasi-score by the same number. The score's magnitude changes, but its zero set does not. Thus the fitted mean parameters can be unchanged even though the uncertainty attached to those estimates changes. The distinction matters: scale invariance is a property of a particular estimating equation, not a claim that dispersion is irrelevant to inference.

Definition: Suppose $Y_1,\ldots,Y_n$ are independent responses with $E[Y_i]=\mu_i(\beta)$, where $\beta$ is a vector of mean-model parameters. A common-dispersion variance model has $$ \operatorname{Var}(Y_i)=\phi V_i(\mu_i(\beta)), \qquad \phi>0, $$ where $V_i(\mu_i)>0$ is a specified variance function and the same scalar $\phi$ applies to every observation. Write $D_i(\beta)=\partial\mu_i(\beta)/\partial\beta$, as a column vector. The quasi-score for $\beta$ is $$ U_\beta(\beta;\phi)=\sum_{i=1}^n \frac{D_i(\beta)\{Y_i-\mu_i(\beta)\}}{\phi V_i(\mu_i(\beta))}. $$

The formula assumes differentiable mean functions and positive variance functions on the parameter region under consideration. Fixed observation weights can also be included in the variance function; the key condition is that $\phi$ multiplies every observation's variance by the same factor. The quasi-score uses the mean and variance assumptions, not a full probability law. This is why the mean-variance models and quasi-score construction developed earlier are relevant here.

The Scale-Invariance Result

Define the score without its common scale factor by $$ U_0(\beta)=\sum_{i=1}^n \frac{D_i(\beta)\{Y_i-\mu_i(\beta)\}}{V_i(\mu_i(\beta))}. $$ Then $U_\beta(\beta;\phi)=U_0(\beta)/\phi$. This simple factorization gives an exact result, with no large-sample approximation required.

Theorem: For fixed data and fixed mean and variance functions, if $\phi>0$ is common to all observations, then $$ \{\beta:U_\beta(\beta;\phi)=0\}=\{\beta:U_0(\beta)=0\}. $$ If the estimating equation has a unique solution, that solution is identical for every $\phi>0$. If a quasi-likelihood criterion is defined by $Q_\phi(\beta)=Q_0(\beta)/\phi$, its maximizers are also unchanged when $\phi>0$.

Proof. For any $\beta$, $U_\beta(\beta;\phi)=U_0(\beta)/\phi$. Since $\phi>0$, division by $\phi$ cannot turn a nonzero vector into the zero vector, or a zero vector into a nonzero one. Therefore $U_\beta(\beta;\phi)=0$ exactly when $U_0(\beta)=0$, which proves equality of the solution sets. Also, multiplication of all criterion values by the same positive number preserves their ordering: $Q_0(\beta_1)\geq Q_0(\beta_2)$ if and only if $Q_0(\beta_1)/\phi\geq Q_0(\beta_2)/\phi$. Hence the maximizers are unchanged. $\square$

The maximizer statement applies when the criterion is scaled in this way. It does not assert that every procedure called an estimator, or every likelihood built from a related variance model, is automatically invariant. For an estimating equation, the test is direct: a common positive factor must multiply the entire equation without changing its other terms.

Worked Example: A Poisson-Type Mean With an Unknown Scale

Suppose $Y_1,Y_2,Y_3$ have a common mean $\mu>0$ and variance $\phi\mu$. Use the log link $\log\mu=\beta$, so $d\mu/d\beta=\mu$. The quasi-score for $\beta$ is $$ U_\beta(\beta;\phi)=\sum_{i=1}^3 \frac{\mu(Y_i-\mu)}{\phi\mu} =\frac{1}{\phi}\sum_{i=1}^3(Y_i-\mu). $$

For observed values $1,2,5$, the score equation is $$ \frac{1}{\phi}(8-3\mu)=0. $$ Because $\phi>0$, this gives $\hat\mu=8/3$, and therefore $\hat\beta=\log(8/3)$. For example, with $\phi=1$ the score is $8-3\mu$; with $\phi=4$ it is $(8-3\mu)/4$. Both have the same root. The variance at the fitted mean is nevertheless four times as large under $\phi=4$ as under $\phi=1$.

Worked Examples Across Mean Models

Worked Example: Gaussian Identity-Link Regression

Consider three independent responses with $E[Y_i]=\alpha+\beta x_i$ and $\operatorname{Var}(Y_i)=\phi$. The observed data are $(x_i,Y_i)=(0,1),(1,2),(2,2)$. With $D_i=(1,x_i)^\mathsf{T}$, the score equation is $$ U_{(\alpha,\beta)}(\alpha,\beta;\phi) =\frac{1}{\phi} \begin{pmatrix} \sum_i(Y_i-\alpha-\beta x_i)\\ \sum_i x_i(Y_i-\alpha-\beta x_i) \end{pmatrix} =0. $$

The common factor can be removed, leaving the two equations $$ 3\alpha+3\beta=5,\qquad 3\alpha+5\beta=6. $$ Subtracting the first equation from the second gives $2\beta=1$, so $\hat\beta=1/2$. Substitution gives $3\alpha+3/2=5$, hence $\hat\alpha=7/6$. The fitted values are $7/6,5/3,13/6$, and the residuals are $-1/6,1/3,-1/6$. Their sum is zero, and their $x_i$-weighted sum is $1/3-2/6=0$, checking both equations. Changing the common $\phi$ leaves these fitted coefficients unchanged.

Worked Example: Grouped Binomial Means With a Logit Link

For two groups, suppose there are $m_1=5$ and $m_2=10$ trials, with observed successes $r_1=2$ and $r_2=6$. Let each group have its own logit parameter $\alpha_j$, so $p_j=\exp(\alpha_j)/\{1+\exp(\alpha_j)\}$ and $\mu_j=m_jp_j$. Assume $$ \operatorname{Var}(Y_j)=\phi m_jp_j(1-p_j). $$ Since $d\mu_j/d\alpha_j=m_jp_j(1-p_j)$, the score component for $\alpha_j$ is $$ U_{\alpha_j}(\alpha_j;\phi) =\frac{m_jp_j(1-p_j)(r_j-m_jp_j)} {\phi m_jp_j(1-p_j)} =\frac{r_j-m_jp_j}{\phi}. $$

Setting each component to zero gives $p_1=2/5=0.4$ and $p_2=6/10=0.6$, for every $\phi>0$. The fitted logits are $\hat\alpha_1=\log(0.4/0.6)=\log(2/3)$ and $\hat\alpha_2=\log(0.6/0.4)=\log(3/2)$. Their values do not depend on the common scale. The variance of each group response does: the fitted variances are $\phi(5)(0.4)(0.6)=1.2\phi$ and $\phi(10)(0.6)(0.4)=2.4\phi$.

Why Standard Errors Still Depend on Dispersion

The score's roots and its variability answer different questions. To see how the scale enters uncertainty, let $U_0$ be the unscaled score above, and define its sensitivity matrix by $$ H_0=-E\left[\frac{\partial U_0}{\partial\beta^\mathsf{T}}\right]. $$ Assume differentiability and finite second moments. Under the stated mean-variance model with independent observations, differentiating each score term and taking expectations removes the terms multiplied by $Y_i-\mu_i$, because $E[Y_i-\mu_i]=0$. Thus $$ H_0=\sum_{i=1}^n \frac{D_iD_i^\mathsf{T}}{V_i(\mu_i)}. $$ If this matrix is nonsingular, the usual first-order estimating-equation expansion gives covariance $H^{-1}JH^{-\mathsf{T}}$, where $H$ is the score sensitivity and $J$ is the score variability.

For the scaled score, $H=H_0/\phi$. Its variability is $J=\operatorname{Var}(U_0)/\phi^2$. Independence and $\operatorname{Var}(Y_i)=\phi V_i(\mu_i)$ give $$ \operatorname{Var}(U_0)=\phi H_0,\qquad J=\frac{H_0}{\phi}. $$ Consequently, $$ H^{-1}JH^{-\mathsf{T}} =(\phi H_0^{-1})\left(\frac{H_0}{\phi}\right)(\phi H_0^{-\mathsf{T}}) =\phi H_0^{-1}, $$ using the symmetry of $H_0$. This is the model-based first-order covariance when $H_0$ is nonsingular. In a regular large-sample setting—where the estimator is consistent, the mean functions are differentiable, the score satisfies a central limit theorem, and the Taylor remainder is negligible—the same expression is the leading covariance in the asymptotic distribution. Thus larger dispersion leaves the fitted parameters unchanged but increases their variance in proportion to $\phi$, and their standard errors in proportion to $\sqrt{\phi}$.

Worked Example: Unequal Scales Do Not Cancel as One Common Factor

Suppose two independent observations have a common unknown mean $\mu$, with variances $\phi c_1$ and $\phi c_2$, where $c_1=1$ and $c_2=4$. For this mean model, the quasi-score equation is $$ \frac{1}{\phi}\left\{\frac{Y_1-\mu}{1}+\frac{Y_2-\mu}{4}\right\}=0. $$ With observed values $Y_1=0$ and $Y_2=10$, cancelling the common $\phi$ leaves $$ -\mu+\frac{10-\mu}{4}=0. $$ Multiplying by four gives $-4\mu+10-\mu=0$, so $\hat\mu=2$. If the two relative scale factors were both one instead, the equation would be $(0-\mu)+(10-\mu)=0$, giving $\hat\mu=5$. The common factor $\phi$ cancels in either model, but the unequal relative variances change the weighted estimate.

Common Mistake and Practical Interpretation

A common mistake is to reason, “Dispersion divides the score, so it has no effect on the analysis.” The correction is that it has no effect on the roots of this score equation when it is common and positive. It does affect the variance model and hence the uncertainty calculation. It can also affect a fitted result when it is not a single common multiplier: observation-specific scale factors change the relative weights, as the last example showed.

Before claiming scale invariance, check three features of the estimating equation:

The scale parameter is positive and common to all observations.
It multiplies every component of the estimating equation, rather than entering the mean function or changing relative weights.
The target is a solution of that equation (or a criterion scaled by the same positive factor), not an unrelated estimator or inferential quantity.

This separates estimation of the mean parameters from estimation and use of dispersion. The subsequent Pearson-based approach addresses how to estimate the scale from residual information; the point here is that an estimated common scale can be important for uncertainty even when it does not move the quasi-score root.

Key takeaway: In a quasi-score with variance $\phi V_i(\mu_i)$, a common $\phi>0$ scales the whole score by $1/\phi$. Its roots are unchanged, while the model-based covariance scales with $\phi$. Unequal observation-specific multipliers generally change relative weights and can change the estimate.

Check Your Understanding

Use the factorization and examples above to answer these questions.

Which condition on the dispersion parameter allows it to be removed from a quasi-score equation without changing its roots?
In the Gaussian example, what are the fitted intercept and slope, and why are they unchanged when the common dispersion changes?
If the common dispersion is multiplied by four, what happens to the model-based standard errors under the covariance result?
For two observations with different relative variance factors, why might the estimate of a shared mean differ from the ordinary average?
Does an unchanged quasi-score root imply that dispersion is irrelevant to inference? Explain.