Why Quasi and Ordinary GLMs Often Share Coefficients

Why the Coefficients Can Stay the Same

Overdispersion means that the variability in the responses is larger than the variance assumed by an ordinary binomial or Poisson model. A common response is to use a quasi-model with a dispersion multiplier $\phi$. This changes the variance used in the quasi-score, so it is natural to wonder whether it also changes the fitted regression coefficients.

When the multiplier is the same positive constant for every observation, it only rescales the coefficient estimating equations. Multiplying an equation by a nonzero constant does not change its roots. This is why quasi-binomial and quasi-Poisson fits often have the same coefficients as their ordinary GLM counterparts. “Often” matters: the equality relies on the same mean model and weights, and on the quasi-score being proportional to the ordinary likelihood score.

The Score Equations and the Cancellation

Let $Y_1,\ldots,Y_n$ be independent responses with means $\mu_i$. Suppose the mean model is $\eta_i=x_i^\mathsf{T}\beta$, with $\mu_i=g^{-1}(\eta_i)$, where $x_i$ is the design vector and $g$ is the link. Write $V_i(\mu_i)$ for the variance function at unit dispersion, including any trial-count or weight factor used in the model. The quasi-score for a common dispersion $\phi>0$ is

U_Q(\beta) = \sum_{i=1}^n \frac{Y_i-\mu_i}{\phi V_i(\mu_i)} \frac{\partial \mu_i}{\partial \beta}.

Here $\partial\mu_i/\partial\beta$ is a column vector, so $U_Q(\beta)$ has one component for each regression coefficient. Since $\partial\mu_i/\partial\beta=(d\mu_i/d\eta_i)x_i$, the link determines how each residual contributes to the coefficient score.

Result: If $\phi$ is a common positive scalar, then $U_Q(\beta)=U_0(\beta)/\phi$, where $U_0$ is the same quasi-score with $\phi=1$. Therefore $U_Q(\beta)=0$ if and only if $U_0(\beta)=0$. A common dispersion multiplier does not change the set of coefficient roots.

Proof. Factor $1/\phi$ out of the sum:

U_Q(\beta) = \frac{1}{\phi} \sum_{i=1}^n \frac{Y_i-\mu_i}{V_i(\mu_i)} \frac{\partial \mu_i}{\partial \beta} = \frac{1}{\phi}U_0(\beta).

Because $\phi>0$, it is nonzero. Thus $U_Q(\beta)=0$ exactly when $U_0(\beta)=0$. This establishes equality of the root sets, not by itself uniqueness or convergence of a particular numerical fitting algorithm. $\square$

When the Ordinary GLM Score Has the Same Form

The cancellation result concerns quasi-score roots. To conclude that they match ordinary GLM coefficients, we also need the ordinary likelihood score to equal the unit-dispersion score $U_0$. That identity holds for binomial and Poisson GLMs with their usual likelihoods and variance functions.

Result: For independent observations from a regular one-parameter exponential family with log density or mass function $y_i\theta_i-b_i(\theta_i)+c_i(y_i)$, mean $\mu_i=b_i'(\theta_i)$, and variance $V_i(\mu_i)=b_i''(\theta_i)$, the likelihood score for $\beta$ is $$ U_L(\beta) = \sum_{i=1}^n \frac{Y_i-\mu_i}{V_i(\mu_i)} \frac{\partial\mu_i}{\partial\beta}. $$ This formula applies where the parameterization is differentiable and the means are in the interior of the model’s parameter space.

Derivation. The log-likelihood contribution from observation $i$ is $Y_i\theta_i-b_i(\theta_i)+c_i(Y_i)$. Differentiating with respect to $\theta_i$ gives $Y_i-b_i'(\theta_i)=Y_i-\mu_i$. Also $d\mu_i/d\theta_i=b_i''(\theta_i)=V_i(\mu_i)$, so $d\theta_i/d\mu_i=1/V_i(\mu_i)$. Applying the chain rule with respect to $\beta$ gives

\frac{\partial \ell_i}{\partial\beta} = (Y_i-\mu_i)\frac{\partial\theta_i}{\partial\beta} = \frac{Y_i-\mu_i}{V_i(\mu_i)} \frac{\partial\mu_i}{\partial\beta}.

Summing these contributions proves the score identity. For a binomial response with $m_i$ trials, $V_i(\mu_i)=m_i p_i(1-p_i)$, where $\mu_i=m_i p_i$. For a Poisson response, $V_i(\mu_i)=\mu_i$. Thus, with these variance functions and the same mean model in both fits, the ordinary likelihood score is $U_0$. Consequently, the quasi-score with common $\phi$ is the ordinary score divided by $\phi$. $\square$

The result does not require a canonical link: the chain-rule factor $\partial\mu_i/\partial\beta$ appears in both scores for any differentiable link. It does require the ordinary likelihood and the quasi-score to use the same mean structure, variance function, and observation weighting. If the coefficient equations have a unique relevant solution, the resulting coefficient estimates match. If there are multiple roots, the equations have the same roots, but a fitting procedure’s choice of root still matters.

Worked Examples

Worked Example: A Grouped Quasi-Binomial Fit

Two groups have $6$ successes in $10$ trials and $12$ successes in $20$ trials. Fit a grouped binomial model with a separate probability $p_j$ for each group $j$, using a logit link. Compare its coefficient equations with those of a quasi-binomial model with common $\phi=2$.

For group $j$, let $y_j$ be the success count and $m_j$ the trial count. The quasi-score for its group-specific logit coefficient is

\frac{y_j-m_jp_j}{\phi m_jp_j(1-p_j)} \frac{\partial (m_jp_j)}{\partial \eta_j} = \frac{y_j-m_jp_j}{\phi m_jp_j(1-p_j)} m_jp_j(1-p_j) = \frac{y_j-m_jp_j}{\phi}.

Setting this to zero gives $p_j=y_j/m_j$, because $\phi\ne0$. The fitted probabilities are $p_1=6/10=0.6$ and $p_2=12/20=0.6$. With group 1 as the reference, the fitted coefficients are $\hat\beta_0=\log(0.6/0.4)=\log(3/2)$ and $\hat\beta_1=0$. The ordinary binomial score gives the same equations without the factor $1/2$, so it gives the same coefficients.

The equality is not because the two groups have equal trial counts; they do not. It follows because the same trial-count factors are used in both models and the dispersion multiplier is common.

Worked Example: A Quasi-Poisson Rate With Exposure

Suppose three independent counts are $4,9,11$, observed over exposures $1,2,3$. Use a log link and a common rate $r=\exp(\beta_0)$, so $\mu_i=t_i\exp(\beta_0)$, where $t_i$ is the exposure. Compare the ordinary Poisson and quasi-Poisson coefficient equations with $\phi=3$.

For the log link, $\partial\mu_i/\partial\beta_0=\mu_i$. The Poisson variance function is $V_i(\mu_i)=\mu_i$, so the quasi-score is

U_Q(\beta_0) = \sum_{i=1}^3 \frac{Y_i-\mu_i}{3\mu_i}\mu_i = \frac{1}{3}\sum_{i=1}^3(Y_i-\mu_i).

Setting this to zero requires $\sum_iY_i=\sum_i\mu_i$. Since the total count is $4+9+11=24$ and the total exposure is $1+2+3=6$, this gives $24=6\exp(\beta_0)$. Therefore $\hat r=4$, $\hat\beta_0=\log 4$, and the fitted means are $4,8,12$. They sum to $24$, as required.

The ordinary Poisson score is $\sum_i(Y_i-\mu_i)$, which is three times the quasi-score. Its root is also $\hat\beta_0=\log 4$. The exposure changes each fitted mean, but it does not interfere with cancellation of the common multiplier.

Worked Example: Unequal Multipliers Need Not Cancel

Consider two independent Poisson counts, $Y_1=0$ and $Y_2=10$, with a common mean $\mu=\exp(\beta_0)$. Suppose, for illustration, the score uses observation-specific multipliers $\phi_1=1$ and $\phi_2=4$, rather than one common dispersion. Its equation is

\frac{Y_1-\mu}{\phi_1}+\frac{Y_2-\mu}{\phi_2} = -\mu+\frac{10-\mu}{4} = 2.5-1.25\mu = 0.

Solving gives $\hat\mu=2$ and $\hat\beta_0=\log 2$. The ordinary Poisson score is $(0-\mu)+(10-\mu)=10-2\mu$, whose root is $\hat\mu=5$, or $\hat\beta_0=\log 5$. The estimates differ because the two residuals receive different weights; there is no single scalar to factor out. This example is not the standard common-dispersion quasi-Poisson model. It isolates why the common-multiplier condition matters.

When Matching Coefficients Is a Safe Conclusion

For standard quasi-binomial and quasi-Poisson models, coefficient matching with an ordinary GLM follows when the following conditions hold:

The responses use the same independent-observation setup and mean model in both fits.
The link, design matrix, offsets, trial counts, exposures, and observation weights are the same.
The quasi variance function is the ordinary binomial or Poisson variance function multiplied by one common positive $\phi$.
The ordinary GLM uses the corresponding binomial or Poisson likelihood, so its score is the unit-dispersion score.
The coefficient estimates refer to the same solution of the score equations.

A common $\phi$ can be unknown: its numerical value is irrelevant to the location of a score root, as long as it is positive and shared by all observations. Also, the algebraic equality of roots does not establish that the assumed mean model is correct. It is a statement about the estimating equations, not a guarantee about the scientific adequacy of the model.

What Does Not Stay the Same

Equal coefficient estimates do not imply that the two analyses are identical. The quasi-model scales the score and uses a larger variance when $\phi>1$. As a result, uncertainty calculations based on that variance differ from those based on the ordinary binomial or Poisson model. The likelihood itself is also not changed into a quasi-likelihood by merely multiplying a score by a constant.

A common mistake is to reason, “The variance is larger, so the fitted coefficients must move toward zero.” That conclusion does not follow. In these models the common multiplier rescales the entire coefficient score, leaving its roots fixed. Coefficients can change if the mean model, weights, or variance pattern changes; they need not change simply because a common overdispersion multiplier is used.

Key takeaway: Standard quasi-binomial and quasi-Poisson coefficient estimates match ordinary GLM estimates when their coefficient scores differ only by one common, nonzero scalar and the same score root is selected. The multiplier affects the score’s scale, not its zeros.

Check Your Understanding

Use the score equations and assumptions above to answer these questions.

Why does multiplying every component of a coefficient score by the same positive $\phi^{-1}$ leave its roots unchanged?
Which variance functions appear in the ordinary binomial and Poisson score identities?
Does the score cancellation require a canonical link? Explain briefly.
Name two changes to the model setup that could invalidate the conclusion that quasi and ordinary coefficients match.
In the unequal-multiplier example, why does the quasi-score root differ from the ordinary Poisson root?