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Quasi-Likelihood Theory · Tutorial 675 of 981

Mean–Variance Specification for Quasi-Likelihood

Learn to state a quasi-likelihood mean–variance specification precisely, interpret its dispersion parameter, and distinguish it from a complete probability model.

Advanced 9 min read

What You'll Learn

  • State conditional mean and variance assumptions for a quasi-likelihood model
  • Interpret the roles of the variance function and dispersion parameter
  • Express Poisson-type and grouped-binomial variance specifications
  • Explain why matching the first two moments does not determine a response distribution
  • Derive the variance of an average under conditional independence

Why Specify Moments Without a Full Distribution?

A generalized linear model often begins with a probability distribution for each response. But in practice, the conditional mean may be scientifically credible while the familiar distributional variance is too small, or the response law may not be known well enough to justify a full distributional assumption. A quasi-likelihood model makes a narrower commitment: it specifies how the conditional mean depends on predictors and how the conditional variance depends on that mean.

This distinction is useful when assessing overdispersion. A variance larger than the Poisson or binomial variance does not, by itself, identify a particular alternative distribution. A mean–variance specification can represent the extra variability without claiming more about the response than the available assumptions support.

The Mean–Variance Specification

Definition: Let \(Y_i\) be a response and let \(\mathcal C_i\) denote the information being conditioned on, including the observed predictors. A quasi-likelihood mean–variance specification consists of

\(E[Y_i\mid\mathcal C_i]=\mu_i\),
\(\operatorname{Var}(Y_i\mid\mathcal C_i)=\phi V(\mu_i)\),

where \(V\) is a specified, nonnegative variance function on the allowed mean space and \(\phi>0\) is a dispersion parameter. These equations require the indicated conditional first and second moments to exist. They do not specify a conditional probability mass function or density.

The mean may be related to predictors through a regression structure, for example \(\mu_i=g^{-1}(\mathbf{x}_i^{T}\boldsymbol{\beta})\), where \(g\) is a link function. The central object here is the conditional variance relationship. The function \(V\) describes how the variance changes as the mean changes; \(\phi\) multiplies that variance relationship by a common scale factor. A conventional normalization of \(V\) matters: replacing \(V\) by \(cV\) and \(\phi\) by \(\phi/c\), for \(c>0\), leaves the product unchanged. Thus the variance function is normally fixed by the chosen model family or convention.

For a baseline Poisson model, \(V(\mu)=\mu\) and \(\phi=1\). Writing \(\phi V(\mu)\) allows a Poisson-type mean–variance relationship with a different variance scale. For a binomial count with a fixed number of trials \(m\), the baseline variance function in terms of the count mean is \(V(\mu)=\mu(1-\mu/m)\). The specified function must be compatible with the response and its possible means; for example, a binomial count mean must lie between 0 and \(m\).

The conditional formulation is important. It describes the variation remaining after conditioning on the stated predictors and other information in \(\mathcal C_i\). If important predictors or dependence structure are omitted, the resulting conditional moments need not describe the process of interest. The specification also does not imply that different responses are independent; conditional independence is an additional assumption when needed.

Worked Example: A Poisson-Type Count Model With Exposure

Suppose \(Y_i\) is a count observed over exposure \(t_i\), and the proposed mean is \(\mu_i=t_i\exp(\mathbf{x}_i^T\boldsymbol{\beta})\). Use the variance function \(V(\mu)=\mu\), and suppose \(\phi=1.6\). For an observation with \(t_i=2\) and \(\exp(\mathbf{x}_i^T\boldsymbol{\beta})=3\), calculate the specified mean and variance.

First, the conditional mean is \( \mu_i=2(3)=6 \). The variance function evaluated at that mean is \(V(6)=6\). Therefore,

$$ \operatorname{Var}(Y_i\mid\mathcal C_i)=\phi V(\mu_i)=1.6(6)=9.6. $$

The specification says that the conditional mean is 6 and the conditional variance is 9.6. A Poisson distribution with mean 6 would instead have variance 6. The quasi specification does not assert a particular count distribution with mean 6 and variance 9.6; it records the moments being assumed.

Worked Example: Extra Variation in a Grouped Binomial Response

Let \(Y_i\) count successes among \(m=10\) trials, with conditional success probability \(p_i=0.3\). Consider the grouped-binomial variance function \(V(\mu)=\mu(1-\mu/10)\) and dispersion \(\phi=1.5\). Find the conditional mean and specified variance.

The conditional mean count is \(\mu_i=mp_i=10(0.3)=3\). Hence

$$ V(\mu_i)=3\left(1-\frac{3}{10}\right)=3(0.7)=2.1, \qquad \operatorname{Var}(Y_i\mid\mathcal C_i)=1.5(2.1)=3.15. $$

The ordinary binomial variance at these values is \(mp_i(1-p_i)=10(0.3)(0.7)=2.1\). The specified variance is larger by the factor 1.5. This states a mean–variance relationship; it does not say that the response has a binomial distribution with a noninteger effective number of trials, nor does it identify another full distribution.

Two Consequences of the Specification

Result: If \(Y_1,\ldots,Y_n\) are conditionally independent given \(\mathcal C\), and each has a finite conditional variance \(\phi V(\mu_i)\), then

\(E[\bar Y\mid\mathcal C]=\frac{1}{n}\sum_{i=1}^{n}\mu_i\),
\(\operatorname{Var}(\bar Y\mid\mathcal C)=\frac{\phi}{n^2}\sum_{i=1}^{n}V(\mu_i)\),

where \(\bar Y=n^{-1}\sum_{i=1}^{n}Y_i\).

Derivation. Conditional linearity of expectation gives \(E[\bar Y\mid\mathcal C]=n^{-1}\sum_i E[Y_i\mid\mathcal C]=n^{-1}\sum_i\mu_i\). For the variance, conditional independence makes every conditional covariance between distinct responses zero. Therefore,

$$ \begin{aligned} \operatorname{Var}(\bar Y\mid\mathcal C) &=\frac{1}{n^2}\operatorname{Var}\left(\sum_{i=1}^{n}Y_i\middle|\mathcal C\right)\\ &=\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(Y_i\mid\mathcal C)\\ &=\frac{\phi}{n^2}\sum_{i=1}^{n}V(\mu_i). \end{aligned} $$

If the responses are not conditionally independent, covariance terms must be included, so the displayed variance formula does not follow from the individual mean–variance specifications alone. \(\square\)

Result: A conditional mean and variance do not, in general, uniquely determine a response distribution. In particular, the specification \(E[Y\mid\mathcal C]=2\) and \(\operatorname{Var}(Y\mid\mathcal C)=4\) is compatible with two different nonnegative discrete distributions.

Verification. In the first distribution, assign probabilities \(1/2\) to each of 0 and 4. Its mean is \((0+4)/2=2\), and its variance is \[ \frac{1}{2}(0-2)^2+\frac{1}{2}(4-2)^2=4. \] In the second distribution, assign probabilities \(1/3\), \(1/2\), and \(1/6\) to 0, 2, and 6, respectively. These probabilities sum to 1, and its mean is \[ 0\left(\frac13\right)+2\left(\frac12\right)+6\left(\frac16\right)=2. \] Its second moment is \(0^2(1/3)+2^2(1/2)+6^2(1/6)=8\), so its variance is \(8-2^2=4\). Yet the first distribution assigns probability \(1/2\) to zero, while the second assigns probability \(1/3\). Thus the distributions differ despite having the same mean and variance. \(\square\)

Worked Example: Distinct Distributions With the Same Specified Moments

Take the variance function \(V(\mu)=\mu\), set \(\phi=2\), and consider a response whose conditional mean is \(\mu=2\). The specification gives conditional variance \(\phi V(\mu)=2(2)=4\). The two distributions just verified both satisfy these moments: one puts probability \(1/2\) on each of 0 and 4, while the other puts probabilities \(1/3,1/2,1/6\) on 0, 2, and 6.

The example separates the moment assumption from a distributional assumption. Neither distribution is being selected by the mean–variance equations, and the equations alone cannot determine probabilities such as \(P(Y=0\mid\mathcal C)\). A full distributional model would need to specify those probabilities as well.

What the Dispersion Parameter Does—and Does Not Do

For fixed \(V\) and \(\mu_i\), multiplying \(\phi\) by a constant multiplies each specified conditional variance by that same constant, without changing the specified conditional mean. When \(\phi=1\), the variance follows the chosen baseline function \(V\); \(\phi>1\) gives a larger variance than that baseline, while \(0<\phi<1\) gives a smaller one. The terms “overdispersion” and “underdispersion” therefore depend on which baseline variance function is being used.

A single common \(\phi\) makes a particular claim: all conditional variances share the same multiplicative adjustment. It does not allow an arbitrary variance at each mean. If the observed variance pattern changes shape rather than simply changing scale, the chosen \(V\) may not be adequate. Nor does a scalar dispersion parameter correct for dependence between responses.

The moment specification also does not automatically supply every feature needed for distribution-based calculations. It does not determine tail probabilities, the probability of a zero, or the shape of a sampling distribution. These require further assumptions or methods. The strength of a quasi-likelihood approach is precisely that it can use a specified mean and variance without pretending those moments identify a complete response law.

Common Mistake

Common mistake: “The response has mean \(\mu\) and variance \(\phi V(\mu)\), so it follows the familiar distribution associated with \(V\), with its dispersion adjusted.”

Correction: The equations specify only the first two conditional moments. Unless a complete distribution is separately assumed, they do not determine the response probabilities or density. Treating the moment specification as a full distribution can lead to unjustified probability calculations.

A related modeling error is to attribute every variance excess to \(\phi>1\). A misspecified mean structure or dependence among observations can also produce extra variation relative to a proposed baseline. The conditional mean, variance function, and any dependence assumptions must be considered together.

Check Your Understanding

Use the definitions and derivations above to answer the following questions.

  1. For a response with mean \(\mu_i=4\), variance function \(V(\mu)=\mu\), and \(\phi=1.25\), what conditional variance is specified?
  2. For a grouped count with \(m=8\) and mean \(\mu=2\), write its binomial variance function and evaluate it.
  3. Why is conditional independence needed for the stated variance formula for \(\bar Y\)?
  4. Can a mean–variance specification determine \(P(Y_i=0\mid\mathcal C_i)\)? Explain.
  5. What does \(\phi>1\) mean relative to the baseline variance function, and what does it not imply about the response distribution?
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