Why a Correct Model Does Not Give a Dispersion Ratio of Exactly One
A Pearson dispersion ratio or a deviance dispersion ratio is a diagnostic summary, not a constraint imposed by a correctly specified model. Even when the mean structure and response distribution are right, a particular sample will usually have more or less residual variation than its average reference level. The ratios therefore fluctuate around one. Their approximate spread follows from the reference distribution of the statistics in their numerators.
The chi-square reference is an approximation, not a universal finite-sample identity. Its quality depends on the sampling regime. In this tutorial, the derivation is for a fixed number \(m\) of independent response cells, with adequate information in each cell—for example, grouped binomial cell sizes or Poisson expected counts increasing—while the number \(p\) of fitted regression parameters remains fixed. The true parameter is assumed to be an interior point, the model is identifiable and regular, and the design has full column rank. These conditions rule out boundary estimates and sparse-cell regimes in which the usual approximation may be poor.
For Poisson and binomial GLMs, the conventional null model has unit dispersion. In models with a known common dispersion \(\phi\), the Pearson statistic is instead scaled by \(\phi\), and the corresponding null comparison is \(X_P^2/\phi\) against its degrees of freedom. Below, the unit-dispersion case makes the reasoning most direct.
How Fitting Parameters Changes the Pearson Statistic
If the means were known rather than estimated, standardized residuals would, under suitable large-information conditions, behave approximately like independent standard normal variables. Their squared sum would then have approximately \(m\) degrees of freedom. Estimating \(p\) parameters uses information in the response to fit the means. To first order, this removes \(p\) directions of variation. The residual quadratic form therefore has \(m-p\), rather than \(m\), degrees of freedom.
Proof. Write \(\mu\) for the vector of true means, \(\Sigma=\operatorname{diag}\{V(\mu_1),\ldots,V(\mu_m)\}\), and \(D\) for the \(m\)-by-\(p\) matrix whose \(i\)th row is the derivative of \(\mu_i\) with respect to the parameter vector. Let \(z=\Sigma^{-1/2}(Y-\mu)\). By assumption, \(z\xrightarrow{d}N(0,I_m)\).
A first-order expansion of the score equation around the true parameter gives \(\hat\beta-\beta \approx (D'\Sigma^{-1}D)^{-1}D'\Sigma^{-1}(Y-\mu)\). This expression is the first-order adjustment to the parameter estimate needed to make the fitted residuals satisfy the score equations. Consequently, the fitted standardized residual vector is approximately \(z-A(A'A)^{-1}A'z\), where \(A=\Sigma^{-1/2}D\). The matrix \(P=A(A'A)^{-1}A'\) is the orthogonal projection onto the column space of \(A\). Its rank is \(p\), by full column rank of the design derivatives.
The Pearson statistic is the squared length of the fitted standardized residual vector, up to a term tending to zero from evaluating the variances at fitted rather than true means. Thus \(X_P^2\) is asymptotically \(z'(I_m-P)z\). The matrix \(I_m-P\) is also an orthogonal projection; its rank is \(m-p\). Orthogonally diagonalizing it transforms \(z\) into a vector that still converges to \(N(0,I_m)\), and the quadratic form becomes the sum of squares of \(m-p\) standard normal coordinates. That sum has a \(\chi^2_{m-p}\) distribution. Therefore \(X_P^2\xrightarrow{d}\chi^2_{m-p}\). \(\square\)
This derivation explains the degrees of freedom, but also its limitations. It relies on a normal approximation for standardized errors and on a regular first-order fit. A large number of individual Bernoulli or Poisson observations does not by itself guarantee the stated regime: if each observation is its own cell and expected counts stay small, the fixed-cell, high-information argument does not apply.
Worked Example: The Amount of Random Fluctuation at 20 Degrees of Freedom
Suppose a fitted model has \(\nu=20\) residual degrees of freedom and the chi-square approximation is reasonable. Under the null, \(X_P^2\) is approximately \(\chi^2_{20}\). A chi-square variable with \(\nu\) degrees of freedom has mean \(\nu\) and variance \(2\nu\), so here \(E[X_P^2]\approx20\), \(\operatorname{Var}(X_P^2)\approx40\), and \(\operatorname{SD}(X_P^2)\approx\sqrt{40}=6.325\).
The dispersion ratio is \(R_P=X_P^2/20\). Dividing the approximate mean by 20 gives \(E[R_P]\approx1\), and dividing the approximate standard deviation by 20 gives \(\operatorname{SD}(R_P)\approx6.325/20=0.316\). Thus a ratio noticeably above or below one is compatible with ordinary sampling variation. For example, if \(X_P^2=25.4\), then \(R_P=25.4/20=1.27\), only \(5.4/6.325\approx0.85\) approximate standard deviations above its reference mean. The ratio is not expected to equal one in every sample.
Why Deviance Has the Same Approximate Reference
The residual deviance compares the fitted model with the cell-by-cell saturated model. When the model is regular and the true parameter is interior, Wilks' theorem says that twice the log-likelihood difference between nested models is asymptotically chi-square, with degrees of freedom equal to the difference in their parameter dimensions. Here the saturated model has \(m\) cell means and the fitted model has \(p\) parameters, so the dimension difference is \(m-p\).
The local expansion can be seen cell by cell. For a Poisson response with fitted mean \(\mu\) and observed value \(y=\mu+\delta\), its deviance contribution is \(2\{y\log(y/\mu)-(y-\mu)\}\). Writing \(x=\delta/\mu\), this is \(2\mu\{(1+x)\log(1+x)-x\}\). The Taylor expansion \((1+x)\log(1+x)-x=x^2/2-x^3/6+O(x^4)\) gives a leading contribution \(\delta^2/\mu\), the Poisson Pearson contribution. For grouped binomial data, expansion of the two-category log likelihood around its fitted success probability gives the leading term \((y-np)^2/\{np(1-p)\}\), also the Pearson contribution. When the model is fitted, the same parameter-estimation adjustment acts on these local quadratic terms. The chi-square result for deviance also follows directly from Wilks' theorem; the expansion explains why it tracks Pearson's statistic near the fitted model.
Worked Example: One Grouped Binomial Cell
Suppose a binomial cell has \(n=100\), observed successes \(y=45\), and fitted success probability \(p=0.40\). The fitted counts are 40 successes and 60 failures. The cell's Pearson contribution is \((45-40)^2/40+(55-60)^2/60=25/40+25/60=0.625+0.4167\approx1.0417\).
The binomial deviance contribution is \(2\{45\log(45/40)+55\log(55/60)\}\). Using \(\log(1.125)\approx0.117783\) and \(\log(11/12)\approx-0.0870114\), the expression inside braces is \(45(0.117783)+55(-0.0870114)\approx5.3002-4.7856=0.5146\). Therefore the deviance contribution is approximately \(2(0.5146)=1.0292\). Both contributions are of similar size, as the quadratic approximation suggests; they are not identical for this finite observation. A single cell's contribution is not itself a residual dispersion ratio, since that ratio sums contributions over the model's residual degrees of freedom.
Worked Example: One Poisson Cell Near Its Fitted Mean
For a Poisson cell with fitted mean \(\mu=10\) and observed count \(y=12\), the Pearson contribution is \((12-10)^2/10=4/10=0.4\).
The deviance contribution is \(2\{12\log(12/10)-(12-10)\}=2\{12\log(1.2)-2\}\). Since \(\log(1.2)\approx0.1823216\), this is \(2\{2.1878592-2\}=0.3757184\), or approximately \(0.3757\). The difference from 0.4 reflects higher-order terms in the likelihood expansion. As counts and information increase under the stated conditions, the quadratic term governs the approximation more closely.
Why the Ratios Fluctuate Around One
If \(Q\) has a \(\chi^2_\nu\) distribution, it is a sum of squares of \(\nu\) independent standard normal variables. Each squared standard normal has mean 1 and variance 2. Hence \(E[Q]=\nu\), \(\operatorname{Var}(Q)=2\nu\), and for \(R=Q/\nu\), \[ E[R]=1,\qquad \operatorname{Var}(R)=\frac{2}{\nu},\qquad \operatorname{SD}(R)=\sqrt{\frac{2}{\nu}}. \]
These identities give the basic explanation: a statistic with \(\nu\) residual degrees of freedom averages \(\nu\) units of squared standardized error, but that average remains random. For example, the approximate standard deviation is \(\sqrt{2/8}=0.5\) with eight degrees of freedom and \(\sqrt{2/50}=0.2\) with fifty. More degrees of freedom usually make the ratio less variable relative to one. They do not force a finite-sample ratio to equal one.
The expectation statements for an actual GLM statistic are approximate, not automatically exact. Convergence in distribution alone does not guarantee convergence of expectations; an additional moment condition, such as uniform integrability of the relevant statistics, is needed to infer convergence of their means. The chi-square mean and variance are exact for the reference distribution itself. In routine use, they describe the approximate null center and spread when the approximation is adequate.
Common Mistakes and Practical Meaning
- Treating one as a hard cutoff. A ratio of 1.2 is not automatically evidence against the model. Its interpretation depends partly on the residual degrees of freedom and on whether the chi-square approximation is credible.
- Forgetting that parameters are estimated. Comparing a residual sum of squares to \(m\) degrees of freedom ignores the fitted directions. In the regular full-rank setting, the residual reference uses \(m-p\).
- Assuming Pearson and deviance must match. They share a leading quadratic approximation, not an exact identity. Their individual contributions can differ, particularly when a response is far from its fitted mean.
- Using the approximation with sparse cells without checking. The normal and likelihood approximations can be poor when expected counts are small, estimates approach a boundary, or the design is nearly rank deficient. A ratio near one does not repair a weak reference approximation.
The practical value of this derivation is calibration: it prevents ordinary sampling fluctuation from being mistaken for a defect in the model. A dispersion ratio is evidence to interpret in context, not a direct estimate that must equal one under the null. Its approximate chi-square reference explains both its null center and why the amount of variation depends on residual degrees of freedom.
Check Your Understanding
Use the definitions and approximations developed above to answer these questions.
- Why does fitting \(p\) regular, identifiable parameters lead to \(m-p\), rather than \(m\), approximate residual degrees of freedom?
- If a Pearson statistic is approximately \(\chi^2_{18}\), what are the approximate mean and standard deviation of its dispersion ratio?
- Explain why the Pearson and deviance contributions in a single cell can differ even when both use the same fitted model.
- Give one feature of a data set or fitted model that can make the chi-square approximation unreliable.
- Why does convergence in distribution to a chi-square variable not, by itself, establish that the finite-sample dispersion ratio has expectation exactly one?