Comparing Pearson and Deviance Evidence

Why Compare Two Dispersion Diagnostics?

For a fitted binomial or Poisson generalized linear model, the Pearson statistic and residual deviance are both used to assess whether the data vary more than the fitted mean–variance model would suggest. Each statistic summarizes departures between observed responses and fitted means, but they measure those departures on different scales. Consequently, their dispersion ratios need not agree, and neither ratio alone identifies which observations are responsible.

The useful comparison is therefore two-part: compare the overall ratios, then examine the per-observation contributions that add up to each statistic. This separates a general pattern of extra variation from a discrepancy driven by a few observations, such as a very low count or a response near a binomial boundary.

Definitions and Comparison

For independent grouped binomial responses \(Y_i\sim\operatorname{Binomial}(m_i,p_i)\), write \(\hat{\mu}_i=m_i\hat p_i\), where \(0<\hat p_i<1\). The corresponding contributions are

A term of the form \(0\log 0\) is defined to be zero. The binomial formula includes individual Bernoulli observations as the special case \(m_i=1\). These definitions use the fitted mean structure; they do not presume that a dispersion parameter has already been estimated.

If there are \(n\) response units and the fitted mean model has rank \(r\), the commonly reported dispersion ratios are \(X_P^2/\nu\) and \(D/\nu\), with residual degrees of freedom \(\nu=n-r\). Their interpretation as diagnostics is approximate: fitted parameters affect the residual statistics, and the ratios are not exact measures of a population dispersion parameter. Since both use the same \(\nu\), their difference is determined by the difference between the two totals.

Worked Comparisons

Why the Contributions May Disagree

Both contributions are approximately quadratic when an observation lies close to its fitted mean. Farther away, the deviance reflects the asymmetry of the binomial or Poisson likelihood, while the Pearson contribution remains a squared residual divided by the fitted variance. The direction of the difference depends on the response and fitted mean; there is no universal rule that one statistic must be larger.

Justification. For Poisson data, treat the deviance contribution as a function of a real argument \(y>0\), with \(\mu\) fixed. Its value and first derivative at \(y=\mu\) are zero; its second and third derivatives there are \(2/\mu\) and \(-2/\mu^2\). Taylor's theorem through the cubic term gives \(\delta^2/\mu-\delta^3/(3\mu^2)\). The fourth derivative is continuous and bounded on a sufficiently small closed interval around \(\mu>0\), so the remainder is \(O(\delta^4)\).

For the binomial expression, write \(q=y/m\). Its value as a function of \(q\) is \(2m[q\log(q/p)+(1-q)\log((1-q)/(1-p))]\). At \(q=p\), its value and first derivative are zero, and its second and third derivatives are \(2m/[p(1-p)]\) and \(2m(2p-1)/[p^2(1-p)^2]\). Substituting \(q-p=\delta/m\) into Taylor's theorem yields the stated expansion. Since \(p\) is strictly between zero and one, the function has bounded fourth derivative on a sufficiently small closed interval around \(p\), which justifies the remainder order. \(\square\)

The expansions explain local agreement, not agreement for every observation. For example, in the Poisson expansion the cubic term changes sign with \(\delta\): a large positive residual tends to make deviance smaller than Pearson locally, while a negative residual tends to make it larger. For binomial data, the fitted probability also affects the cubic term, and the exact contribution can be especially important near zero or \(m\). The Taylor approximations should not be used as numerical substitutes for the exact formulas when residuals are large or fitted probabilities are near the boundary.

Proof. By definition, \(D=\sum_{i=1}^nD_i\) and \(X_P^2=\sum_{i=1}^nP_i\). Subtracting the sums gives \(D-X_P^2=\sum_i(D_i-P_i)\). Dividing both sides by the same positive \(\nu\) gives the ratio identity. Thus an overall difference is exactly the sum of the observation-level differences; a few contributions can dominate it, or positive and negative differences can offset one another. \(\square\)

How to Use the Comparison

A practical comparison should preserve the information that the totals discard. For each observation, calculate or inspect \(P_i\), \(D_i\), and \(D_i-P_i\), then identify the largest contributions and the largest signed differences. In grouped binomial data, retain the group sizes and fitted probabilities in the inspection; a response count cannot be interpreted apart from its denominator. In Poisson data with unequal exposure, compare counts with their fitted means, not with one another without accounting for exposure.

Common Mistake

A common error is to treat Pearson and deviance ratios as interchangeable readings of the same quantity, or to explain their difference using only the observation with the largest Pearson contribution. The ratios summarize sums, and individual deviations \(D_i-P_i\) can have either sign. A large contribution to one statistic does not automatically account for the difference between the totals; other observations may offset it or enlarge it.

The correction is to calculate both sets of contributions and add their signed differences. In the first worked example, the high-count observation has Pearson contribution \(4\) and deviance contribution about \(3.09035\). Those are that observation’s contributions, not the total Pearson statistic. The total Pearson statistic is \(8\), and the total deviance is about \(11.09035\), because the low count contributes \(8\) to deviance. The direction of the overall comparison is explained by the complete contribution sum.