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Hermite count regression explains overdispersed counts by separating ordinary singleton events from pair-producing doublet events. The decomposition is intuitive, but it has a strict variance range and should not be treated as a universal alternative to Poisson regression.
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Chapter contents
Choose the route closest to the question, or read from Chapter 1 for the complete argument. The teaching chapters remain open; these grouped routes form the index.
Route 1 · Build the count mechanism
Route 2 · Work the dual-Poisson ledger
Route 3 · Fit and test the range
Route 4 · Check limits and evidence
Route 5 · Ask and continue
CHAPTER 1 OF 12 · BUILD THE COUNT MECHANISM
1. Why Hermite regression exists
Ordinary Poisson regression assumes that the conditional variance equals the conditional mean. That equality is useful when events arrive independently and one at a time, but some processes also generate pairs: two defects from one disturbance, two claims from one incident, or two detections from one underlying burst.
The Hermite distribution represents that mechanism with two independent Poisson components. Let X₁ count singleton events and X₂ count doublet-producing events. The observed count is Y = X₁ + 2X₂. This does more than inflate a variance parameter: it gives the extra variation a concrete event-size interpretation.
The model belongs beside the Probability root, double-Poisson approximation checks and COM-Poisson flexibility. It is useful when pair production is plausible; it is not a generic label for every overdispersed count.
CHAPTER 2 OF 12 · BUILD THE COUNT MECHANISM
2. The dual-Poisson mechanism determines the moments
Write X₁ ~ Poisson(λ₁) and X₂ ~ Poisson(λ₂), independently, with λ₁ ≥ 0 and λ₂ ≥ 0. Then E(Y) = λ₁ + 2λ₂ and Var(Y) = λ₁ + 4λ₂. The second component contributes twice its rate to the mean but four times its rate to the variance because every occurrence adds two counts.
The probability-generating function is exp{λ₁(s−1) + λ₂(s²−1)}. This is the form analysed by Kemp and Kemp. It immediately shows that setting λ₂ = 0 recovers an ordinary Poisson distribution.
The first three probabilities also expose the mechanism. P(Y=0)=exp{−(λ₁+λ₂)}; P(Y=1)=P(Y=0)λ₁; and P(Y=2)=P(Y=0)(λ₁²/2+λ₂). A count of two can arise either from two singleton events or one doublet event.
CHAPTER 3 OF 12 · WORK THE DUAL-POISSON LEDGER
3. Worked ledger: four singleton events and one-and-a-half doublets
Take λ₁ = 4 and λ₂ = 1.5. The mean is 4 + 2(1.5) = 7, while the variance is 4 + 4(1.5) = 10. The variance-to-mean ratio is 10/7 = 1.429.
| Quantity | Calculation | Result |
|---|---|---|
| Singleton rate | λ₁ | 4.000 |
| Doublet rate | λ₂ | 1.500 |
| Mean | λ₁ + 2λ₂ | 7.000 |
| Variance | λ₁ + 4λ₂ | 10.000 |
| Dispersion index | variance / mean | 1.429 |
| Probability of zero | exp{−5.5} | 0.00409 |
| Probability of one | exp{−5.5}×4 | 0.01635 |
| Probability of two | exp{−5.5}(4²/2+1.5) | 0.03882 |
The ledger explains why the model is overdispersed even though both components are Poisson. It also warns against interpreting λ₂ as the expected number of observed pairs. The observed data contain only totals; identifying the components relies on the distributional shape and covariates.
CHAPTER 4 OF 12 · WORK THE DUAL-POISSON LEDGER
4. Recovering component rates from a mean and variance
For the ordinary Hermite model, the moment equations can be inverted. Subtracting the mean from the variance gives 2λ₂, so λ₂ = (Var−Mean)/2. Then λ₁ = 2Mean−Var.
Using mean 7 and variance 10 gives λ₂ = (10−7)/2 = 1.5 and λ₁ = 14−10 = 4. These are method-of-moments values, not automatically maximum-likelihood estimates, but they make the parameter constraints visible.
| Observed moment pattern | Implied λ₁ | Implied λ₂ | Hermite verdict |
|---|---|---|---|
| Mean 7, variance 7 | 7.0 | 0.0 | Poisson boundary |
| Mean 7, variance 10 | 4.0 | 1.5 | Feasible interior |
| Mean 7, variance 14 | 0.0 | 3.5 | All-doublet boundary |
| Mean 7, variance 15 | −1.0 | 4.0 | Infeasible |
| Mean 7, variance 5 | 9.0 | −1.0 | Infeasible |
CHAPTER 5 OF 12 · WORK THE DUAL-POISSON LEDGER
5. The dispersion range is narrow and testable
Nonnegative component rates require Mean ≤ Var ≤ 2Mean. Equivalently, the variance-to-mean ratio must lie between one and two. That is a model property, not a numerical inconvenience.
If residual variance exceeds twice the fitted mean, a two-component Hermite model cannot reproduce it without a negative singleton rate. If residual variance falls below the mean, it cannot represent the underdispersion either. A generalized Hermite distribution can add larger event groups, but that changes the scientific mechanism and should be justified.
This range check should be performed over fitted covariate values, not only on one overall sample mean. A regression can enter an invalid component region locally even when a global moment ratio looks acceptable if the parameterisation is careless.
CHAPTER 6 OF 12 · FIT AND TEST THE RANGE
6. Regression needs two nonnegative component predictors
A direct construction uses log λ₁ᵢ = xᵢᵀβ and log λ₂ᵢ = zᵢᵀγ. The log links keep both rates positive. Covariates can act on singleton intensity, doublet intensity or both, which is more interpretable than forcing one coefficient to control every source of variation.
When exposure eᵢ changes the opportunity for both mechanisms proportionally, include log eᵢ as an offset in both predictors. If only singleton opportunities scale with exposure, do not copy the offset into the doublet predictor. The physical counting process decides the offset structure.
Coefficients do not translate into one universal rate ratio for E(Y), because the mean is a sum of two differently weighted components. Report predicted means and component contributions at useful covariate settings rather than interpreting exp(β) as the total-count ratio without calculation.
CHAPTER 7 OF 12 · FIT AND TEST THE RANGE
7. Changed context: the doublet mechanism grows faster
Suppose a temperature rise multiplies λ₁ by 1.2 but multiplies λ₂ by 2. Starting from λ₁ = 4 and λ₂ = 1.5, the new rates are 4.8 and 3.0. The mean moves from 7 to 10.8, while the variance moves from 10 to 16.8.
The new dispersion index is 16.8/10.8 = 1.556. A single multiplicative mean model would miss why variance grew more quickly: the pair-producing pathway became a larger share of the total.
The transfer check is to compute component rates, mean, variance and selected probabilities in the changed setting. Then simulate replicated counts and compare zeros, twos and the upper tail. A good fit at the baseline does not establish that the component regressions transfer.
CHAPTER 8 OF 12 · FIT AND TEST THE RANGE
8. Diagnostics must examine parity, zeros and tails
Because doublet events add two, the Hermite mechanism can leave information in even-versus-odd count frequencies. Strong even-count excess can support the model, but parity alone is not proof: rounding, reporting rules or batch sizes may create the same pattern.
Compare observed and replicated frequencies for zero, one, two and larger counts across fitted-mean bands. Plot residual variance against fitted mean and test whether the implied dispersion index remains between one and two. Inspect influence, because a few large counts can inflate λ₂.
Compare predictive performance with generalized-Poisson regression and negative-binomial rate models. The best comparison asks which probability law predicts held-out counts and respects the mechanism, not which one reports the smallest in-sample deviance.
CHAPTER 9 OF 12 · CHECK LIMITS AND EVIDENCE
9. Limits and common misconceptions
Hermite regression does not discover literal pairs automatically. X₁ and X₂ are latent components; the data may support their combined probability law without identifying a causal pair-production process.
The model cannot handle arbitrary overdispersion. Its ordinary two-component form stops at variance twice the mean. It also does not repair serial dependence, clustering across subjects, zero modification, truncation or an incorrect exposure definition.
An estimated λ₂ near zero places the model near the Poisson boundary, where routine large-sample tests may behave poorly. Use likelihood profiles, simulation or a carefully justified boundary test instead of treating a symmetric Wald interval as decisive.
CHAPTER 10 OF 12 · CHECK LIMITS AND EVIDENCE
10. Evidence and reproducibility route
Kemp and Kemp introduced the named distribution, its generating function, moments, parameter conditions and estimation in Some properties of the ‘Hermite’ distribution. A modern computational account is available in the peer-reviewed R Journal article Generalized Hermite Distribution Modelling with the R Package hermite.
Archive the exact probability law, component predictors, offsets, parameter constraints, optimiser, starting values and convergence checks. Save fitted component-rate ranges and the implied dispersion index for every observation. Record whether uncertainty intervals came from a Hessian, profile likelihood, bootstrap or simulation.
Reproduce the λ₁ = 4, λ₂ = 1.5 ledger from the generating mechanism and by direct probability recursion. Agreement between two calculations is a useful implementation check before a regression is trusted.
CHAPTER 11 OF 12 · ASK AND CONTINUE
11. Frequently asked questions
Is Hermite regression the same as double-Poisson regression?
No. Hermite is an exact compound-Poisson construction with singleton and doublet components. Double-Poisson uses a different double-exponential-family probability law and approximate moment relationships.
Can the ordinary Hermite model fit underdispersion?
No. With nonnegative component rates, its variance is at least its mean.
Why not use negative binomial for every overdispersed count?
Negative binomial is often useful, but its mixing interpretation and variance function differ. Hermite is attractive when grouped event sizes, especially pairs, have a plausible role.
What happens when variance exceeds twice the mean?
The ordinary Hermite moment inversion produces a negative λ₁. Choose a better mechanism, a generalized group-size model or another count family rather than hiding the violation.
CHAPTER 12 OF 12 · ASK AND CONTINUE
12. Final transfer check and connected reading
Recalculate the mean, variance, dispersion index and first three probabilities for λ₁ = 4 and λ₂ = 1.5. Invert mean 7 and variance 10 to recover the same component rates. Then test mean 7 and variance 15 and explain why the implied negative singleton rate rejects the ordinary model.
Continue through the Probability root, double-Poisson regression, COM-Poisson regression, generalized-Poisson diagnostics and negative-binomial exposure models. Together they separate a mechanistic count model from a convenient variance adjustment.
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