Why model a count-valued treatment?
A treatment need not be binary or continuous. It may be the number of doses received, visits attended, or units of a treatment delivered during a period. Such a treatment takes values in \(0,1,2,\ldots\). For inverse probability weighting, the relevant quantity is the conditional probability assigned to the count actually observed for each individual, given that individual’s covariates or treatment history.
This is a probability mass, not a continuous density. A Poisson regression models the conditional mean of the count; its probability mass function then supplies the probability of each possible count. A correct conditional mean alone is not enough: the Poisson model also imposes a particular distributional shape.
Assumptions and likelihood
The model is appropriate when the treatment is a nonnegative integer count and the conditional distribution is reasonably represented by a Poisson distribution. In particular, conditional on \(L\), a Poisson variable has both mean and variance \(\lambda(L)\). The log link ensures a positive fitted mean for every finite linear predictor. The covariates must be measured before the treatment being modeled, and the chosen regression terms must adequately represent how the conditional mean varies with those covariates.
For independent observations \((A_i,L_i)\), \(i=1,\ldots,n\), the conditional likelihood under the regression model is $$ L(\beta)=\prod_{i=1}^n \frac{\exp\{-\exp(x_i^\mathsf{T}\beta)\} \exp(x_i^\mathsf{T}\beta A_i)}{A_i!}. $$ Taking logarithms gives $$ \ell(\beta)=\sum_{i=1}^n \left[A_i x_i^\mathsf{T}\beta-\exp(x_i^\mathsf{T}\beta)-\log(A_i!)\right]. $$ If observations are not independent, this likelihood’s usual uncertainty calculations do not automatically apply. The treatment model must also be defined at the correct time point: in a longitudinal setting, \(L_i\) should represent the history available before the treatment count at that time.
Worked Example: Probabilities for Counts Under a Poisson Mean
Suppose the fitted conditional mean for an individual is \(\hat\lambda=1.5\). Calculate the fitted probabilities for counts 0 through 3 and the probability of a count of at least 4.
Using \(f(a)=\exp(-1.5)1.5^a/a!\), the probabilities are $$ \begin{aligned} \hat f(0)&=\exp(-1.5)=0.2231301601,\\ \hat f(1)&=1.5\exp(-1.5)=0.3346952402,\\ \hat f(2)&=\frac{1.5^2}{2}\exp(-1.5)=0.2510214302,\\ \hat f(3)&=\frac{1.5^3}{6}\exp(-1.5)=0.1255107151. \end{aligned} $$ Adding the unrounded probabilities gives \(0.9343575456\), which rounds to \(0.934358\). Therefore, $$ P(A\geq 4)=1-\sum_{a=0}^3 f(a)=1-0.9343575456=0.0656424544, $$ which rounds to \(0.065642\). The complement is calculated from the unrounded sum, rather than from already rounded probabilities.
Deriving and interpreting fitted probabilities
The Poisson mass formula follows from the defining probability mass function: for any fixed \(\lambda>0\), the probability at the observed value \(a\) is \(\exp(-\lambda)\lambda^a/a!\). In regression, the covariates determine \(\lambda\), so the fitted mass is evaluated at \(\hat\lambda_i\) and \(A_i\). The following result also verifies that the formula assigns total probability one.
Proof. Each term is nonnegative because \(\exp(-\lambda)>0\), \(\lambda^a>0\), and \(a!>0\). By the power-series expansion of the exponential function, $$ \sum_{a=0}^{\infty}f(a) =\exp(-\lambda)\sum_{a=0}^{\infty}\frac{\lambda^a}{a!} =\exp(-\lambda)\exp(\lambda)=1. $$ Thus these values form a probability mass function. \(\square\)
In practice, \(\lambda\) is usually estimated by maximizing the conditional likelihood. In the special intercept-only model, the estimate has a simple form.
Proof. The log-likelihood, omitting no terms that depend on \(\lambda\), is $$ \ell(\lambda)=\sum_{i=1}^n\left[A_i\log(\lambda)-\lambda-\log(A_i!)\right]. $$ For \(\lambda>0\), its derivative is $$ \frac{d\ell}{d\lambda}=\frac{\sum_i A_i}{\lambda}-n. $$ When \(\sum_i A_i>0\), setting this derivative to zero yields \(\hat\lambda=\sum_i A_i/n=\bar A\). The second derivative is \(-\sum_i A_i/\lambda^2<0\), so this stationary point is a maximum. If \(\sum_i A_i=0\), then \(\ell(\lambda)=-n\lambda\) (since all \(A_i=0\)); it decreases as \(\lambda\) increases, so its supremum over positive \(\lambda\) is approached as \(\lambda\) tends to zero. \(\square\)
Worked Example: A Covariate-Dependent Fitted Probability
Suppose the fitted model is \(\log(\lambda)=\log(2)+x\log(1.5)\), where \(x\) is a binary covariate. Find the fitted probability of the observed count \(A=2\) when \(x=1\), and compare it with the fitted probability of \(A=1\) when \(x=0\).
For \(x=1\), the fitted mean is $$ \hat\lambda=\exp\{\log(2)+\log(1.5)\}=2(1.5)=3. $$ Thus, for the observed count 2, $$ \hat f(2\mid x=1)=\frac{\exp(-3)3^2}{2!} =4.5\exp(-3)=0.2240418077. $$ For \(x=0\), the fitted mean is \(\hat\lambda=\exp\{\log(2)\}=2\). The probability of count 1 is $$ \hat f(1\mid x=0)=\frac{\exp(-2)2^1}{1!} =2\exp(-2)=0.2706705665. $$ These are probabilities for different observed counts under different covariate values. A fitted mean is not itself the probability of the observed count.
More generally, if the regression coefficient estimator is consistent for \(\beta_0\), and the specified conditional mean model is correct, the plug-in probability for a fixed covariate vector and count is consistent for the corresponding conditional probability.
Proof. The map from \(\beta\) to the displayed mass is a composition of linear, exponential, and multiplication operations, and is continuous for every finite \(\beta\). The continuous mapping theorem applied to \(\hat\beta\xrightarrow{p}\beta_0\) therefore gives the stated convergence in probability. This conclusion requires the true conditional mass to have the specified Poisson-regression form; consistency of the regression coefficients alone does not make a misspecified Poisson mass correct. \(\square\)
From fitted masses to treatment weights
For a discrete treatment, the denominator of an inverse probability weight uses the conditional probability of the observed treatment value. If \(A_i=a_i\) is the observed count and \(L_i=l_i\) its covariates or history, a fitted denominator is $$ \hat f(a_i\mid l_i)= \frac{\exp(-\hat\lambda_i)\hat\lambda_i^{a_i}}{a_i!}, \qquad \hat\lambda_i=\exp(x_i^\mathsf{T}\hat\beta). $$ A stabilized weight can use a numerator probability from a less conditional treatment model. The numerator and denominator must both be probabilities for the same observed count; substituting a continuous density or the fitted mean would be incorrect.
Worked Example: A Stabilized Weight for an Observed Count
For illustration, suppose an individual has observed treatment count \(A=3\). A numerator model assigns a Poisson mean of 1.5, while the conditional denominator model for this individual assigns a mean of 2. Calculate the ratio of numerator to denominator probabilities.
The numerator probability is $$ \hat f(3)=\frac{\exp(-1.5)1.5^3}{3!} =0.1255107151. $$ The denominator probability is $$ \hat f(3\mid L)=\frac{\exp(-2)2^3}{3!} =0.1804470443. $$ Therefore the stabilized ratio is $$ \frac{0.1255107151}{0.1804470443} =\frac{\exp(-1.5)(1.5^3/6)}{\exp(-2)(2^3/6)} =\exp(0.5)\frac{27}{64} =0.695554286\ \text{(approximately)}. $$ The ratio is formed from probabilities of the observed count 3, not from the ratio \(1.5/2\) of the two means.
Diagnostics, alternatives, and limitations
A Poisson model is compact and often useful when the conditional count distribution is not too irregular. Its central restriction is equidispersion: conditional variance equals conditional mean. Useful checks include comparing observed and model-implied counts across covariate patterns, examining residual patterns, and assessing whether zeros or large counts occur much more often than the fitted model predicts. In weighted analyses, unusually small fitted probabilities for observed counts can create very large weights; inspect the fitted probabilities and the resulting weight distribution rather than treating a successful optimizer as evidence that the treatment model is adequate.
If the conditional mean changes nonlinearly with covariates, an inadequate regression specification can be addressed by revising the mean model, for example by including appropriate nonlinear terms or interactions. If the conditional count distribution remains substantially more variable than the Poisson distribution allows, the Poisson mass may assign poor probabilities even when its mean model is useful. Negative binomial models are a standard alternative for overdispersed counts. Empirical probabilities can be useful when treatment histories have ample observations, but sparse covariate patterns can make them unstable. A multinomial model can represent a finite set of count categories without imposing the Poisson shape, but requires a decision about category handling and can use many parameters.
A model can also fail because the treatment support or measurement process is inconsistent with a Poisson count: for example, a count capped by design has a different possible support, and a treatment recorded only above a threshold is not an ordinary Poisson observation. Finally, even a well-fitting treatment model does not by itself establish causal identification. The causal interpretation of inverse probability weights still depends on the relevant assumptions for the treatment process, including adequate control of measured confounding and positivity.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Under a Poisson model with mean 2, what is the probability of observing count 0?
- Why is the fitted conditional mean not the fitted probability of the observed treatment count?
- In an intercept-only Poisson model with counts \(0,1,2,3\), what is the maximum likelihood estimate of the mean?
- Name one diagnostic concern that may indicate a Poisson model is inadequate for a treatment count.
- If the observed count is \(a_i\), which count must be used in the denominator probability of an inverse probability weight?
Solutions. (1) \(f(0)=\exp(-2)=0.1353353\). (2) The mean is the parameter \(\lambda\); the probability is the mass \(\exp(-\lambda)\lambda^a/a!\) evaluated at the observed count \(a\). (3) The sample mean is \((0+1+2+3)/4=1.5\). (4) Examples include conditional variance much larger than the fitted mean, excess zeros, or systematic disagreement between observed and fitted counts. (5) Use the individual’s observed count \(a_i\), giving \(\hat f(a_i\mid L_i)\).