Deriving Poisson Count Imputations

From a Fitted Mean to a Count Imputation

Poisson regression for count imputation supplies a conditional mean for each record with a missing count. That mean is not itself an imputed count: it can be fractional, while a count must be a nonnegative integer. To create a stochastic imputation, use the fitted mean to define a probability for every possible count, then draw one count from that distribution.

This tutorial derives that calculation and works through the inverse-cumulative-probability method. The focus is the response draw conditional on a specified fitted model. In a full multiple-imputation procedure, uncertainty in the fitted model parameters must also be represented using a suitable proper-imputation procedure; drawing counts while treating estimated parameters as known does not, by itself, account for that uncertainty.

Definition: For a record with covariates $x$, a Poisson count imputation assumes that the missing response $Y$, conditional on $x$ and model parameters, has a Poisson distribution with conditional mean $\lambda(x)>0$. Its probability of taking the count $k$, for $k=0,1,2,\ldots$, is $$ P(Y=k\mid x)=\frac{\exp\{-\lambda(x)\}\lambda(x)^k}{k!}. $$ The imputed value is a random draw from this distribution, not the mean $\lambda(x)$.

Assumptions and the Probability Calculation

The usual Poisson regression specification uses a log link:

\log\{\lambda(x)\}=x^\mathsf{T}\beta,\qquad \lambda(x)=\exp(x^\mathsf{T}\beta).

Here $x$ is the covariate vector for the record and $\beta$ is the coefficient vector. The assumptions relevant to the count draw are that the conditional response follows the Poisson distribution, the log-linear mean model is suitable for the covariates, and the covariate values needed to calculate $\lambda(x)$ are available or have already been imputed. Under the Poisson model, the conditional variance equals the conditional mean: $\operatorname{Var}(Y\mid x)=\lambda(x)$. The model also assigns probability only to nonnegative integer counts.

If records have differing known exposure, such as time at risk, include it as an offset. For exposure $t>0$, a common model is $\log\{\lambda(x,t)\}=\log(t)+x^\mathsf{T}\beta$, so $\lambda(x,t)=t\exp(x^\mathsf{T}\beta)$. The exposure must be meaningful, available for the record, and handled on the same scale used to fit the model. An offset changes the mean used in the probability calculation; it does not change the form of the Poisson probability mass function.

For a fixed fitted mean, probabilities can be computed starting at zero and proceeding upward. The probability at zero is $\exp(-\lambda)$. Each next probability is the previous probability multiplied by $\lambda/k$.

p_0=\exp(-\lambda),\qquad p_k=p_{k-1}\frac{\lambda}{k} \quad\text{for }k=1,2,\ldots

This recursion follows by dividing the Poisson probability at $k$ by the probability at $k-1$:

\frac{P(Y=k\mid x)}{P(Y=k-1\mid x)} = \frac{\exp(-\lambda)\lambda^k/k!} {\exp(-\lambda)\lambda^{k-1}/(k-1)!} = \frac{\lambda}{k}.

The probabilities sum to one because $\exp(-\lambda)\sum_{k=0}^{\infty}\lambda^k/k!=1$, by the power-series expansion of the exponential function. Thus the cumulative probability $F(k)=\sum_{j=0}^{k}p_j$ increases from zero toward one.

Turning Probabilities Into a Stochastic Draw

The inverse-cumulative-probability method uses a uniform random number $U$ between zero and one. Select the smallest nonnegative integer $k$ for which $F(k)\geq U$. Repeating this procedure with new independent uniform draws produces Poisson draws with the specified mean. A software random-number generator usually performs this operation internally, but the cumulative probabilities explain what the draw means.

Key result: Given $\lambda>0$, calculate the Poisson probabilities $p_k$ and cumulative probabilities $F(k)$. For $U\sim\operatorname{Uniform}(0,1)$, the imputed count is the smallest $k\in\{0,1,2,\ldots\}$ such that $F(k)\geq U$. The fitted mean $\lambda$ may be fractional; the resulting imputed count is always a nonnegative integer.

Worked Example: Drawing a Count When the Mean Is 1.2

Suppose the fitted Poisson mean for a hypothetical record is $\lambda=1.2$, with the model and covariates considered appropriate for this illustration. A uniform draw from the random-number generator is $U=0.74$. Find the imputed count.

First calculate the probabilities, using the mass function and its recursion:

p_0=\exp(-1.2)=0.3012,\qquad p_1=p_0\frac{1.2}{1}=0.3614, $$ $$ p_2=p_1\frac{1.2}{2}=0.2169,\qquad p_3=p_2\frac{1.2}{3}=0.08674.

The cumulative probabilities through counts zero, one, two, and three are approximately $0.3012$, $0.6626$, $0.8795$, and $0.9662$, respectively. Since $0.6626<0.74\leq0.8795$, the smallest count whose cumulative probability reaches $U$ is $k=2$. The stochastic imputation is therefore 2 counts, not 1.2.

The draw is random: another uniform value could select zero, one, three, or another nonnegative integer. The conditional mean remains 1.2, which describes the model distribution over repeated draws, not a required value for any single record.

Covariate Patterns and Exposure

The mean must be evaluated separately for each missing record using that record's covariate pattern. A coefficient changes the log mean; exponentiating the linear predictor gives the positive mean required by the Poisson mass function. The next example shows how this pattern-specific mean is carried through to a count draw.

Worked Example: Covariate Pattern, Mean, and Count Draw

Consider the hypothetical model $\log\{\lambda(x)\}=\log(2)+\log(1.5)x$, where $x$ indicates a particular risk factor and is either zero or one. For a record with $x=1$, a random-number generator returns $U=0.80$. Assume the log-linear Poisson model is suitable for this illustration and that this record's covariate is known.

The fitted linear predictor and mean are

\log\{\lambda(1)\}=\log(2)+\log(1.5)=\log(3), \qquad \lambda(1)=3.

Using the Poisson mass function, probabilities for counts zero through four are approximately $0.04979$, $0.1494$, $0.2240$, $0.2240$, and $0.1680$. For example, $p_2=p_1(3/2)=0.2240$, and $p_4=p_3(3/4)=0.1680$. The cumulative probability through count three is $0.6472$, and through count four it is $0.8153$. Since $0.6472<0.80\leq0.8153$, the imputed count is **4**.

For comparison, the record with $x=0$ has mean $\exp\{\log(2)\}=2$, not 3. Its probabilities and possible imputed value therefore differ. Applying one common count draw or one common mean to all records would ignore the fitted conditional model.

Now suppose a record has exposure $t=3$ and a fitted event rate of $0.4$ per exposure unit. Its expected count is $\lambda=t(0.4)=1.2$, equivalently $\log(\lambda)=\log(3)+\log(0.4)$. The count probabilities are calculated with $\lambda=1.2$. The exposure-adjusted mean is still a mean and may be fractional; the stochastic draw from its Poisson distribution is an integer.

Several Missing Counts and Practical Checks

For each missing response, use its own covariates to obtain its conditional mean, then generate a separate stochastic count draw. As a simple hypothetical illustration, suppose two records have fitted means 0.8 and 1.2, and the generated uniform values are 0.60 and 0.90. For the first record, $p_0=\exp(-0.8)=0.4493$ and $p_1=p_0(0.8)=0.3595$, so $F(0)=0.4493<0.60\leq F(1)=0.8088$; its imputation is 1. For the second record, $F(2)=0.8795<0.90\leq F(3)=0.9662$; its imputation is 3. Each result is a nonnegative integer, and neither has to equal its fitted mean.

Useful checks separate model predictions from imputed outcomes. Fitted means should be positive and plausible for the covariate patterns and exposure values. Imputed count values should be nonnegative integers. Compare the distribution of imputed counts with observed counts among records with comparable predictors, while remembering that observed and missing records need not have identical covariate distributions. Check for coding errors, unexpected extreme fitted means, poor fit across common or large counts, and a mismatch between the observed count scale and exposure definition.

A fractional fitted mean is not a failure: it is the expected count under the model. Conversely, an integer imputation does not establish that the model is adequate. The Poisson assumption can be unsuitable when conditional variation or the frequency of zeros is poorly represented. These are reasons to investigate model specification and fit; the separate topic of overdispersed counts considers an alternative model.

Why Stochastic Counts Matter

Replacing a missing count by the fitted mean gives a fractional value in many cases and suppresses the variation that the conditional distribution allows. Rounding that mean repairs the integer format but does not reproduce the Poisson probabilities. For example, at mean 1.2, rounding every prediction to one would omit the nonzero probabilities of zero, two, and larger counts. A Poisson draw instead respects the support and the relative probabilities implied by the fitted mean.

This distinction matters when multiple completed data sets are analyzed: different draws represent uncertainty about the unobserved responses conditional on the model. They do not remove the need for a reasonable imputation model, nor do response draws alone account for uncertainty in estimated coefficients. The assumptions about missingness and the broader rationale for multiple imputation belong to Missing Data (C38); here the specific point is how a Poisson model turns a mean into a count.

Calculate the record-specific mean.
Evaluate the log-linear predictor for the record and exponentiate it; include the known exposure offset if the model uses one.

Construct Poisson probabilities.
Start with $p_0=\exp(-\lambda)$ and use $p_k=p_{k-1}\lambda/k$ to obtain probabilities through a sufficiently large count.

Generate the response draw.
Generate $U\sim\operatorname{Uniform}(0,1)$, then select the first count whose cumulative probability is at least $U$.

Check the result and model.
Confirm the imputation is a nonnegative integer, and separately assess whether fitted means and the Poisson distribution are plausible for the records.

Common Mistakes and Limitations

Using the fitted mean as the imputed value. A mean such as 1.2 is not a count draw. Use the Poisson probabilities to obtain a nonnegative integer.
Expecting a fitted mean to be an integer. Fitted conditional means are positive and can be fractional. It is the imputed response value, not the fitted mean, that must be a nonnegative integer.
Rounding a mean instead of sampling. Rounding is not equivalent to drawing from the model and generally produces the wrong conditional distribution.
Ignoring covariates or exposure. Calculate the mean for each record's pattern and use the same exposure definition as in the fitted model.
Treating a plausible draw as proof of a good model. An integer draw can still come from a misspecified conditional distribution. Check fit and the plausibility of the means as well.
Ignoring coefficient uncertainty. A stochastic response draw conditional on fixed fitted coefficients is only one part of proper multiple imputation; model-parameter uncertainty also needs suitable treatment.

Key takeaway: A Poisson model gives each missing count a record-specific conditional mean. The probability mass function converts that mean into probabilities over nonnegative integers, and inverse-CDF sampling turns those probabilities into a stochastic count imputation.

Check Your Understanding

Answer each question, then compare with the solutions below.

For a hypothetical record with $\lambda=0.5$, calculate $p_0$, $p_1$, and $F(1)$. If $U=0.70$, what count is selected?
A model gives a record a fitted mean of 2.4. Is 2.4 a valid count imputation? Explain what should happen instead.
For the model $\log\{\lambda(x)\}=\log(2)+\log(1.5)x$, what is the fitted mean when $x=0$, and what is it when $x=1$?
A model includes an exposure of 4 units and a fitted rate of 0.25 per unit. Calculate the Poisson mean used to obtain count probabilities. Does the exposure make the imputed count fractional?

Solutions. (1) $p_0=\exp(-0.5)=0.6065$; $p_1=p_0(0.5)=0.3033$; hence $F(1)=0.9098$. Since $F(0)=0.6065<0.70\leq F(1)$, the draw is 1. (2) No. It is a conditional mean and may be fractional; calculate the Poisson probabilities for $\lambda=2.4$ and sample a nonnegative integer using a uniform draw. (3) At $x=0$, $\lambda=\exp\{\log(2)\}=2$. At $x=1$, $\lambda=\exp\{\log(2)+\log(1.5)\}=3$. (4) The mean is $4(0.25)=1.0$ count. The mean is 1.0 in this case, but the sampled imputation is a nonnegative integer because it is drawn from the Poisson distribution; exposure does not make a count draw fractional.