Parameter Uncertainty in Count Predictions

Why a Future Count Has Two Sources of Uncertainty

For a specified covariate profile, a Poisson regression model assigns a future count a Poisson distribution with a mean determined by the coefficients. Even if those coefficients were known, the future count would still vary from one observation to another. In practice, however, the coefficients are estimated from data. The fitted mean is therefore uncertain too.

These are distinct sources of variation. Poisson sampling variation describes variation in a future count conditional on a particular model mean. Parameter uncertainty describes uncertainty about which model mean is appropriate because the coefficients have been estimated. A predictive distribution that includes both averages the conditional Poisson distribution over an uncertainty distribution for the coefficients.

Definition: For a future count $Y^*$ at covariate vector $x^*$, suppose $Y^* \mid \beta,x^* \sim \operatorname{Poisson}(\mu(\beta))$, where $\mu(\beta)=\exp(x^{*T}\beta)$. The plug-in predictive distribution substitutes an estimate $\hat\beta$ for $\beta$, giving $Y^* \mid \hat\beta,x^* \sim \operatorname{Poisson}(\hat\mu)$, where $\hat\mu=\exp(x^{*T}\hat\beta)$. Given a probability distribution $Q$ representing uncertainty about $\beta$, the parameter-integrated predictive distribution averages the conditional Poisson probabilities over $Q$: $$ P_Q(Y^*=k\mid x^*)=\int \frac{\exp\{-\mu(\beta)\}\mu(\beta)^k}{k!}\,Q(d\beta),\qquad k=0,1,2,\ldots. $$

The plug-in distribution is a single Poisson distribution. The integrated distribution is generally a mixture of Poisson distributions and need not itself be Poisson. The integration requires a clearly specified $Q$. In Bayesian prediction, $Q$ is the posterior distribution of the coefficients given the observed data. In a frequentist analysis, a distribution approximating coefficient uncertainty can be constructed, for example, from an asymptotic sampling approximation or a parametric bootstrap. Such a frequentist distribution is an approximation for prediction, not automatically a posterior distribution or an exact probability distribution for a fixed parameter.

What Integration Does to a Predictive Distribution

The mixture construction is a valid probability distribution, and its moments show directly how parameter uncertainty enters. Suppose $Q$ is a probability distribution and the relevant expectations exist. Conditional on $\beta$, a Poisson variable has mean and variance $\mu(\beta)$. Averaging these conditional distributions gives the following result.

Theorem: If $Y^*\mid\beta,x^*\sim\operatorname{Poisson}(\mu(\beta))$, then the parameter-integrated predictive distribution has $$ E_Q[Y^*\mid x^*]=E_Q[\mu(\beta)] $$ and $$ \operatorname{Var}_Q(Y^*\mid x^*)=E_Q[\mu(\beta)]+\operatorname{Var}_Q(\mu(\beta)). $$ These statements require $E_Q[\mu(\beta)]<\infty$ for the mean, and $E_Q[\mu(\beta)^2]<\infty$ for the variance.

Proof. For each nonnegative integer $k$, the integrated probability is nonnegative. Moreover, because sums and integrals of nonnegative terms can be interchanged,

\sum_{k=0}^{\infty}P_Q(Y^*=k\mid x^*) =\int\left\{\sum_{k=0}^{\infty}P(Y^*=k\mid\beta,x^*)\right\}Q(d\beta) =\int 1\,Q(d\beta)=1.

Thus the integrated probabilities sum to one. For the mean, the law of total expectation gives

E_Q[Y^*\mid x^*] =E_Q\!\left[E(Y^*\mid\beta,x^*)\right] =E_Q[\mu(\beta)].

For the variance, the law of total variance gives

\operatorname{Var}_Q(Y^*\mid x^*) =E_Q\!\left[\operatorname{Var}(Y^*\mid\beta,x^*)\right] +\operatorname{Var}_Q\!\left(E[Y^*\mid\beta,x^*]\right) =E_Q[\mu(\beta)]+\operatorname{Var}_Q(\mu(\beta)).

The first term is the average Poisson sampling variance. The second is the additional variation in the conditional mean across the coefficient uncertainty distribution. If the mean is not constant under $Q$, this second term is positive. $\square$

A plug-in Poisson prediction instead has mean and variance both equal to $\hat\mu$. It therefore omits variation in the mean across possible coefficient values. Its mean need not equal the integrated predictive mean: in general $E_Q[\exp(x^{*T}\beta)]$ is not $\exp(x^{*T}E_Q[\beta])$. Consequently, comparing the two distributions requires attention to both their means and their variances, not just their labels.

Worked Example: Two Possible Count Means

At a specified profile, suppose an uncertainty distribution assigns probability $1/2$ to a Poisson mean of 2 and probability $1/2$ to a Poisson mean of 6. A plug-in analysis uses the mean 4. Compare the probabilities of zero and one future event and the predictive variances.

For the plug-in distribution, $Y^*\sim\operatorname{Poisson}(4)$. Hence $P(Y^*=0)=e^{-4}\approx0.0183$ and $P(Y^*=1)=4e^{-4}\approx0.0733$. Its variance is 4.

For the integrated prediction, average the two conditional probabilities:

P_Q(Y^*=0)=\frac{e^{-2}+e^{-6}}{2} \approx\frac{0.1353+0.0025}{2}=0.0689,

P_Q(Y^*=1)=\frac{2e^{-2}+6e^{-6}}{2} \approx\frac{0.2707+0.0149}{2}=0.1428.

The integrated mean is $(2+6)/2=4$, matching the plug-in mean in this example. The uncertainty-distribution variance of the mean is $\{(2-4)^2+(6-4)^2\}/2=4$. Therefore the integrated predictive variance is $E_Q[\mu]+\operatorname{Var}_Q(\mu)=4+4=8$, twice the plug-in variance. The two distributions have the same mean but quite different probabilities, especially for zero and one event.

Worked Integration for a Gamma Distribution

When uncertainty is expressed directly about a positive Poisson mean, a Gamma distribution gives a closed-form mixture. Use the shape-rate convention: $\mu\sim\operatorname{Gamma}(a,b)$ has density $b^a\mu^{a-1}e^{-b\mu}/\Gamma(a)$ for $\mu>0$, with $a,b>0$. Conditional on $\mu$, let $Y^*\sim\operatorname{Poisson}(\mu)$. Integrating gives

\begin{aligned} P(Y^*=k) &=\int_0^\infty \frac{e^{-\mu}\mu^k}{k!} \frac{b^a\mu^{a-1}e^{-b\mu}}{\Gamma(a)}\,d\mu\\ &=\frac{b^a}{k!\,\Gamma(a)} \int_0^\infty \mu^{a+k-1}e^{-(b+1)\mu}\,d\mu\\ &=\frac{\Gamma(a+k)}{\Gamma(a)k!} \left(\frac{b}{b+1}\right)^a \left(\frac{1}{b+1}\right)^k . \end{aligned}

The last equality follows by rescaling the Gamma integral: for $c>0$, $\int_0^\infty u^{r-1}e^{-cu}\,du=\Gamma(r)/c^r$. The resulting count distribution is negative binomial under the convention that its probability at $k$ is the expression displayed above. Its mean and variance also follow from the mixture theorem: $E[\mu]=a/b$ and $\operatorname{Var}(\mu)=a/b^2$, so $E[Y^*]=a/b$ and $\operatorname{Var}(Y^*)=a/b+a/b^2$.

Worked Example: A Gamma-Uncertain Poisson Mean

Suppose the uncertainty distribution for the future mean is $\mu\sim\operatorname{Gamma}(3,1)$, in shape-rate form. Compare the integrated probabilities for zero and one event with the plug-in prediction at the mean of this Gamma distribution.

The mean and variance of $\mu$ are $3/1=3$ and $3/1^2=3$. Thus the plug-in distribution is $\operatorname{Poisson}(3)$, with $P(Y^*=0)=e^{-3}\approx0.0498$, $P(Y^*=1)=3e^{-3}\approx0.1494$, and variance 3.

For the integrated distribution, take $a=3$, $b=1$ in the formula above. At $k=0$, $P(Y^*=0)=(1/2)^3=1/8=0.125$. At $k=1$,

P(Y^*=1)=\frac{\Gamma(4)}{\Gamma(3)1!} \left(\frac12\right)^3\left(\frac12\right) =3\left(\frac18\right)\left(\frac12\right) =\frac{3}{16}=0.1875.

The integrated predictive mean is 3, but its variance is $3+3=6$, rather than 3. Uncertainty about the mean has increased the probability of zero and one event relative to the plug-in calculation; the remaining probability is also redistributed across larger counts. This calculation illustrates that uncertainty about a mean can change the full predictive distribution even when the two methods agree on the expected count.

From Coefficient Covariance to Count Uncertainty

In a fitted log-linear count model, the coefficient uncertainty is often summarized by an estimated covariance matrix. At a fixed covariate vector $x^*$, the estimated linear predictor is $x^{*T}\hat\beta$, and its estimated variance is $x^{*T}\widehat{\operatorname{Var}}(\hat\beta)x^*$. A normal approximation for this linear predictor can be used to approximate an uncertainty distribution for the mean. Because the mean is the exponential of the linear predictor, the resulting mean uncertainty is generally asymmetric.

Worked Example: Propagating Coefficient Uncertainty to a Count

At profile $x^*=(1,2)^T$, suppose $\hat\beta=(\log 3,0)^T$ and the estimated covariance matrix is $\widehat{\operatorname{Var}}(\hat\beta)=\begin{pmatrix}0.01&0\\0&0.01\end{pmatrix}$. Use the normal approximation $\beta\sim N_2(\hat\beta,\widehat{\operatorname{Var}}(\hat\beta))$ as an uncertainty distribution for this calculation. Find the plug-in and integrated predictive means and variances.

The fitted linear predictor is $x^{*T}\hat\beta=\log 3+2(0)=\log 3$, so the plug-in mean is $\hat\mu=3$. The variance of the uncertain linear predictor is $0.01+2^2(0.01)=0.05$. Therefore write $\eta=x^{*T}\beta\sim N(\log 3,0.05)$ and $\mu=e^\eta$.

For a normal random variable $\eta\sim N(m,v)$, completing the square in its density gives $E[e^{t\eta}]=e^{tm+t^2v/2}$. Applying this identity with $t=1$ and $t=2$,

E_Q[\mu]=3e^{0.05/2}=3e^{0.025}\approx3.076,

\operatorname{Var}_Q(\mu) =E_Q[\mu^2]-\{E_Q[\mu]\}^2 =9e^{0.10}-9e^{0.05} =9e^{0.05}(e^{0.05}-1)\approx0.485.

The integrated predictive mean is therefore about 3.076 and its variance is about $3.076+0.485=3.561$. The plug-in prediction has mean and variance both 3. Here the integrated mean exceeds the fitted mean because exponentiation is nonlinear; the variance is also larger because it includes variation in the mean.

Choosing and Interpreting the Prediction

A plug-in predictive distribution can be appropriate when the goal is explicitly to predict under the fitted coefficients as if they were known, or when coefficient uncertainty is negligible for the intended use. It should not be confused with a confidence interval for the fitted mean. A confidence interval for a mean describes uncertainty about that mean; a predictive distribution describes possible values of a future count. Integrating coefficient uncertainty produces a predictive distribution that carries both sources of variation.

The choice of uncertainty distribution matters. A Bayesian integrated prediction uses the posterior distribution conditional on the observed data and the specified prior and likelihood. A frequentist approximation can use an estimated sampling distribution for $\hat\beta$, but its construction and calibration must be stated. One should not silently treat a standard error as if it were itself a probability distribution for the fixed coefficient.

Check the target. A predicted conditional mean is not a future count and has no Poisson sampling variation by itself.
Check the uncertainty distribution. State whether coefficient uncertainty is represented by a posterior, an approximation, or another explicitly defined distribution.
Check the scale. Uncertainty in the linear predictor is transformed through the exponential link; averaging coefficients and then exponentiating generally differs from averaging the means.
Check what the reported probabilities represent. Plug-in probabilities condition on one fitted mean; integrated probabilities average over possible means.

A common mistake is to claim that a plug-in Poisson prediction already accounts for coefficient uncertainty because its variance equals its mean. That equality accounts only for Poisson variation conditional on the fitted mean. For an integrated prediction, the variance is $E_Q[\mu]+\operatorname{Var}_Q(\mu)$; the second term is absent from the plug-in calculation. Another mistake is to assume every integrated probability must be larger than its plug-in counterpart. Probabilities across count values must sum to one, so changes in some probabilities are balanced by changes elsewhere; the direction depends on the count and the uncertainty distribution.

Key takeaway: A plug-in prediction uses one estimated count mean. A parameter-integrated prediction averages conditional Poisson probabilities over uncertainty in that mean, adding variation across possible fitted means and potentially changing both the predictive variance and individual count probabilities.

Check Your Understanding

Use the definitions and derivations above to answer these questions.

At a fixed profile, what uncertainty does a plug-in Poisson predictive distribution include, and what uncertainty does it leave out?
If $E_Q[\mu]=5$ and $\operatorname{Var}_Q(\mu)=2$, what are the mean and variance of the integrated predictive count?
Why can two predictive distributions have the same expected count but different probabilities of zero events?
For a Gamma distribution in shape-rate form, what roles do $a$ and $b$ play in its mean $a/b$ and variance $a/b^2$?
Why is a frequentist approximation for coefficient uncertainty not automatically a posterior distribution?