Predictive Claim Counts Under Gamma-Poisson Mixing

Why Predictive Counts Are Not Simply Poisson

A claim count for a future period is uncertain for two reasons: claims vary randomly even if the underlying rate is fixed, and the rate itself is uncertain. A Poisson model with an unknown rate captures both sources when prediction averages over the posterior distribution of that rate. The resulting predictive count distribution is negative binomial, not generally Poisson.

We use a hypothetical portfolio segment in which each risk has its own constant claim rate $\lambda$. For the calculations below, the observed claim count is recorded over one unit of exposure, and the future count is also for one unit of exposure. Conditional on $\lambda$, the observed and future counts are independent Poisson variables. There is no interest or mortality assumption in this frequency model.

Definition: A gamma distribution with shape $\alpha>0$ and rate $\beta>0$, written $\operatorname{Gamma}(\alpha,\beta)$, has density $$ \pi(\lambda)=\frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda},\qquad \lambda>0. $$ Here $\beta$ is a rate parameter, not a scale parameter. If the observed count is $n$ over one unit of exposure, the posterior distribution is $\operatorname{Gamma}(r,b)$, where $r=\alpha+n$ and $b=\beta+1$. The future count $M$, over one unit of exposure, has a negative binomial posterior predictive distribution obtained by averaging its conditional Poisson distribution over this posterior.

From the Posterior Rate to a Predictive Distribution

Once the data have been observed, the posterior parameters $r$ and $b$ summarize the information about $\lambda$. Conditional on $\lambda$, the future count satisfies $M\mid\lambda\sim\operatorname{Poisson}(\lambda)$, so $P(M=k\mid\lambda)=e^{-\lambda}\lambda^k/k!$. The predictive probability of $k$ claims is the average of this conditional probability over the posterior density of $\lambda$.

\begin{aligned} P(M=k\mid n) &=\int_0^\infty P(M=k\mid\lambda)\,\pi(\lambda\mid n)\,d\lambda\\ &=\int_0^\infty \frac{e^{-\lambda}\lambda^k}{k!} \frac{b^r}{\Gamma(r)}\lambda^{r-1}e^{-b\lambda}\,d\lambda\\ &=\frac{b^r}{\Gamma(r)k!} \int_0^\infty \lambda^{r+k-1}e^{-(b+1)\lambda}\,d\lambda\\ &=\frac{\Gamma(r+k)}{\Gamma(r)k!} \left(\frac{b}{b+1}\right)^r \left(\frac{1}{b+1}\right)^k,\qquad k=0,1,2,\ldots. \end{aligned}

The last equality follows from the gamma integral $\int_0^\infty x^{a-1}e^{-cx}\,dx=\Gamma(a)/c^a$, for $a,c>0$. This integration explains why uncertainty about the rate changes the predictive model: the posterior rate is not fixed at one value but is averaged over all its possible values.

Result: The predictive probability function is $$ P(M=k\mid n)=\frac{\Gamma(r+k)}{\Gamma(r)k!}p^r(1-p)^k,\qquad p=\frac{b}{b+1}. $$ This is the negative binomial convention in which $k$ counts the failures before the $r$th success, with success probability $p=b/(b+1)$. In this application, “successes” and “failures” are just the standard negative binomial parameterization; they are not claims and nonclaims. For integer $r$, the leading coefficient can also be written $\binom{r+k-1}{k}$.

In particular, $P(M=0)=p^r$. For integer $r$, probabilities can be calculated successively using $P(M=k+1)/P(M=k)=(r+k)(1-p)/(k+1)$. This recurrence can be useful when several adjacent count probabilities are needed.

Predictive Mean and Variance

The predictive mean and variance account for both count randomness conditional on the rate and uncertainty about the rate. Given the posterior, a gamma variable with shape $r$ and rate $b$ has mean $r/b$ and variance $r/b^2$. Applying conditional expectation and the law of total variance gives the results directly.

\begin{aligned} E[M\mid n] &=E\{E[M\mid\lambda,n]\mid n\} =E[\lambda\mid n] =\frac{r}{b},\\ \operatorname{Var}(M\mid n) &=E\{\operatorname{Var}(M\mid\lambda,n)\mid n\} +\operatorname{Var}\{E[M\mid\lambda,n]\mid n\}\\ &=E[\lambda\mid n]+\operatorname{Var}(\lambda\mid n) =\frac{r}{b}+\frac{r}{b^2}. \end{aligned}

The first variance term, $r/b$, is the expected conditional Poisson variance. The additional term, $r/b^2$, is the posterior variance of the rate. It is positive for a proper gamma posterior, so the predictive count is more variable than a Poisson count with its mean fixed at $r/b$. A Poisson model using only the posterior mean rate would miss this extra predictive uncertainty.

Update the rate distribution.
For a gamma prior $\operatorname{Gamma}(\alpha,\beta)$ and an observed count $n$ over one exposure unit, use the posterior $\operatorname{Gamma}(\alpha+n,\beta+1)$.

Identify the predictive parameters.
Set $r=\alpha+n$, $b=\beta+1$, and $p=b/(b+1)$ for a future count over one exposure unit.

Calculate the requested quantity.
Use the negative binomial probability function for count probabilities, $r/b$ for the mean, and $r/b+r/b^2$ for the variance.

Check the model and interpretation.
Confirm that the future count is conditional on the same risk rate and is for the stated one-unit exposure. Do not substitute a plug-in Poisson variance for the predictive variance.

Worked Predictive Calculations

Worked Example: A Zero-Claim History and Two Future Claims

A hypothetical risk has claim rate $\lambda$ with prior distribution $\operatorname{Gamma}(2,3)$, using shape-rate parameters. It has one claim during one unit of exposure. For a future one-unit period, calculate the probability of no claims, the probability of exactly two claims, and the predictive mean and variance.

Method and update. The observed count is $n=1$, so the posterior shape is $r=2+1=3$ and the posterior rate is $b=3+1=4$. Thus $\lambda\mid n\sim\operatorname{Gamma}(3,4)$, and the negative binomial success probability is $p=4/(4+1)=4/5$. The probability of a failure is $1-p=1/5$.

No claims. For $k=0$, the coefficient is 1: $P(M=0\mid n)=(4/5)^3=64/125=0.5120$.

Exactly two claims. The negative binomial coefficient is $\Gamma(3+2)/(\Gamma(3)2!)=24/(2\cdot2)=6$. Therefore, $P(M=2\mid n)=6(4/5)^3(1/5)^2=6(64/125)(1/25)=384/3125=0.12288$, or $0.1229$ to four decimal places.

Mean and variance. The mean is $r/b=3/4=0.75$. The variance is $3/4+3/4^2=3/4+3/16=15/16=0.9375$.

Check and trap avoided. The probabilities are between 0 and 1, and the variance exceeds the mean by $3/16$, the posterior rate variance. Treating the rate as fixed at $0.75$ would incorrectly give a Poisson variance of $0.75$ and would not give the predictive probabilities above.

Worked Example: Exactly One Future Claim After Two Observed Claims

A hypothetical risk has prior $\lambda\sim\operatorname{Gamma}(1,1)$. It produces two claims over one unit of exposure. Find the probability of exactly one claim in the next one-unit period, along with the predictive mean and variance.

Method and posterior. With $n=2$, the posterior shape is $r=1+2=3$ and rate is $b=1+1=2$. Thus $p=b/(b+1)=2/3$, while $1-p=1/3$.

Probability. For $k=1$, the coefficient is $\Gamma(4)/(\Gamma(3)1!)=6/2=3$. Hence $P(M=1\mid n)=3(2/3)^3(1/3)=3(8/27)(1/3)=8/27=0.296296\ldots$, or $0.2963$ to four decimal places.

Mean and variance. The predictive mean is $r/b=3/2=1.5$. The predictive variance is $3/2+3/2^2=3/2+3/4=9/4=2.25$.

Check and trap avoided. Here $P(M=0)=(2/3)^3=8/27$, which happens to equal $P(M=1)$; the probability of one claim is not $0.3164$. The coefficient for $k=1$ is 3, and the factor $(1-p)^k$ is $1/3$. Omitting either factor gives an incorrect result.

Worked Example: Probability of at Most One Claim

A hypothetical risk has prior $\lambda\sim\operatorname{Gamma}(3,2)$ and records zero claims in one unit of exposure. Calculate the probability of at most one claim during the next one-unit period, and give the predictive mean and variance.

Method and posterior. Since $n=0$, the posterior parameters are $r=3+0=3$ and $b=2+1=3$. Consequently $p=3/4$ and $1-p=1/4$.

Probability of at most one. The zero-claim probability is $P(M=0\mid n)=(3/4)^3=27/64=0.421875$. For exactly one claim, the coefficient is $\Gamma(4)/(\Gamma(3)1!)=6/2=3$, so $P(M=1\mid n)=3(3/4)^3(1/4)=81/256=0.31640625$. Adding the disjoint outcomes gives $P(M\le1\mid n)=27/64+81/256=108/256+81/256=189/256=0.73828125$, or $0.7383$ to four decimal places.

Mean and variance. The mean is $r/b=3/3=1$. The variance is $3/3+3/3^2=1+1/3=4/3=1.3333$ to four decimal places.

Check and trap avoided. The two probabilities sum to less than 1, as they should because counts of two or more remain possible. “At most one” includes both zero and one; reporting only $P(M=1)$ would answer a different question.

Scope, Interpretation, and Common Exam Traps

The calculation depends on the stated model: the rate is constant for the risk, the prior is gamma with the rate parameterization shown, and counts are Poisson conditional on that rate. The posterior predictive result integrates over uncertainty about the risk’s rate; it is not the same as conditioning on a single estimated rate. It also describes one future unit of exposure in these examples. Different exposure amounts require the corresponding Poisson exposure term in both updating and prediction; do not silently apply the one-unit formulas to a different period.

Confusing rate and scale. In $\operatorname{Gamma}(\alpha,\beta)$, $\beta$ is the rate, so the mean is $\alpha/\beta$, not $\alpha\beta$.
Updating the posterior but not the prediction. The future count adds one unit to the exponential term when the predictive probability is integrated. That is why the negative binomial probability uses $b+1$, not $b$, in its denominator.
Reversing the negative binomial convention. In the probability function used here, $k$ is the number of failures before the $r$th success, and $p=b/(b+1)$ is the success probability.
Using the wrong variance. The predictive variance is $r/b+r/b^2$. The first term alone is the expected conditional Poisson variance; the second reflects uncertainty in $\lambda$.
Misreading a count event. “At most one” includes 0 and 1; “at least one” is $1-P(M=0)$. Translate the wording into the count event before calculating.

Key takeaway: With a gamma posterior $\operatorname{Gamma}(r,b)$, a future one-unit Poisson count has negative binomial probabilities with $p=b/(b+1)$, mean $r/b$, and variance $r/b+r/b^2$. The extra variance term is the predictive cost of rate uncertainty.

Check Your Understanding

For each problem, use the shape-rate gamma parameterization and assume one unit of observed exposure and one unit of future exposure. Give probabilities to four decimal places.

A risk has prior $\operatorname{Gamma}(2,3)$ and observes zero claims. Find the probability of zero future claims and the predictive mean. (Answer: $P(M=0)=0.6400$; mean $=0.5000$.)
A risk has prior $\operatorname{Gamma}(1,1)$ and observes two claims. Find the probability of exactly one future claim. (Answer: $8/27=0.2963$.)
A risk has prior $\operatorname{Gamma}(2,2)$ and observes one claim. Find the probability of exactly two future claims. (Answer: $0.1582$.)
A risk has prior $\operatorname{Gamma}(3,1)$ and observes one claim. Find the probability of at most one future claim and the predictive variance. (Answer: $P(M\le1)=112/243=0.4609$; variance $=3.0000$.)
A risk has prior $\operatorname{Gamma}(2,3)$ and observes one claim. Find the probability of at least two future claims, then give the predictive mean and variance. (Answer: $P(M\ge2)=0.1808$; mean $=0.7500$; variance $=0.9375$.)

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