Beta-Binomial Prediction of Claim Incidence

Why Predict Claim Incidence With a Distribution?

A forecast for future claim incidence must account for two sources of variation: whether each future risk produces a claim, and uncertainty about the underlying claim probability. If that probability were known, a binomial model would describe the first source. When it is not known, a beta prior updated with observed experience provides a distribution for the probability itself. Averaging the binomial forecast over that posterior gives the beta-binomial predictive distribution.

This model is useful when each exposure unit has a binary outcome over a specified period: claim or no claim. It predicts the number of claims among a fixed number of future units. It does not describe claim severity, multiple claims per unit, or differing probabilities across future risks.

Definition: Suppose the claim probability $p$ has a beta prior with parameters $\alpha>0$ and $\beta>0$, and, conditional on $p$, each of $n$ observed risks independently has a Bernoulli claim outcome. If $k$ claims are observed, the posterior distribution is $\operatorname{Beta}(a,b)$, where $a=\alpha+k$ and $b=\beta+n-k$. For $m$ future risks with the same conditional independence and probability assumptions, their total claim count $K$, after averaging over this posterior, has a beta-binomial predictive distribution.

From Observed Experience to a Predictive Distribution

Let $p$ be the unknown probability of a claim for one risk during the specified period. The prior $\operatorname{Beta}(\alpha,\beta)$ has density proportional to $p^{\alpha-1}(1-p)^{\beta-1}$ for $0<p<1$. Given $k$ claims among $n$ observed risks, the binomial likelihood is proportional to $p^k(1-p)^{n-k}$. Multiplying the prior density by the likelihood gives a posterior density proportional to

p^{\alpha+k-1}(1-p)^{\beta+n-k-1}.

This is the kernel of a beta density, so the posterior parameters are $a=\alpha+k$ and $b=\beta+n-k$. Conditional on a particular value of $p$, the future count has a binomial distribution: $K\mid p\sim\operatorname{Binomial}(m,p)$. To obtain the predictive probability of $j$ future claims, average that conditional probability over the posterior distribution of $p$.

P(K=j\mid \text{observed data}) =\binom{m}{j}\frac{B(a+j,b+m-j)}{B(a,b)},\qquad j=0,1,\ldots,m,

where $B(r,s)=\int_0^1 p^{r-1}(1-p)^{s-1}\,dp$ is the beta function. This formula is the beta-binomial probability mass function. Equivalently, writing $(x)_r=x(x+1)\cdots(x+r-1)$ for a rising factorial, the probability is

P(K=j\mid \text{observed data}) =\binom{m}{j}\frac{(a)_j(b)_{m-j}}{(a+b)_m}.

The beta-function form follows directly by integrating the conditional binomial probability. The binomial coefficient does not depend on $p$, so it remains outside the integral:

\begin{aligned} P(K=j\mid \text{data}) &=\int_0^1 \binom{m}{j}p^j(1-p)^{m-j} \frac{p^{a-1}(1-p)^{b-1}}{B(a,b)}\,dp\\ &=\binom{m}{j}\frac{1}{B(a,b)} \int_0^1 p^{a+j-1}(1-p)^{b+m-j-1}\,dp\\ &=\binom{m}{j}\frac{B(a+j,b+m-j)}{B(a,b)}. \end{aligned}

The integral is a beta function with parameters $a+j$ and $b+m-j$, which proves the result. This integration is essential: the predictive probability is not generally the binomial probability evaluated at the posterior mean of $p$.

Predictive Mean and Variance

The posterior mean of $p$ is $a/(a+b)$. For the future count, the predictive mean is $m$ times that posterior mean. Its variance includes both binomial variation conditional on $p$ and uncertainty about $p$.

Theorem: If $p\mid\text{data}\sim\operatorname{Beta}(a,b)$ and $K\mid p\sim\operatorname{Binomial}(m,p)$, then $$ E[K\mid\text{data}]=\frac{ma}{a+b},\qquad \operatorname{Var}(K\mid\text{data})= \frac{mab(a+b+m)}{(a+b)^2(a+b+1)}. $$

To justify the variance, set $s=a+b$. The law of total variance gives $\operatorname{Var}(K\mid\text{data})=E[\operatorname{Var}(K\mid p,\text{data})\mid\text{data}]+\operatorname{Var}(E[K\mid p,\text{data}]\mid\text{data})$. Since the conditional mean and variance are $mp$ and $mp(1-p)$, respectively, and a $\operatorname{Beta}(a,b)$ variable has $E[p]=a/s$, $\operatorname{Var}(p)=ab/[s^2(s+1)]$, and $E[p(1-p)]=ab/[s(s+1)]$, we obtain

\begin{aligned} \operatorname{Var}(K\mid\text{data}) &=m\frac{ab}{s(s+1)}+m^2\frac{ab}{s^2(s+1)}\\ &=\frac{mab(s+m)}{s^2(s+1)}. \end{aligned}

The second variance component reflects uncertainty in the shared claim probability. Conditional on $p$, future outcomes are independent; after $p$ is averaged out, the outcomes are positively associated. That is why predictive variance can exceed the variance from a binomial model using the posterior mean as if it were known.

Worked Predictive Calculations

Worked Example: Probability of Exactly One Future Claim

A hypothetical insurer models whether each small commercial building has a claim during the next year. The prior claim probability is $\operatorname{Beta}(2,8)$. Among 10 observed buildings, 3 had claims. For 4 future buildings assumed to have the same claim probability, find the probability of exactly one claim.

Method and posterior. The beta-binomial method applies because each risk has a binary outcome and the risks are conditionally independent given a common $p$. Add the observed claims and nonclaims to the prior parameters: $a=2+3=5$ and $b=8+(10-3)=15$. Thus the posterior is $\operatorname{Beta}(5,15)$.

For $m=4$ and $j=1$, use the rising-factorial formula:

P(K=1\mid\text{data}) =\binom{4}{1}\frac{(5)_1(15)_3}{(20)_4} =\frac{4(5)(15)(16)(17)}{(20)(21)(22)(23)}.

The numerator is $4\cdot5\cdot15\cdot16\cdot17=81{,}600$. The denominator is $20\cdot21\cdot22\cdot23=212{,}520$. Therefore the probability is $81{,}600/212{,}520=0.383963\ldots$, or 0.3840 to four decimals.

Check: The result is between 0 and 1, and exactly one claim is one of the possible counts from 0 through 4. The common mistake avoided is using the original prior parameters rather than the posterior parameters after observing the 3 claims and 7 nonclaims.

Worked Example: Probability of at Least Two Claims

A hypothetical insurer starts with a $\operatorname{Beta}(1,9)$ prior for the annual claim probability among a defined group of devices. It observes 2 claims among 6 devices. For 3 future devices from the same group, calculate the probability of at least 2 claims.

Method and posterior. The posterior parameters are $a=1+2=3$ and $b=9+(6-2)=13$, giving $\operatorname{Beta}(3,13)$. At least 2 claims means $K=2$ or $K=3$, so calculate and add those two beta-binomial probabilities.

\begin{aligned} P(K=2\mid\text{data}) &=\binom{3}{2}\frac{(3)_2(13)_1}{(16)_3} =\frac{3(3)(4)(13)}{16(17)(18)} =\frac{468}{4896},\\ P(K=3\mid\text{data}) &=\binom{3}{3}\frac{(3)_3(13)_0}{(16)_3} =\frac{3(4)(5)}{16(17)(18)} =\frac{60}{4896}. \end{aligned}

The denominator is $16\cdot17\cdot18=4{,}896$. Adding the numerators gives $468+60=528$, so the requested probability is $528/4{,}896=11/102=0.107843\ldots$, or 0.1078 to four decimals.

Check: The event includes both possible counts 2 and 3; omitting the probability of 3 would understate the answer. The result is also below 1, as required for a probability.

Worked Example: An Exact Count and Predictive Variance

For a hypothetical policy group, the prior claim probability is $\operatorname{Beta}(4,6)$. The experience is 5 claims among 8 policies. For 5 future policies, calculate the probability of exactly 2 claims and the predictive variance of the count.

Method and posterior. Updating gives $a=4+5=9$ and $b=6+(8-5)=9$. The posterior is $\operatorname{Beta}(9,9)$, with $s=a+b=18$. For exactly 2 claims:

P(K=2\mid\text{data}) =\binom{5}{2}\frac{(9)_2(9)_3}{(18)_5} =\frac{10(9)(10)(9)(10)(11)} {18(19)(20)(21)(22)} =\frac{891{,}000}{3{,}160{,}080} =\frac{825}{2{,}926} =0.281954\ldots.

Thus $P(K=2\mid\text{data})=\mathbf{0.2820}$ to four decimals. For the variance, substitute $m=5$, $a=b=9$, and $s=18$:

\operatorname{Var}(K\mid\text{data}) =\frac{5(9)(9)(18+5)}{18^2(19)} =\frac{9{,}315}{6{,}156} =1.51267\ldots.

The predictive mean is $5(9/18)=2.5$. The variance is about $1.5127$ claims squared. For comparison, treating the posterior mean $p=0.5$ as known would give binomial variance $5(0.5)(0.5)=1.25$, which is smaller because it omits uncertainty about $p$.

Check: The posterior is symmetric, so the expected future count is half of 5, consistent with 2.5. The method avoids using a fixed-probability binomial variance as the full predictive variance.

Common Exam Trap: Plugging In the Posterior Mean

A tempting shortcut is to calculate the posterior mean $a/(a+b)$, then use a binomial distribution with that fixed value. This can give a reasonable mean forecast, but it is not the beta-binomial predictive distribution: it discards uncertainty about the probability. The distinction can change a tail probability.

In the second worked example, the posterior mean is $3/(3+13)=3/16=0.1875$. The plug-in calculation for at least 2 claims among 3 would be

\binom{3}{2}(0.1875)^2(0.8125)+(0.1875)^3 =0.092285\ldots.

The correct beta-binomial probability is $0.107843\ldots$, not $0.092285\ldots$. The correction is to average the conditional binomial probability over the full beta posterior, as in the beta-binomial formula. A plug-in binomial forecast is appropriate only when the claim probability is treated as known, rather than uncertain under the posterior.

Model Conditions and Interpretation

The predictive distribution relies on a specific structure. The historical and future risks must be represented by binary outcomes over the same defined period, and the model assumes a common unknown claim probability $p$. Conditional on $p$, outcomes are independent with that probability. The beta prior is updated using the observed number of claims and nonclaims. If future risks have different probabilities, or if the outcomes are claim counts that can exceed one per risk, this beta-binomial model does not directly apply.

The model makes the future outcomes exchangeable after $p$ is integrated out: before seeing their outcomes, no future risk is favored over another, but their outcomes are positively associated through uncertainty about the shared $p$. In ratemaking, that distinction matters when a forecast concerns an entire group. A posterior mean alone describes an average incidence level; a posterior predictive distribution also assigns probabilities to possible group claim counts.

Check Your Understanding

For each problem, update the beta parameters first, then use the predictive distribution for the stated number of future risks.

A hypothetical portfolio has prior $\operatorname{Beta}(2,5)$ claim probability. It observes 1 claim among 4 risks. For 2 future risks, find the probability of exactly 1 claim. (Answer: $4/11=0.3636$)
A prior is $\operatorname{Beta}(3,7)$, and the observed experience is 0 claims among 5 risks. For 4 future risks, find the probability of no claims. (Answer: $91/204=0.4461$)
A hypothetical device group has prior $\operatorname{Beta}(1,4)$ and observes 2 claims among 5 devices. For 3 future devices, find the probability of exactly 2 claims. (Answer: $21/110=0.1909$)
After updating, a portfolio's claim probability has posterior $\operatorname{Beta}(4,6)$. For 4 future risks, find the probability of at least one claim. (Answer: $1-\frac{6\cdot7\cdot8\cdot9}{10\cdot11\cdot12\cdot13}=0.8238$)
A prior is $\operatorname{Beta}(1,3)$, and a hypothetical group has 1 claim among 2 observed risks. For 2 future risks, find the probability both have claims. (Answer: $1/7=0.1429$)

What You'll Learn