Beta-Binomial Predictive Counts and Premiums

Why the Predictive Distribution Matters

A policyholder’s claim probability is generally not known with certainty. A beta prior represents uncertainty about that probability, and observed claim indicators update the prior. When predicting future claims, the updated probability remains uncertain: the predictive distribution must average over its posterior distribution, rather than treat one estimated probability as known.

This distinction matters for both probabilities and premiums. A predictive probability such as the chance of at least two claims accounts for uncertainty about the underlying claim probability. The expected future count also averages over that uncertainty. In a simple hypothetical product with a fixed payment for each claim, multiplying the expected count by the payment per claim gives the expected claim payment.

Model and assumptions: For one risk, let $\theta$ be its unknown probability of a claim in each specified trial period. Conditional on $\theta$, claim indicators in distinct periods are independent Bernoulli random variables with probability $\theta$. The prior is $\theta\sim\operatorname{Beta}(\alpha,\beta)$, where $\alpha>0$ and $\beta>0$. The past and future periods use the same claim probability, and future indicators are conditionally independent of past indicators given $\theta$.

If $n$ past periods contain $k$ claims, then there are $n-k$ nonclaims. The beta-binomial conjugate update gives the posterior $\theta\mid\text{data}\sim\operatorname{Beta}(a,b)$, where $a=\alpha+k$ and $b=\beta+n-k$. This posterior update is the starting point for prediction; the remaining task is to average the future binomial probabilities over it.

Deriving the Beta-Binomial Predictive Distribution

Let $K$ be the number of claims in $m$ future periods. Conditional on $\theta$, $K\mid\theta$ has a binomial distribution with $m$ trials and success probability $\theta$. The predictive probability is obtained by integrating that conditional probability against the posterior density of $\theta$.

P(K=j\mid\text{data}) =\int_0^1 {m\choose j}\theta^j(1-\theta)^{m-j} \frac{\theta^{a-1}(1-\theta)^{b-1}}{B(a,b)}\,d\theta ={m\choose j}\frac{B(a+j,b+m-j)}{B(a,b)}, \quad j=0,1,\ldots,m.

Here $B(a,b)=\int_0^1 t^{a-1}(1-t)^{b-1}\,dt$ is the beta function. The integral follows by collecting powers of $\theta$: the integrand’s beta-kernel powers become $\theta^{a+j-1}(1-\theta)^{b+m-j-1}$. Its integral is $B(a+j,b+m-j)$, yielding the displayed result after division by $B(a,b)$.

Definition: If the posterior claim probability is $\operatorname{Beta}(a,b)$, the future count $K$ has a beta-binomial predictive distribution with parameters $m,a,b$: $P(K=j\mid\text{data})={m\choose j}B(a+j,b+m-j)/B(a,b)$ for $j=0,\ldots,m$. The parameters $a,b$ describe the posterior, not the original prior, unless no data have been observed.

For hand calculations, define the rising factorial $(x)^{\overline r}=x(x+1)\cdots(x+r-1)$, with $(x)^{\overline 0}=1$. The beta-function ratio can then be written as $B(a+j,b+m-j)/B(a,b)=(a)^{\overline j}(b)^{\overline{m-j}}/(a+b)^{\overline m}$. This makes small predictive probabilities straightforward to evaluate without computing beta integrals.

Theorem: Given a $\operatorname{Beta}(a,b)$ posterior and $m$ future periods, $E[K\mid\text{data}]=m a/(a+b)$, and $\operatorname{Var}(K\mid\text{data})=mab(a+b+m)/((a+b)^2(a+b+1))$.

For the mean, conditional expectation gives $E[K\mid\theta]=m\theta$. Averaging over the posterior therefore gives $E[K\mid\text{data}]=mE[\theta\mid\text{data}]=ma/(a+b)$. For the variance, the law of total variance gives $\operatorname{Var}(K\mid\text{data})=E[m\theta(1-\theta)\mid\text{data}]+\operatorname{Var}(m\theta\mid\text{data})$. Using $E[\theta]=a/(a+b)$ and $\operatorname{Var}(\theta)=ab/((a+b)^2(a+b+1))$ for a beta distribution and simplifying gives the stated expression. The additional variation reflects uncertainty about $\theta$; conditional independence of future periods does not make them independent after that uncertainty is averaged out.

Worked Predictive Calculations

Worked Example: Predicting a Short-Term Claim Count

A hypothetical insurer models whether a small commercial device has a covered incident in each year. The incident probability $\theta$ has prior distribution $\operatorname{Beta}(2,3)$. Over four observed years there is one claim. Predict the number of claims in the next three years, including the probability of no claims and the expected claim count. If the product pays a fixed $120 per claim, calculate the expected claim payment for those three years.

Method and update. There are $n=4$ observed years, $k=1$ claim and $4-1=3$ nonclaims. The posterior parameters are $a=2+1=3$ and $b=3+3=6$, so $\theta\mid\text{data}\sim\operatorname{Beta}(3,6)$.

Probability of no future claims. For $m=3$ and $j=0$, the beta-binomial formula gives $P(K=0\mid\text{data})=B(3,9)/B(3,6)=(6)^{\overline 3}/(9)^{\overline 3}$. The numerator is $6\cdot7\cdot8=336$, and the denominator is $9\cdot10\cdot11=990$. Thus the probability is $336/990=56/165=0.3394$ to four decimal places.

Expected count and payment. The posterior mean claim probability is $3/(3+6)=1/3$. The expected claim count is $3(1/3)=1$. At $120 per claim, the expected payment is $1\cdot\$120=\$120.00$.

Check and trap avoided. The count is between zero and three, and the expected count of one is consistent with an average future probability of $1/3$. A plug-in binomial calculation would give $(2/3)^3=0.2963$ for no claims, not the correct predictive probability $0.3394$. The beta-binomial calculation averages over uncertainty in $\theta$; treating its posterior mean as a known probability does not.

Worked Example: Probability of Exactly One Claim

A hypothetical roadside-assistance plan assigns a claim probability $\theta$ with prior distribution $\operatorname{Beta}(1,4)$ for each covered month. A customer has two claims in five observed months. Find the probability of exactly one claim in the next two months and the expected number of claims in those months. If each claim incurs a fixed $250 payment, find the expected payment.

Method and update. The observed record has $k=2$ claims and $5-2=3$ nonclaims. The posterior is $\operatorname{Beta}(a,b)$, with $a=1+2=3$ and $b=4+3=7$.

Exactly one future claim. Set $m=2$ and $j=1$: $P(K=1\mid\text{data})={2\choose1}B(4,8)/B(3,7)$. Using rising factorials, this is $2(3)^{\overline1}(7)^{\overline1}/(10)^{\overline2}$. The numerator is $2\cdot3\cdot7=42$, and the denominator is $10\cdot11=110$. Therefore the probability is $42/110=21/55=0.3818$ to four decimal places.

Expected count and payment. The posterior mean probability is $3/(3+7)=0.3$. Thus $E[K\mid\text{data}]=2(0.3)=0.6$ claims. The expected fixed claim payment is $0.6\cdot\$250=\$150.00$.

Check and trap avoided. The probability of exactly one claim is below one, and the expected count is less than the maximum of two. Do not interpret $0.6$ expected claims as a 60% probability of exactly one claim: the count can also be zero or two.

Worked Example: At Least Two Claims and a Count-Based Premium

For a hypothetical equipment-protection contract, let the annual incident probability have prior distribution $\operatorname{Beta}(2,2)$. The first three years have no claims. Calculate the probability of at least two claims in the next four years, the expected number of those claims, and the expected payment if the contract pays a fixed $90 per claim.

Method and update. Here $n=3$ and $k=0$, so the posterior is $\operatorname{Beta}(2+0,2+3)=\operatorname{Beta}(2,5)$. For $m=4$, use the complement $P(K\ge2)=1-P(K=0)-P(K=1)$.

Zero claims. $P(K=0\mid\text{data})=(5)^{\overline4}/(7)^{\overline4}$. The numerator is $5\cdot6\cdot7\cdot8=1680$, and the denominator is $7\cdot8\cdot9\cdot10=5040$. Thus $P(K=0)=1/3$.

Exactly one claim. $P(K=1\mid\text{data})={4\choose1}(2)^{\overline1}(5)^{\overline3}/(7)^{\overline4}$. The numerator is $4\cdot2\cdot(5\cdot6\cdot7)=1680$, and the denominator is $5040$. Thus $P(K=1)=1/3$. Consequently, $P(K\ge2)=1-1/3-1/3=1/3=0.3333$ to four decimal places.

Expected count and payment. The posterior mean probability is $2/(2+5)=2/7$. The expected count is $4(2/7)=8/7$. The expected payment is $(8/7)\cdot\$90=\$720/7=\$102.86$, rounded to the nearest cent.

Check and trap avoided. The expected count $8/7$ is between zero and four, and the probability of at least two claims is between zero and one. The expected payment uses the expected count, not the probability of at least two claims: it includes payments for one-claim outcomes as well as multiple-claim outcomes.

Using the Predictive Count in a Premium

If each future claim generates a fixed payment $c$, with no other payments or expenses included, the total payment is $cK$. Linearity of expectation gives $E[cK\mid\text{data}]=c\,ma/(a+b)$. This is a count-based expected payment under the stated fixed-payment assumption, not a general premium formula for variable claim severities or other expenses.

The predictive distribution is also useful when a decision depends on a threshold, such as the chance of at least two claims. In that case, calculate the relevant probability from the beta-binomial probabilities, often by summing the appropriate terms or using a complement. Do not substitute the expected count for a threshold probability: an expected value and a probability answer different questions.

Key takeaway: Update the beta parameters using the observed claims and nonclaims, then mix the future binomial distribution over that posterior. The resulting beta-binomial distribution gives predictive probabilities; its mean gives expected future claims and, under a fixed-payment assumption, a count-based expected payment.

Check Your Understanding

For each problem, use the posterior beta distribution as the mixing distribution for the stated future periods. Round probabilities to four decimal places and dollar amounts to the nearest cent.

A hypothetical risk has prior $\operatorname{Beta}(2,3)$. In two observed periods there is one claim. Find the posterior parameters, the expected count in the next two periods, and the probability of no claims in those two periods. (Answer: posterior $\operatorname{Beta}(3,4)$; expected count $6/7=0.8571$; $P(K=0)=5/14=0.3571$.)
A hypothetical monthly policy has prior $\operatorname{Beta}(1,2)$, and records two claims in three months. Find the posterior probability of a claim and the probability of exactly one claim in the next month. (Answer: posterior $\operatorname{Beta}(3,3)$; probability $0.5000$.)
A hypothetical account has prior $\operatorname{Beta}(2,5)$, with one claim in four observed periods. Find the probability of no claims in the next three periods. (Answer: posterior $\operatorname{Beta}(3,8)$; $P(K=0)=60/143=0.4196$.)
A hypothetical warranty risk has prior $\operatorname{Beta}(3,2)$, with no claims in two observed periods. Find the probability of two claims in the next two periods and the expected count in those periods. (Answer: posterior $\operatorname{Beta}(3,4)$; $P(K=2)=3/14=0.2143$; expected count $6/7=0.8571$.)
A hypothetical contract has prior $\operatorname{Beta}(2,2)$, with one claim in two observed periods. If the contract pays a fixed $75 per claim, calculate the expected payment for the next four periods. (Answer: posterior $\operatorname{Beta}(3,3)$; expected count $4(3/6)=2$; expected payment $150.00.)

What You'll Learn