Deriving Beta-Binomial Credibility Estimates

Why the Beta-Binomial Update Is a Credibility Estimate

For a risk with a fixed number of opportunities to claim, the observed claim proportion is informative but can be unstable when there are few trials. A prior distribution represents the starting uncertainty about the risk’s claim probability. Updating that prior with the observed outcomes produces a posterior estimate that combines the two sources of information.

The Beta-Binomial model makes this update especially transparent: the posterior distribution is another Beta distribution, and its mean is a weighted average of the observed claim proportion and the prior mean. The weights depend on the number of trials and the prior’s concentration. This result is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Model and assumptions: Let $\theta$ be a risk’s unknown probability of a claim on one opportunity. Before observing outcomes, assume $\theta\sim\operatorname{Beta}(\alpha,\beta)$, where $\alpha>0$ and $\beta>0$. Conditional on $\theta$, there are $n$ fixed, comparable trials, each with an independent Bernoulli outcome and claim probability $\theta$. Let $K$ be the number of claims observed, so $K=k$ and the number of nonclaims is $n-k$. The model does not directly apply when trial probabilities differ, outcomes are conditionally dependent, or the observation process changes the likelihood.

From Prior and Likelihood to Posterior

The Beta prior has density proportional to $\theta^{\alpha-1}(1-\theta)^{\beta-1}$ for $0<\theta<1$. Under the stated fixed-trial model, the likelihood of observing $k$ claims in $n$ trials is

P(K=k\mid\theta)=\binom{n}{k}\theta^k(1-\theta)^{n-k}.

Bayes’ rule says that the posterior density is proportional to the prior density multiplied by the likelihood. The binomial coefficient does not depend on $\theta$, so it is part of the normalizing constant rather than the posterior’s parameter-dependent kernel.

\begin{aligned} f(\theta\mid K=k) &\propto \theta^{\alpha-1}(1-\theta)^{\beta-1} \theta^k(1-\theta)^{n-k}\\ &= \theta^{\alpha+k-1}(1-\theta)^{\beta+n-k-1}. \end{aligned}

This has the form of a Beta density. Claims add to the first parameter; nonclaims add to the second. Therefore the posterior distribution is

\theta\mid K=k\sim\operatorname{Beta}(\alpha+k,\ \beta+n-k).

Posterior mean: If $\theta\mid K=k\sim\operatorname{Beta}(a,b)$, then $E[\theta\mid K=k]=a/(a+b)$. Thus, for the fixed-trial Beta-Binomial model, $$ E[\theta\mid K=k]=\frac{\alpha+k}{\alpha+\beta+n}. $$

Why the Posterior Mean Is a Weighted Average

Let $s=\alpha+\beta$ denote the prior parameter total. The prior mean is $\alpha/s$, and the observed claim proportion, when $n>0$, is $k/n$. Split the posterior mean’s numerator into its prior and observed parts:

\begin{aligned} E[\theta\mid K=k] &=\frac{\alpha+k}{s+n}\\ &=\frac{s}{s+n}\frac{\alpha}{s} +\frac{n}{s+n}\frac{k}{n}. \end{aligned}

The equality follows because the right-hand numerator, after using denominator $s+n$, is $\alpha+k$. The weights are nonnegative and sum to one: $s/(s+n)+n/(s+n)=1$. Hence the posterior mean lies between the prior mean and observed proportion. More trials increase the observed proportion’s weight; a more concentrated prior increases the prior mean’s weight.

The Beta parameters can be understood as prior claim and nonclaim counts in this algebraic sense: $\alpha$ contributes to the claim component, and $\beta$ to the nonclaim component. They need not be literal historical observations. The ratio $s=\alpha+\beta$ acts as the prior weight in this mean formula; it is not a claim count by itself.

For $n=0$, the observed proportion $k/n$ is undefined, so the weighted-proportion expression is not used. The posterior is simply the prior, $\operatorname{Beta}(\alpha,\beta)$, and its mean remains $\alpha/(\alpha+\beta)$.

Worked Applications

Worked Example: A Small Block of Claim Trials

A hypothetical equipment warranty has claim probability $\theta$ per covered unit over a specified inspection interval. The prior is $\operatorname{Beta}(2,10)$. Among 24 independent, comparable units, 3 generate a claim. Find the posterior distribution and posterior mean, then verify the credibility-weighted form. No interest or mortality assumption is relevant; the outcome model is Bernoulli conditional on $\theta$, with a Beta prior as stated.

Method. Apply the fixed-trial Beta-Binomial update. There are $n=24$ trials, $k=3$ claims, and $24-3=21$ nonclaims. Thus

\theta\mid K=3\sim\operatorname{Beta}(2+3,\ 10+21) =\operatorname{Beta}(5,31).

The posterior parameter total is $5+31=36$, so the posterior mean is $5/36=0.138888\ldots$, or $0.1389$ to four decimal places. The observed proportion is $3/24=0.125$, and the prior mean is $2/(2+10)=1/6$.

The prior weight is $12/(12+24)=1/3$; the observed-data weight is $24/36=2/3$. The weighted estimate is

\frac{1}{3}\left(\frac{1}{6}\right) +\frac{2}{3}\left(\frac{3}{24}\right) =\frac{1}{18}+\frac{1}{12} =\frac{2}{36}+\frac{3}{36} =\frac{5}{36}.

The two calculations agree, checking the update. Common mistake avoided: adding only the 3 claims to the prior and forgetting the 21 nonclaims would give the wrong second posterior parameter. The nonclaims update $\beta$, not $\alpha$.

Worked Example: Unequal Prior and Observed Claim Proportions

A hypothetical roadside-assistance program models whether a service call occurs during one defined coverage window. Its prior claim probability is represented by $\operatorname{Beta}(3,9)$. In a fixed sample of 40 comparable coverage windows, 6 have a claim. Find the posterior mean and quantify the contribution from each proportion. Assume conditional independence and a common $\theta$ for all 40 windows; no interest or mortality assumption is involved.

Method. The number of nonclaims is $40-6=34$. The posterior distribution is therefore

\theta\mid K=6\sim\operatorname{Beta}(3+6,\ 9+34) =\operatorname{Beta}(9,43).

The parameter total is $9+43=52$, giving posterior mean $9/52=0.1730769\ldots$, or $0.1731$ to four decimal places. The prior mean is $3/12=0.25$, and the observed claim proportion is $6/40=0.15$. The prior weight is $12/52=3/13$, while the data weight is $40/52=10/13$.

\frac{3}{13}(0.25)+\frac{10}{13}(0.15) =\frac{3}{52}+\frac{6}{52} =\frac{9}{52} =0.1730769\ldots

The posterior mean is closer to the observed proportion because the 40 trials receive more weight than the prior total of 12. The arithmetic checks against the posterior parameters. Common mistake avoided: taking the unweighted average $(0.25+0.15)/2=0.20$ incorrectly gives equal influence to the prior and the 40 observations.

Worked Example: Combining Two Groups of Comparable Trials

A hypothetical insurer reviews two batches of the same type of scheduled device inspection, with one binary claim outcome per inspection. The common claim probability has prior $\operatorname{Beta}(4,16)$. The first batch has 30 inspections and 2 claims; the second has 50 inspections and 6 claims. Assuming all 80 outcomes are conditionally independent with the same $\theta$, find the posterior mean. No interest or mortality assumption applies.

Method. Add the trial counts and claim counts before updating: $n=30+50=80$, and $k=2+6=8$. The total number of nonclaims is $80-8=72$. The likelihood contributions multiply, producing a parameter-dependent factor $\theta^{2+6}(1-\theta)^{28+44}=\theta^8(1-\theta)^{72}$. The posterior is

\theta\mid\text{both batches} \sim\operatorname{Beta}(4+8,\ 16+72) =\operatorname{Beta}(12,88).

Its parameter total is $12+88=100$, so the posterior mean is $12/100=0.12$. The prior mean is $4/20=0.20$, and the pooled observed proportion is $8/80=0.10$. The prior and data weights are $20/100=0.20$ and $80/100=0.80$, respectively:

$$ 0.20(0.20)+0.80(0.10)=0.04+0.08=0.12. $$

The group counts and the posterior parameters give the same answer. Common mistake avoided: averaging the two batch proportions equally would ignore their different sizes. Here the pooled proportion is $(2+6)/(30+50)=0.10$, not $\big((2/30)+(6/50)\big)/2$.

Conditions, Interpretation, and Exam Traps

The weighted-average result depends on the model, not just on the appearance of the formula. All trials must be comparable conditional on the same $\theta$, and the outcomes must have the specified Bernoulli likelihood. If one group has a different underlying probability, treating all its trials as sharing one $\theta$ can be inappropriate. If the number of trials is not fixed by the design, the likelihood for the observation process may also need attention; the fixed-$n$ derivation above should not be applied automatically.

The posterior mean is a summary of uncertainty about the risk’s probability. It is not the posterior distribution itself. Two different Beta posteriors can have the same mean but different parameter totals and therefore different spreads. For the estimate requested here, however, the posterior mean is exactly the credibility-weighted blend derived above.

A frequent exam error is to use $n$ as the denominator for the prior weight or to count only claims in the posterior. A reliable sequence is: identify $n$, identify $k$, calculate $n-k$, update both Beta parameters, then compute the posterior mean. Finally, compare it with the weighted-average expression. The mean must be between the prior mean and the observed claim proportion when $n>0$; a result outside that interval signals an arithmetic or modeling error.

Record the trial totals.
Write $n$ for the fixed number of trials, $k$ for claims, and $n-k$ for nonclaims.

Update each prior parameter.
From $\operatorname{Beta}(\alpha,\beta)$, form $\operatorname{Beta}(\alpha+k,\beta+n-k)$.

Calculate and check the mean.
Use $(\alpha+k)/(\alpha+\beta+n)$, then verify it as the weighted average of prior and observed proportions.

Key takeaway: Under a fixed number of conditionally independent Bernoulli trials with a shared claim probability and a Beta prior, claims increment $\alpha$, nonclaims increment $\beta$, and the posterior mean weights the observed proportion by $n/(\alpha+\beta+n)$ and the prior mean by $(\alpha+\beta)/(\alpha+\beta+n)$.

Check Your Understanding

For each problem, use the fixed-trial assumptions stated and check the posterior mean using both the parameter update and the weighted-average form.

A hypothetical portfolio segment has prior $\operatorname{Beta}(2,8)$. In 30 comparable independent trials, 3 claims occur. Give the posterior distribution and mean. (Answer: $\operatorname{Beta}(5,35)$; mean $5/40=0.1250$.)
A prior is $\operatorname{Beta}(4,12)$. A fixed set of 20 trials produces 5 claims. What is the posterior mean? (Answer: $\operatorname{Beta}(9,27)$; mean $9/36=0.2500$.)
For prior $\operatorname{Beta}(3,7)$, no claims occur in 25 trials. Find the posterior and posterior mean. (Answer: $\operatorname{Beta}(3,32)$; mean $3/35=0.0857$.)
A hypothetical inspection process has prior $\operatorname{Beta}(2,10)$. One batch has 12 trials and 1 claim; another has 28 trials and 5 claims. Assuming a shared conditional claim probability, find the posterior distribution and mean. (Answer: $\operatorname{Beta}(8,44)$; mean $8/52=2/13\approx0.1538$.)
A risk has prior $\operatorname{Beta}(1,9)$, followed by 19 fixed trials with 4 claims. Find the posterior mean and its prior-versus-data weights. (Answer: posterior $\operatorname{Beta}(5,24)$, mean $5/29\approx0.1724$; prior weight $10/29$, data weight $19/29$.)

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