Beta Priors for Binomial Claim Probabilities

Why Update a Claim Probability?

Suppose a hypothetical insurer classifies each eligible exposure as having either a claim or no claim during a fixed observation period. The underlying claim probability may differ across risks, and the insurer represents uncertainty about one risk’s probability by a probability distribution. A beta prior is useful because it can be updated directly from binomial observations.

The key calculation is simple once the assumptions are clear: add observed claims to the first beta shape parameter and add observed nonclaims to the second. The posterior distribution then summarizes uncertainty about the risk’s claim probability after observing the data. This tutorial develops that update, calculates posterior moments, and explains how to interpret the posterior mode, including cases where the density is unbounded at an endpoint. The treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: A random probability $p$ has a beta distribution with shape parameters $a>0$ and $b>0$, written $p\sim\operatorname{Beta}(a,b)$, if its density for $0<p<1$ is $$ f(p)=\frac{p^{a-1}(1-p)^{b-1}}{B(a,b)}, $$ where $B(a,b)=\int_0^1p^{a-1}(1-p)^{b-1}\,dp=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. The parameters $a$ and $b$ are positive shape parameters; they need not be integers.

Model and Posterior Update

Let $p$ be the fixed but unknown claim probability for a particular risk during the observation period. Conditional on $p$, suppose there are $n$ trials, each with a claim indicator equal to 1 for a claim and 0 for no claim. Assume the indicators are conditionally independent and each has probability $p$ of a claim. If $X$ is the number of claims, then $X\mid p\sim\operatorname{Binomial}(n,p)$. The model requires a common claim probability across these trials and a consistent definition of a trial and claim.

Theorem: If $p\sim\operatorname{Beta}(a,b)$ before observation and $X=x$ claims are observed in $n$ binomial trials, then $$ p\mid X=x\sim\operatorname{Beta}\bigl(a+x,\ b+n-x\bigr). $$ The first posterior shape adds the number of claims; the second adds the number of nonclaims.

To see why, the binomial likelihood for the observed count is proportional, as a function of $p$, to $p^x(1-p)^{n-x}$. Multiplying it by the prior density adds the exponents of matching powers:

f(p\mid X=x)\ \propto\ p^x(1-p)^{n-x}p^{a-1}(1-p)^{b-1} =p^{a+x-1}(1-p)^{b+n-x-1}.

This is the kernel of a beta density with shape parameters $a+x$ and $b+n-x$. Dividing by the appropriate beta function makes the density integrate to 1. The binomial coefficient $\binom{n}{x}$ does not affect the posterior because it is constant with respect to $p$.

Write the prior.
Record the prior as $\operatorname{Beta}(a,b)$, using the stated parameter order.

Separate the observations.
Count $x$ claims and $n-x$ nonclaims. Check that each trial belongs to the same modeled risk and period.

Update both shapes.
Use posterior parameters $a+x$ and $b+n-x$, respectively.

Calculate the requested summary.
Use the posterior distribution, not the prior, for posterior moments or the mode.

Worked Example: Updating From Claims and Nonclaims

A hypothetical risk has prior claim probability $p\sim\operatorname{Beta}(2,6)$. During a defined period, 2 claims are observed among 10 independent, equally defined trials. Find the posterior distribution, mean, variance, and mode.

Method and assumptions. Apply beta-binomial conjugate updating. The observations are treated as conditionally independent Bernoulli trials with a common fixed probability $p$. There are $x=2$ claims and $n-x=10-2=8$ nonclaims.

Posterior parameters. The updated shapes are $a'=2+2=4$ and $b'=6+8=14$. Thus $p\mid X=2\sim\operatorname{Beta}(4,14)$.

Posterior mean. For a beta distribution, the mean is the first shape divided by the sum of the shapes: $$ E[p\mid X=2]=\frac{4}{4+14}=\frac{4}{18}=\frac{2}{9}\approx0.2222. $$

Posterior variance. The beta variance is $a'b'/[(a'+b')^2(a'+b'+1)]$, so $$ \operatorname{Var}(p\mid X=2)=\frac{4(14)}{18^2(19)} =\frac{56}{324(19)} =\frac{56}{6156}\approx0.0091. $$

Posterior mode. Since both posterior shapes exceed 1, the density has an interior mode: $$ \frac{a'-1}{a'+b'-2}=\frac{4-1}{4+14-2}=\frac{3}{16}=0.1875. $$ The mean $0.2222$ and mode $0.1875$ are different summaries, not competing calculations. The common mistake avoided here is adding all 10 trials to the first shape: only the 2 claims go to that shape; the 8 nonclaims go to the second.

Check. The posterior shapes sum to $18$, which is the prior shape sum $2+6=8$ plus the 10 observed trials. The posterior mean is between 0 and 1, and the variance is positive.

Posterior Moments and Their Meaning

For $p\sim\operatorname{Beta}(a,b)$, the first two moments give the mean and variance directly:

E[p]=\frac{a}{a+b}, \qquad \operatorname{Var}(p)=\frac{ab}{(a+b)^2(a+b+1)}.

One justification follows from beta-function ratios. For a nonnegative integer $r$, integrating $p^r$ against the beta density gives $E[p^r]=B(a+r,b)/B(a,b)$. In particular,

E[p]=\frac{B(a+1,b)}{B(a,b)}=\frac{a}{a+b}, \qquad E[p^2]=\frac{B(a+2,b)}{B(a,b)} =\frac{a(a+1)}{(a+b)(a+b+1)}.

Subtracting the square of the mean gives $$ \operatorname{Var}(p)=E[p^2]-E[p]^2 =\frac{a(a+1)}{(a+b)(a+b+1)}-\frac{a^2}{(a+b)^2} =\frac{ab}{(a+b)^2(a+b+1)}. $$

The posterior mean is the probability-weighted average of possible values of $p$, after observing the data. The variance describes uncertainty about that probability under the posterior distribution. It is not the binomial variance of a future count: that is a different quantity and depends on a future number of trials as well as $p$.

Worked Example: A Posterior With a Boundary Mode

A hypothetical small portfolio has prior $p\sim\operatorname{Beta}(0.5,2)$. Four independent trials are observed, with no claims. Find the posterior distribution, mean, variance, and describe its mode.

Method and assumptions. Use the same conditional binomial model. Here $n=4$, $x=0$, and the number of nonclaims is $4-0=4$.

Posterior parameters. They are $a'=0.5+0=0.5$ and $b'=2+4=6$, so $p\mid X=0\sim\operatorname{Beta}(0.5,6)$.

Posterior mean. $$ E[p\mid X=0]=\frac{0.5}{0.5+6}=\frac{0.5}{6.5}=\frac{1}{13}\approx0.0769. $$

Posterior variance. $$ \operatorname{Var}(p\mid X=0) =\frac{(0.5)(6)}{(6.5)^2(7.5)} =\frac{3}{42.25(7.5)} =\frac{3}{316.875}\approx0.0095. $$

Mode and interpretation. Because the first shape $0.5$ is less than 1, the density is unbounded as $p$ approaches 0. Its mode is at the boundary $p=0$ in the sense of a density that increases without bound near that endpoint. This does not mean the posterior assigns probability 1 to $p=0$; a continuous beta distribution assigns probability 0 to any single point. The common mistake avoided is applying the interior-mode formula when a shape parameter is below 1.

Check. The posterior shapes sum to $6.5$, equal to the prior shape sum $2.5$ plus four trials. The posterior mean is small after four nonclaims, but remains positive.

Posterior Mode: Interior and Boundary Cases

The mode is a value at which the density is largest, when such a finite maximum exists. For $p\sim\operatorname{Beta}(a,b)$, the interior-mode formula applies only when $a>1$ and $b>1$. It follows by differentiating the log density:

\frac{d}{dp}\log f(p)=\frac{a-1}{p}-\frac{b-1}{1-p}.

Setting this derivative equal to zero gives $(a-1)(1-p)=(b-1)p$, hence $p=(a-1)/(a+b-2)$. When both shapes exceed 1, the log density is strictly concave, so this stationary point is the unique interior maximum. If a shape is 1 or below, the density’s behavior at the endpoints must be checked instead.

Shape conditions	Density behavior and mode
$a>1,\ b>1$	Unique interior mode at $(a-1)/(a+b-2)$.
$a=1,\ b=1$	Uniform density; every $p$ in $[0,1]$ is a mode.
$a=1,\ b>1$	Finite maximum at $p=0$.
$a>1,\ b=1$	Finite maximum at $p=1$.
$a<1,\ b\geq1$	Density is unbounded as $p\to0$; this includes $a<1,\ b=1$.
$a\geq1,\ b<1$	Density is unbounded as $p\to1$; this includes $a=1,\ b<1$.
$a<1,\ b<1$	Density is unbounded at both endpoints.

The listed cases cover all positive shape parameters. In particular, the equal-to-one cases matter: with $a<1,b=1$, the density is unbounded at zero; with $a=1,b<1$, it is unbounded at one. An unbounded endpoint density is not a point mass at that endpoint.

Worked Example: Comparing Posterior Mean and Mode

A hypothetical risk has prior $p\sim\operatorname{Beta}(4,6)$. In 15 trials, 9 claims and 6 nonclaims are observed. Calculate the posterior parameters, mean, variance, and mode, and explain why the mean and mode are close but not identical.

Method and assumptions. Apply the beta-binomial update, assuming conditionally independent trials with one common $p$. Here $x=9$ and $n-x=15-9=6$.

Posterior. The shapes are $a'=4+9=13$ and $b'=6+6=12$. Therefore $p\mid X=9\sim\operatorname{Beta}(13,12)$.

Mean. $$ E[p\mid X=9]=\frac{13}{13+12}=\frac{13}{25}=0.5200. $$

Variance. $$ \operatorname{Var}(p\mid X=9)=\frac{13(12)}{25^2(26)} =\frac{156}{625(26)} =\frac{156}{16250}=0.0096. $$

Mode. Both shapes exceed 1, so $$ \operatorname{Mode}(p\mid X=9)=\frac{13-1}{13+12-2} =\frac{12}{23}\approx0.5217. $$ The mean and mode are close because the posterior density is concentrated around values near $0.52$; they are not required to be equal. The common mistake avoided is reporting the mean formula as the mode formula. Here the observed fraction $9/15=0.6$ is also not the posterior mean: the prior influences the posterior.

Check. The posterior shape sum is $25=(4+6)+15$, accounting for all prior shapes and all observed trials. The variance calculation uses $a'+b'+1=26$.

Assumptions and Common Exam Traps

The conjugate update is compact, but it does not correct a mismatched model. A trial must have a meaningful claim-or-nonclaim outcome, and the common-$p$ assumption must be defensible for the observations combined. If probabilities vary by exposure, the exposures have different observation periods, or outcomes are dependent conditional on $p$, the simple binomial likelihood may not apply.

Reversing the shapes. The first shape counts claims; the second counts nonclaims. Confirm this against the density, where $p$ has exponent $a-1$.
Using the number of claims for both updates. The second shape increases by $n-x$, not by $x$. Claims and nonclaims together must account for all $n$ trials.
Treating $a$ and $b$ as literal observed counts. They are shape parameters. Although the update adds counts, the prior shapes need not be integer counts or actual observations.
Confusing a posterior mean with a mode. The mean is $a/(a+b)$; the interior mode $(a-1)/(a+b-2)$ requires both shapes greater than 1. Boundary and unbounded cases must be handled separately.
Interpreting a boundary mode as certainty. A density that grows without bound near an endpoint still describes a continuous distribution; it does not put positive probability on that exact endpoint.

For a calculation check, add the posterior shapes: $(a+x)+(b+n-x)=a+b+n$. This identity catches many count-allocation errors. Then verify that the posterior mean lies strictly between 0 and 1 and that the variance is positive.

Key takeaway: Under a common-probability binomial model, a $\operatorname{Beta}(a,b)$ prior and $x$ claims in $n$ trials produce a $\operatorname{Beta}(a+x,b+n-x)$ posterior. Calculate moments from those updated shapes, and use the posterior mode formula only when both updated shapes exceed 1.

Check Your Understanding

For each hypothetical risk, assume conditionally independent binomial trials with one fixed claim probability. Give the posterior distribution and requested summary.

A risk has prior $\operatorname{Beta}(3,7)$; 2 claims occur in 6 trials. Find the posterior mean and variance. (Answer: $\operatorname{Beta}(5,11)$; mean $5/16=0.3125$; variance $55/4352\approx0.0126$.)
A risk has prior $\operatorname{Beta}(1,1)$; there are no claims in 9 trials. Find the posterior distribution, mean, and mode. (Answer: $\operatorname{Beta}(1,10)$; mean $1/11\approx0.0909$; finite mode at $p=0$.)
A risk has prior $\operatorname{Beta}(2.5,5.5)$; 4 claims occur in 12 trials. Find the posterior mean, variance, and mode. (Answer: $\operatorname{Beta}(6.5,13.5)$; mean $0.3250$; variance $87.75/8400\approx0.0104$; mode $5.5/18\approx0.3056$.)
A risk has prior $\operatorname{Beta}(0.8,1.4)$; all 5 of 5 trials produce claims. Find the posterior parameters and mode. (Answer: $\operatorname{Beta}(5.8,1.4)$; interior mode $4.8/5.2\approx0.9231$.)
A risk has prior $\operatorname{Beta}(3,4)$; 1 claim occurs in 10 trials. Find the posterior mean and variance, and check the posterior shape sum against the total trial count. (Answer: $\operatorname{Beta}(4,13)$; mean $4/17\approx0.2353$; variance $52/5202\approx0.0100$; shape sum $17=3+4+10$.)

Shape conditions	Density behavior and mode
\(a>1,\ b>1\)	Unique interior mode at \((a-1)/(a+b-2)\).
\(a=1,\ b=1\)	Uniform density; every \(p\) in \([0,1]\) is a mode.
\(a=1,\ b>1\)	Finite maximum at \(p=0\).
\(a>1,\ b=1\)	Finite maximum at \(p=1\).
\(a<1,\ b\geq1\)	Density is unbounded as \(p\to0\); this includes \(a<1,\ b=1\).
\(a\geq1,\ b<1\)	Density is unbounded as \(p\to1\); this includes \(a=1,\ b<1\).
\(a<1,\ b<1\)	Density is unbounded at both endpoints.

What You'll Learn