Why Compare Forecasts Across Priors?
A Bayesian claim forecast combines observed experience with assumptions about the range of plausible underlying risks before the experience was observed. Even when two analysts use the same likelihood and the same claims data, different prior distributions can produce different posterior distributions and, in turn, different forecasts. A sensitivity analysis makes that dependence visible.
The practical question is not simply whether two forecasts differ. It is which conclusion changes. A forecast mean may move slightly while the probability of exceeding a severe-claim threshold changes materially; alternatively, several numerical summaries may barely move. For an actuarial decision, it is useful to compare the same quantities under each prior: posterior predictive means, intervals, and probabilities of events relevant to the decision.
This comparison does not determine which prior is correct. It shows how much the forecast depends on the prior assumptions being considered. The earlier material on prior predictive checks concerns whether prior assumptions generate plausible data before observing the experience; here the emphasis is on how alternative priors affect forecasts after the same experience has been observed.
A Consistent Comparison Procedure
A fair comparison changes the prior and holds the rest of the calculation constant. For each prior, update it using the same likelihood and data, then calculate the same predictive summaries for the same future exposure. State clearly whether an apparent difference comes from a changed prior mean, a changed prior concentration, or both.
State the likelihood, its parameterization, and the prior distribution. In particular, distinguish a Gamma rate parameter from a scale parameter.
Use the same observed claims and exposure to obtain a separate posterior under each prior.
Use the same future exposure and calculate matching predictive summaries, such as a mean, an interval, or a threshold probability.
Describe their size and practical relevance. Do not attribute a difference to prior concentration alone if the priors also have different means.
For discrete predictive distributions, a central equal-tail interval must be calculated from the cumulative probabilities, not inferred from a normal approximation unless that approximation is justified. For a 90% equal-tail interval, choose the smallest integer lower endpoint whose cumulative probability is at least 0.05 and the smallest integer upper endpoint whose cumulative probability is at least 0.95. Because the distribution is discrete, the achieved coverage can exceed 90%.
Worked Example: Incidence Forecast With Different Prior Strengths
Worked Example: Incidence Forecast With Different Prior Strengths
A hypothetical portfolio has 3 claims among 20 independent, equal-exposure policies. Let each policy have claim probability \(p\), and, conditional on \(p\), let the claim indicators be independent Bernoulli variables. Compare priors \(\operatorname{Beta}(1,9)\) and \(\operatorname{Beta}(2,18)\) for \(p\). Forecast the number of claims \(K\) among 4 new policies. Find the predictive mean and the probability of at least one future claim under each prior.
Method and assumptions. With a Beta prior and binomial observations, the posterior is Beta with its first parameter increased by the number of claims and its second increased by the number of nonclaims. The two priors have the same mean, \(0.1\), but concentrations 10 and 20. Thus this comparison isolates prior strength, with the observed data and prior mean held fixed. There is no interest-rate or mortality assumption in this claim-incidence model; no discounting is applied.
Under \(\operatorname{Beta}(1,9)\), there are 3 claims and 17 nonclaims, so the posterior is \(\operatorname{Beta}(1+3,9+17)=\operatorname{Beta}(4,26)\). Its posterior mean is \(4/(4+26)=4/30\). Therefore:
The probability of no claims in four future trials is the beta-binomial probability \(E[(1-p)^4\mid\text{data}]\). For a \(\operatorname{Beta}(a,b)\) posterior, this equals \(b(b+1)(b+2)(b+3)/[(a+b)(a+b+1)(a+b+2)(a+b+3)]\). Thus:
Under \(\operatorname{Beta}(2,18)\), the posterior is \(\operatorname{Beta}(5,35)\), with mean \(5/40=0.125\). The predictive mean is \(4(0.125)=0.5000\). The no-claim probability and its complement are:
The forecast means differ by about \(0.0333\) claims, and the probability of at least one claim differs by about 0.0177. Since the priors have equal means, those changes reflect their different concentrations and their effects on the posterior after the same data. Here the forecasts are fairly similar, although they are not identical.
Deriving the Gamma-Poisson Predictive Distribution
For claim counts observed over exposure, a common model assumes a Poisson claim count conditional on an unknown claim rate \(\lambda\), with a Gamma prior. Use the rate parameterization \(\lambda\sim\operatorname{Gamma}(\alpha,\beta)\), where the density is proportional to \(\lambda^{\alpha-1}e^{-\beta\lambda}\); its mean is \(\alpha/\beta\). If \(n\) claims are observed over exposure \(t\), the posterior is \(\operatorname{Gamma}(A,B)\), where \(A=\alpha+n\) and \(B=\beta+t\). Given \(\lambda\), a future count \(N\) over exposure \(s\) is Poisson with mean \(s\lambda\).
The posterior predictive probability is obtained by integrating the conditional Poisson probability over the Gamma posterior. This integration gives:
Dividing the expression at \(k+1\) by the expression at \(k\) cancels the common terms. Using \(\Gamma(A+k+1)=(A+k)\Gamma(A+k)\) and \((k+1)!=(k+1)k!\) gives the useful recurrence:
The denominator is \((k+1)(B+s)\). The \(B+s\) term reflects both the posterior rate \(B\) and the future exposure \(s\); omitting \(s\) produces incorrect probabilities.
Worked Example: Predictive Intervals Under Different Prior Means
Worked Example: Predictive Intervals Under Different Prior Means
A hypothetical equipment pool produces 2 claims over 2 exposure units. Conditional on the unknown claim rate \(\lambda\), the count is Poisson with mean equal to exposure times \(\lambda\). Compare priors \(\operatorname{Gamma}(2,2)\) and \(\operatorname{Gamma}(2,1)\), both in shape-rate form, for a forecast over 2 future exposure units. Find the predictive mean, the equal-tail 90% interval, and \(P(N\ge6)\) under each prior.
Method and assumptions. The Poisson counts are conditionally independent given \(\lambda\), and the stated exposure is the multiplier of \(\lambda\). The prior means are 1 and 2, respectively; the shape is held fixed, but the prior means differ. There is no interest-rate or mortality assumption in this count model.
For \(\operatorname{Gamma}(2,2)\), the posterior is \(\operatorname{Gamma}(2+2,2+2)=\operatorname{Gamma}(4,4)\). For \(\operatorname{Gamma}(2,1)\), it is \(\operatorname{Gamma}(4,3)\). The predictive mean is \(As/B\), so the two means are \(4(2)/4=2.0000\) and \(4(2)/3=2.6667\).
For the first posterior, the recurrence has \(A=4\), \(B+s=6\), and \(P(N=0)=(4/6)^4=0.197531\). Applying the recurrence successively gives probabilities and cumulative probabilities through 5 as follows; displayed values are rounded to six decimals.
| Count \(k\) | \(P(N=k)\), \(\operatorname{Gamma}(4,4)\) | Cumulative probability |
|---|---|---|
| 0 | 0.197531 | 0.197531 |
| 1 | 0.263374 | 0.460905 |
| 2 | 0.219479 | 0.680384 |
| 3 | 0.146319 | 0.826703 |
| 4 | 0.085353 | 0.912056 |
| 5 | 0.045522 | 0.957578 |
The 5th percentile is 0 because the cumulative probability at 0 already exceeds 0.05. The first cumulative probability at least 0.95 occurs at 5, so the equal-tail 90% interval is \([0,5]\). Also, \(P(N\ge6)=1-P(N\le5)=1-0.957578=0.042422\), or 0.0424.
For \(\operatorname{Gamma}(4,3)\), \(P(N=0)=(3/5)^4=0.129600\), and the recurrence uses \(B+s=5\). It gives the following cumulative probabilities:
| Count \(k\) | \(P(N=k)\), \(\operatorname{Gamma}(4,3)\) | Cumulative probability |
|---|---|---|
| 0 | 0.129600 | 0.129600 |
| 1 | 0.207360 | 0.336960 |
| 2 | 0.207360 | 0.544320 |
| 3 | 0.165888 | 0.710208 |
| 4 | 0.116122 | 0.826330 |
| 5 | 0.074318 | 0.900647 |
| 6 | 0.044591 | 0.945238 |
| 7 | 0.025480 | 0.970719 |
Again, the lower endpoint is 0. The cumulative probability at 6 is below 0.95, while that at 7 exceeds 0.95, giving the equal-tail 90% interval \([0,7]\). The tail probability is \(P(N\ge6)=1-P(N\le5)=1-0.900647=0.099353\), or 0.0994. In this comparison, the prior mean changes, and both the predictive mean and the upper interval endpoint move. The probability of at least 6 claims is more than twice as large under the second prior.
Check. The two tail probabilities agree with the cumulative tables: \(1-0.957578=0.042422\) and \(1-0.900647=0.099353\). The second forecast has both the larger predictive mean and the larger probability of reaching 6 claims.
Worked Example: A Threshold Probability Can Move Even When Means Barely Change
Worked Example: A Threshold Probability Can Move Even When Means Barely Change
A hypothetical small commercial portfolio has 1 claim among 8 policies. Let the claim probability \(p\) be common across independent policies conditional on \(p\). Compare priors \(\operatorname{Beta}(1,9)\) and \(\operatorname{Beta}(2,18)\), and forecast the number \(K\) of claims among 3 future policies. Calculate the predictive mean and the probability of at least 2 claims.
Method and assumptions. These Beta priors have the same mean, 0.1, but different concentrations, 10 and 20. Updating on 1 claim and 7 nonclaims gives posterior distributions \(\operatorname{Beta}(2,16)\) and \(\operatorname{Beta}(3,25)\). There is no interest-rate or mortality assumption; this is a binomial incidence model.
The predictive means are \(3(2/18)=0.3333\) and \(3(3/28)=0.3214\). To calculate the threshold probability, subtract the probabilities of zero and one claim from 1. For a \(\operatorname{Beta}(a,b)\) posterior with \(T=a+b\), the beta-binomial probabilities for three future policies are:
For \(\operatorname{Beta}(2,16)\), \(T=18\). Therefore:
Thus \(P(K\ge2)=1-0.715789-0.238596=0.045614\), or 0.0456. For \(\operatorname{Beta}(3,25)\), \(T=28\):
The second tail probability is \(P(K\ge2)=1-0.720443-0.240148=0.039409\), or 0.0394. The predictive means differ by only about 0.0119, and the probability of at least 2 claims changes by about 0.0062. The threshold probability is therefore worth checking directly rather than assuming it must change in proportion to the mean.
Check. For each posterior, the probabilities of 0, 1, and at least 2 claims sum to 1, subject to rounding. For example, \(0.720443+0.240148+0.039409=1.000000\).
Common Mistakes and Interpretation
- Using the prior mean as the forecast rate after observing data. In the first worked example, using \(4(0.1)=0.4000\) as the forecast mean ignores the 3 observed claims among 20 policies. The posterior mean under \(\operatorname{Beta}(1,9)\) is \(4/30\), so the correct predictive mean is \(4(4/30)=0.5333\).
- Using the wrong recurrence denominator. For the \(\operatorname{Gamma}(4,4)\) posterior and future exposure \(s=2\), the correct ratio at \(k=0\) is \(4(2)/(1(4+2))=4/3\). Replacing \(B+s\) with \(B\) gives \(8/4=2\), which is incorrect and generates an inconsistent predictive distribution.
- Changing more than one assumption without saying so. If two priors have different means and concentrations, the forecast difference cannot be attributed only to concentration. State which prior features differ.
- Comparing unlike forecasts. A probability for at least 6 claims over 2 exposure units cannot be compared directly with a probability for at least 6 claims over a different exposure period. Keep the target and exposure fixed.
- Rounding too early. Calculate the predictive probabilities at full precision, sum them, and round the final reported probabilities. Small rounding errors can change a discrete interval endpoint near a quantile threshold.
Check Your Understanding
For each problem, state the posterior update and keep the future exposure fixed when comparing or calculating forecasts.
- A claim-incidence probability has prior \(\operatorname{Beta}(1,9)\). Observe 2 claims among 10 policies and forecast 3 policies. Find the predictive mean and the probability of at least one claim. (Answer: Posterior \(\operatorname{Beta}(3,17)\); mean \(0.4500\); \(P(K\ge1)=0.3708\).)
- A claim rate has prior \(\operatorname{Gamma}(2,2)\) in shape-rate form. Observe 3 claims over exposure 2 and forecast exposure 1. Find the posterior predictive mean and the probability of at least one future claim. (Answer: Posterior \(\operatorname{Gamma}(5,4)\); mean \(1.2500\); \(P(N\ge1)=0.6723\).)
- The posterior claim rate is \(\operatorname{Gamma}(3,2)\) in shape-rate form. For future exposure \(s=1\), calculate the predictive probability of at least 2 claims. (Answer: \(P(N\ge2)=11/27=0.4074\).)
- Compare priors \(\operatorname{Beta}(1,4)\) and \(\operatorname{Beta}(2,8)\) after observing 2 claims among 10 policies. For 2 future policies, find each predictive mean and probability of at least one claim. (Answer: Posterior distributions \(\operatorname{Beta}(3,12)\) and \(\operatorname{Beta}(4,16)\); both means \(0.4000\); probabilities \(0.3500\) and \(0.3524\), respectively.)