Constructing Posterior Predictive Intervals

Why Predictive Intervals Need Both Sources of Uncertainty

A point forecast summarizes the center of a future claim count or loss, but it does not show how much the outcome may vary. A posterior predictive interval describes a range of future outcomes under the model and the observed experience. Its width reflects two distinct sources of uncertainty: process variation, because future claims are random even if the risk parameter were known, and parameter uncertainty, because the risk parameter is not known exactly.

The interval is predictive: it concerns a future observation, not an interval for an unknown parameter. This distinction matters in ratemaking. An interval for a parameter alone omits the random variation in future claims, while an interval calculated from a single estimated parameter usually omits uncertainty about that parameter.

Definition: Let $Y_{\mathrm{new}}$ denote a future claim count or loss and let $\theta$ denote an unknown model parameter. A posterior predictive interval is an interval for $Y_{\mathrm{new}}$ calculated from its distribution after averaging over the posterior uncertainty in $\theta$: $$ P(Y_{\mathrm{new}}\in I\mid \text{data}) =\int P(Y_{\mathrm{new}}\in I\mid \theta)\,\pi(\theta\mid \text{data})\,d\theta. $$ Here $\pi(\theta\mid \text{data})$ is the posterior distribution of the parameter.

For a continuous predictive distribution, an equal-tail interval with nominal coverage $1-\alpha$ has endpoints at predictive quantiles $\alpha/2$ and $1-\alpha/2$. A highest-density interval (HDI) is a region containing the desired predictive probability while including the outcomes with the greatest predictive density first. In a continuous distribution with a smooth density, an HDI typically has equal density at its two endpoints. In a discrete distribution, density is replaced by probability mass, ties may produce more than one valid region, and the achieved coverage may exceed the nominal level.

Equal-Tail Intervals for Predictive Claim Counts

A useful count model is the Poisson-gamma model. Suppose the claim count in a future exposure $t$, conditional on a claim rate $\lambda$, is Poisson with mean $t\lambda$. Suppose the posterior distribution of $\lambda$, after observing the data, is Gamma with shape $a$ and rate $b$. The shape-rate convention means the posterior density is proportional to $\lambda^{a-1}e^{-b\lambda}$, and its mean is $a/b$. Integrating over this posterior gives a negative binomial predictive distribution:

P(N=n\mid \text{data}) = \frac{\Gamma(n+a)}{\Gamma(a)n!} \left(\frac{b}{b+t}\right)^a \left(\frac{t}{b+t}\right)^n,\qquad n=0,1,2,\ldots.

This formula assumes a conditionally Poisson count, a Gamma posterior for the rate, and a fixed future exposure $t$. If exposure itself is uncertain, or the conditional count model is not Poisson, this predictive distribution need not apply. For a discrete equal-tail interval, use the predictive quantile convention $Q(p)=\min\{n:P(N\le n\mid\text{data})\ge p\}$; discreteness means the resulting coverage need not equal exactly $1-\alpha$.

Worked Example: A 90% Equal-Tail Interval for Future Claims

A hypothetical portfolio has a posterior claim-rate distribution $\lambda\mid\text{data}\sim\operatorname{Gamma}(2,1)$, using shape-rate parameters per exposure-year. Forecast the claim count $N$ over one future exposure-year, and construct a 90% equal-tail predictive interval. No interest or mortality assumption enters this count calculation.

Method and parameters. Here $a=2$, $b=1$, and $t=1$. Thus $b/(b+t)=1/2$ and $t/(b+t)=1/2$. The predictive probability is $$ P(N=n\mid\text{data})=(n+1)\left(\frac12\right)^{n+2}. $$

Compute the probabilities and cumulative probabilities until both tail quantiles are located:

\begin{array}{c|rrrrrrr} n&0&1&2&3&4&5&6\\ \hline P(N=n)&0.250000&0.250000&0.187500&0.125000&0.078125&0.046875&0.027344\\ P(N\le n)&0.250000&0.500000&0.687500&0.812500&0.890625&0.937500&0.964844 \end{array}

For the lower endpoint, $P(N\le -1)=0<0.05$, while $P(N\le0)=0.25\ge0.05$, so $Q(0.05)=0$. For the upper endpoint, $P(N\le5)=0.9375<0.95$, while $P(N\le6)=0.96484375\ge0.95$, so $Q(0.95)=6$. The 90% equal-tail interval is therefore $[0,6]$ claims. Its actual predictive coverage is $0.96484375$, with lower-tail probability $0$ and upper-tail probability $P(N>6)=1-0.96484375=0.03515625$. The coverage exceeds 90% because the count is discrete. The common mistake is to report exactly 90% coverage merely because the requested nominal level is 90%; the cumulative probabilities show the actual coverage.

Check. The interval includes all counts from 0 through 6, and their cumulative probability is $0.96484375$, which is at least 0.90.

Why Predictive Variance Includes Two Components

The law of total variance gives a direct explanation of the two sources of predictive uncertainty. Conditional on $\theta$, a future outcome has conditional mean $E[Y_{\mathrm{new}}\mid\theta]$ and conditional variance $\operatorname{Var}(Y_{\mathrm{new}}\mid\theta)$. Averaging these conditional distributions over the posterior yields:

\operatorname{Var}(Y_{\mathrm{new}}\mid\text{data}) = E[\operatorname{Var}(Y_{\mathrm{new}}\mid\theta)\mid\text{data}] + \operatorname{Var}(E[Y_{\mathrm{new}}\mid\theta]\mid\text{data}).

The first term is expected process variance: the future outcome varies around its conditional mean even at a fixed parameter. The second term is parameter uncertainty: the conditional mean itself varies across plausible parameter values. To justify the identity, write $m(\theta)=E[Y_{\mathrm{new}}\mid\theta]$ and $\bar m=E[Y_{\mathrm{new}}\mid\text{data}]$. Then $Y_{\mathrm{new}}-\bar m=(Y_{\mathrm{new}}-m(\theta))+(m(\theta)-\bar m)$. Squaring and taking the posterior predictive expectation, the cross-product is zero because $E[Y_{\mathrm{new}}-m(\theta)\mid\theta]=0$. The two remaining squared terms are precisely the two terms in the variance formula.

For a Poisson future count conditional on $\lambda$, both its conditional mean and variance are $t\lambda$. If the posterior is Gamma with shape $a$ and rate $b$, then $E[\lambda\mid\text{data}]=a/b$ and $\operatorname{Var}(\lambda\mid\text{data})=a/b^2$. Therefore:

E[N\mid\text{data}]=\frac{ta}{b},\qquad \operatorname{Var}(N\mid\text{data}) =\frac{ta}{b}+\frac{t^2a}{b^2}.

The first variance term is the expected conditional Poisson variance. The second is the variance of the conditional mean $t\lambda$. Ignoring the second term produces a plug-in-style uncertainty calculation that is too small whenever the posterior rate is uncertain.

Worked Example: A Normal Predictive Interval for Future Loss

For a hypothetical book of policies, a model gives a posterior predictive normal distribution for the next period’s aggregate loss, with mean 80 thousand dollars. Conditional process variance is 100 squared thousands of dollars, and parameter uncertainty contributes 144 squared thousands of dollars. Construct a central 90% interval using $z=1.645$. No interest or mortality assumption applies; the stated normal model and variance components are the assumptions for this calculation.

Method. By the total-variance decomposition, the predictive variance is the sum of the two variance components: $$ \sigma_{\mathrm{pred}}^2=100+144=244,\qquad \sigma_{\mathrm{pred}}=\sqrt{244}=15.620499\ldots $$

Under the stated normal predictive model, a central 90% interval is the predictive mean plus or minus $1.645$ predictive standard deviations. The margin is $$ 1.645\sqrt{244}=25.695722\ldots $$ Thus the unrounded endpoints are $80-25.695722\ldots=54.304277\ldots$ and $80+25.695722\ldots=105.695722\ldots$. Rounding only at the final step to two decimal places gives $[54.30,105.70]$ thousand dollars.

Check and common mistake. The interval is symmetric about 80, since the lower and upper margins are both $25.695722\ldots$. Using only the process variance would give $80\pm1.645(10)$, which omits parameter uncertainty and is too narrow. Also, when using the stated $z=1.645$, rounding the endpoints to $54.31$ and $105.69$ is incorrect: the computed endpoints round to $54.30$ and $105.70$.

Highest-Density Intervals and Discrete Outcomes

An equal-tail interval allocates probability to the two tails according to a fixed rule. An HDI instead prioritizes the most probable outcomes. For a discrete predictive distribution, one practical construction is to rank outcomes from largest to smallest probability mass and include them until the accumulated mass reaches or exceeds the target coverage. If the selected outcomes form a consecutive set, they can be reported as an interval. When there are ties, different sets may be equally valid; when there are gaps, an HDI region need not be a single interval.

Definition: A discrete highest-density predictive region at target coverage $1-\alpha$ is a set of outcomes with total predictive probability at least $1-\alpha$, formed by including outcomes in nonincreasing order of predictive probability mass. Ties can make the region nonunique, and discreteness can make its coverage exceed $1-\alpha$.

Worked Example: A Highest-Density Interval for Future Incidence

A hypothetical risk group has posterior incidence probability $p\mid\text{data}\sim\operatorname{Beta}(1,3)$. For the next four independent exposures, conditional on $p$, the number $K$ with a claim is binomial with parameters $4$ and $p$. Find a highest-density predictive interval with at least 90% coverage. No interest or mortality assumption is involved.

Method and distribution. Integrating the binomial likelihood over the Beta posterior gives the beta-binomial predictive mass $$ P(K=k\mid\text{data}) ={4\choose k}\frac{B(k+1,\,4-k+3)}{B(1,3)},\qquad k=0,1,2,3,4. $$

Since $B(1,3)=1/3$, evaluating the five masses gives $P(K=0)=3/7$, $P(K=1)=2/7$, $P(K=2)=6/35$, $P(K=3)=3/35$, and $P(K=4)=1/35$. In decimal form these are approximately $0.4286,0.2857,0.1714,0.0857,0.0286$, respectively. The probabilities sum to $(15+10+6+3+1)/35=1$.

Rank the outcomes by probability mass. Include $K=0$ and $K=1$, giving $3/7+2/7=5/7=0.7143$. Next include $K=2$, giving $5/7+6/35=31/35=0.8857$, still below 0.90. Including $K=3$ gives $34/35=0.9714$, which reaches the target. The selected outcomes are consecutive, so the 90% highest-density interval is $[0,3]$ claims, with actual coverage $34/35$, or approximately $0.9714$.

Check and common mistake. The remaining probability is $P(K=4)=1/35$, so the reported coverage $34/35$ is consistent. Stopping after $K=2$ would give only $31/35$, less than 90%. Unlike an equal-tail interval, the HDI does not require the excluded probability to be divided evenly between the two tails.

Practical Choices and Common Traps

Choose the interval construction to match the purpose. Equal-tail intervals are straightforward to explain and place prescribed probability in each tail. HDIs focus on the most probable outcomes, but they may be nonunique for discrete models and may not be a single connected interval. Whichever construction is used, report the nominal target and the actual predictive coverage when discreteness causes them to differ.

Do not confuse a parameter interval with a predictive interval. The former describes uncertainty about a model parameter; the latter describes a future observation and includes process variation.
Do not substitute a point estimate for posterior averaging. A plug-in distribution conditions on one estimated parameter value and generally misses parameter uncertainty.
Do not force exact nominal coverage in a discrete model. Use the stated quantile convention or HDI rule, and calculate the achieved coverage from the predictive probabilities.
Keep the forecast horizon consistent. In a Poisson-gamma model, future exposure $t$ affects both the predictive mass and the variance; changing the exposure unit without changing $t$ changes the answer.

Key takeaway: A posterior predictive interval is built from the full predictive distribution, which averages over posterior parameter uncertainty and retains randomness in future outcomes. State the interval rule, calculate its endpoints or included outcomes, and verify the resulting predictive coverage.

Check Your Understanding

For each problem, identify the predictive distribution or variance calculation needed, then check the interval and coverage stated in the answer.

A posterior rate has distribution $\lambda\mid\text{data}\sim\operatorname{Gamma}(1,1)$ in shape-rate form. Future count $N$, over exposure $t=1$, is conditionally Poisson with mean $t\lambda$. Find the 80% equal-tail predictive interval using $Q(p)=\min\{n:P(N\le n)\ge p\}$. (Answer: $[0,3]$; the predictive mass is $P(N=n)=2^{-(n+1)}$, and the coverage is $15/16=0.9375$.)
A future loss has a normal predictive model with mean 42 thousand dollars, process variance 25, and parameter uncertainty variance 11, both in squared thousands of dollars. Find the central 90% interval using $z=1.645$. (Answer: predictive standard deviation $6$, margin $9.87$, interval $[32.13,51.87]$ thousand dollars.)
Suppose $p\mid\text{data}\sim\operatorname{Beta}(2,1)$, and the next two exposures have conditionally independent Bernoulli outcomes given $p$. Find a highest-density interval for the predictive number of claims with at least 80% coverage. (Answer: $[1,2]$, with coverage $5/6\approx0.8333$.)
A posterior rate is $\operatorname{Gamma}(3,2)$ in shape-rate form, and a future Poisson count has exposure $t=2$. Find the 90% equal-tail predictive interval under the stated lower-quantile convention. (Answer: $[0,8]$; $P(N\le0)=0.125<0.05$ is not true, so the lower quantile is 0 since $P(N\le-1)=0<0.05$, and $P(N\le8)\approx0.9673$ while $P(N\le7)\approx0.9453<0.95$.)

What You'll Learn