Predictive Claim Costs Under a Normal-Normal Model

Why a Predictive Distribution Has More Variance

A posterior estimate of the average claim cost answers a question about an unknown parameter. A prediction for the next claim answers a different question: how much might one future observation vary? Even if the underlying mean were known, individual claims would still fluctuate around it. When the mean is uncertain too, a prediction must account for both sources of uncertainty.

This tutorial develops that distinction for a normal claim-cost model with known process variance. The result is useful for predictive probabilities and intervals, not just for a point estimate of the expected cost. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Model and notation: Let $\theta$ be the unknown mean claim cost. Before observing data, suppose $\theta\sim N(\mu_0,\tau^2)$. Conditional on $\theta$, observed claim costs $X_1,\ldots,X_n$ are independent and identically distributed as $N(\theta,\sigma^2)$, where the process variance $\sigma^2$ is known. A future cost $X_{n+1}$ is conditionally independent of the observed costs given $\theta$, and also has conditional distribution $N(\theta,\sigma^2)$. Here $\tau^2$ describes prior uncertainty about the mean; $\sigma^2$ describes variation among claims around that mean.

The normal assumption permits costs below zero, so it is most suitable when the probability of a negative value is negligible for the application. The model also assumes a common mean and variance for all claims under consideration, and that the variance is known. If these assumptions are unsuitable, the resulting predictive distribution may not describe future costs adequately.

From the Posterior Mean to a Future Claim

Write $\bar{x}=(x_1+\cdots+x_n)/n$ for the observed sample mean. Under the stated model, the posterior distribution of $\theta$, after observing the data, is normal. Its variance and mean are

\tau_n^2=\left(\frac{1}{\tau^2}+\frac{n}{\sigma^2}\right)^{-1}, \qquad \mu_n=\tau_n^2\left(\frac{\mu_0}{\tau^2}+\frac{n\bar{x}}{\sigma^2}\right).

Thus $\theta\mid x_1,\ldots,x_n\sim N(\mu_n,\tau_n^2)$. The posterior mean $\mu_n$ is also the predictive mean for the next claim. But a prediction needs the full predictive distribution, not only its mean.

Theorem: Under the known-variance normal-normal model, the conditional distribution of one future claim, given the observed data, is $$ X_{n+1}\mid x_1,\ldots,x_n\sim N\left(\mu_n,\sigma^2+\tau_n^2\right). $$ The predictive variance is the sum of process variance $\sigma^2$ and posterior variance of the mean $\tau_n^2$.

Why the Variances Add

Conditional on $\theta$, a future claim can be written as $X_{n+1}=\theta+\varepsilon$, where $\varepsilon\sim N(0,\sigma^2)$ is independent of $\theta$ given the observed data. After observing the data, $\theta\sim N(\mu_n,\tau_n^2)$. The future claim is therefore the sum of two independent normal quantities: an uncertain mean and a process deviation.

More explicitly, the conditional density of the future claim is normal with mean $\theta$ and variance $\sigma^2$. Averaging over the posterior distribution of $\theta$ gives a normal distribution whose mean and variance can also be found using conditional expectation and the law of total variance:

E[X_{n+1}\mid x_1,\ldots,x_n] =E[E[X_{n+1}\mid\theta,x_1,\ldots,x_n]\mid x_1,\ldots,x_n] =E[\theta\mid x_1,\ldots,x_n] =\mu_n,

\begin{aligned} \operatorname{Var}(X_{n+1}\mid x_1,\ldots,x_n) &=E[\operatorname{Var}(X_{n+1}\mid\theta,x_1,\ldots,x_n)\mid x_1,\ldots,x_n]\\ &\quad+\operatorname{Var}(E[X_{n+1}\mid\theta,x_1,\ldots,x_n]\mid x_1,\ldots,x_n)\\ &=E[\sigma^2\mid x_1,\ldots,x_n]+\operatorname{Var}(\theta\mid x_1,\ldots,x_n)\\ &=\sigma^2+\tau_n^2. \end{aligned}

The distribution is normal as well: the future claim is the sum of the posterior-normal $\theta$ and an independent normal process deviation. The sum has mean $\mu_n$ and variance $\tau_n^2+\sigma^2$. This argument also identifies when the formula does not apply: for example, if future claims have a different conditional variance, are dependent on the observed claims beyond their shared $\theta$, or the process variance is not known, this derivation cannot be used as written.

Worked Predictions

Worked Example: Probability a Future Cost Exceeds a Threshold

A hypothetical portfolio is modeled with prior mean claim cost $\mu_0=\$1{,}000$ and prior variance $\tau^2=10{,}000$ dollars squared. Conditional claim costs have known process variance $\sigma^2=40{,}000$ dollars squared. Four observed costs have sample mean $\bar{x}=\$1{,}200$. Find the posterior predictive distribution for one future claim and the probability that it exceeds $\$1{,}400$. The method is normal-normal updating followed by posterior prediction.

Step 1: Posterior variance. The prior precision is $1/10{,}000=0.0001$, and the data precision is $4/40{,}000=0.0001$. Therefore,

\tau_n^2=\frac{1}{0.0001+0.0001}=5{,}000.

Step 2: Posterior mean. The prior contribution to the precision-weighted numerator is $1{,}000/10{,}000=0.1$, and the data contribution is $4(1{,}200)/40{,}000=0.12$. Hence,

\mu_n=\frac{0.1+0.12}{0.0002}=\$1{,}100.

Step 3: Predictive distribution and probability. Add the posterior variance of the mean to the process variance:

X_{n+1}\mid x_1,\ldots,x_4\sim N(1{,}100,45{,}000), \qquad \sqrt{45{,}000}=\$212.13\text{ (approximately)}.

Standardizing the threshold gives $z=(1{,}400-1{,}100)/\sqrt{45{,}000}=1.4142$. Thus $P(X_{n+1}>1{,}400\mid x_1,\ldots,x_4)=1-\Phi(1.4142)\approx 0.0787$, to four decimal places. The common mistake avoided here is using $5{,}000$, the posterior variance of $\theta$, as the predictive variance. That would omit process variance; the correct predictive variance is $5{,}000+40{,}000=45{,}000$.

Worked Example: Predictive Tail Probability After Updating

For a hypothetical product, let $\theta\sim N(50,100)$, and let each claim cost conditional on $\theta$ be $N(\theta,100)$. Four observed claim costs have sample mean $\$60$. Find the probability that the next claim cost exceeds $\$70$. The method is to update the mean and then use the predictive, rather than posterior-parameter, variance.

Step 1: Update the mean. The prior precision is $1/100=0.01$, and the data precision is $4/100=0.04$. The posterior variance and mean are

\tau_n^2=\frac{1}{0.01+0.04}=20, \qquad \mu_n=\frac{50(0.01)+60(0.04)}{0.05} =\frac{0.50+2.40}{0.05}=58.

Step 2: Form the predictive distribution. The process variance is $100$, so the predictive variance is $100+20=120$. Therefore,

X_{n+1}\mid x_1,\ldots,x_4\sim N(58,120).

Step 3: Standardize the threshold. The standard deviation is $\sqrt{120}=10.9545$ (approximately), so $z=(70-58)/\sqrt{120}=1.0954$. It follows that $P(X_{n+1}>70\mid x_1,\ldots,x_4)=1-\Phi(1.0954)\approx 0.1367$, to four decimal places. The mistake avoided is treating the posterior mean $58$ as though it were the future cost itself; a future claim varies around that mean.

Worked Example: A Predictive Interval and Its Variance Components

A hypothetical claim model has prior $\theta\sim N(200,900)$ and known process variance $\sigma^2=100$. Two observed costs are $\$180$ and $\$240$. Find a 95% posterior predictive interval for one future cost. Use the normal critical value $1.96$. The method is to calculate the posterior distribution of the mean, add process variance, and apply the normal interval formula.

Step 1: Summarize the data. The sample mean is $\bar{x}=(180+240)/2=210$. The prior precision is $1/900$, and the data precision is $2/100=1/50$. The posterior variance is

\tau_n^2=\left(\frac{1}{900}+\frac{1}{50}\right)^{-1} =\left(\frac{1+18}{900}\right)^{-1} =\frac{900}{19}\approx 47.3684.

Step 2: Find the posterior mean. The precision-weighted numerator is $200/900+2(210)/100=2/9+21/5=199/45$. Thus

\mu_n=\frac{199}{45}\cdot\frac{900}{19} =\frac{3{,}980}{19}\approx \$209.47.

Step 3: Calculate the predictive interval. The predictive variance is $100+900/19=2{,}800/19\approx147.3684$, and its standard deviation is approximately $\$12.1395$. The interval is

209.4737\pm1.96(12.1395) \approx [\$185.68,\ \$233.27].

The endpoints are rounded to the nearest cent. The interval is for a future individual cost, so it includes both process variation and uncertainty in $\theta$. Using only the posterior variance $900/19$ would produce an interval that is too narrow.

Interpreting Predictive Results

The posterior distribution describes uncertainty about the underlying mean after seeing the data. The posterior predictive distribution describes uncertainty about a future observation after seeing the same data. Their means agree, but their variances do not: the predictive variance is larger by exactly the process variance $\sigma^2$. Even with many observations, posterior uncertainty about $\theta$ can shrink while the individual-claim process variation remains.

This distinction matters whenever a calculation concerns a future claim, such as a threshold probability or a range intended to contain a future cost. It is also a useful check on an answer: for positive process variance, predictive variance must exceed posterior variance of the mean. A predictive variance equal to $\tau_n^2$ describes uncertainty about the mean, not about a future claim.

Key takeaway: First update $\theta$ to obtain its posterior mean $\mu_n$ and variance $\tau_n^2$. Then predict one future claim with mean $\mu_n$ and variance $\sigma^2+\tau_n^2$. Do not substitute the posterior variance for the predictive variance.

Check Your Understanding

For each problem, use the known-variance normal-normal model and give the requested predictive result.

A hypothetical claim mean has prior distribution $N(40,25)$. Conditional claim variance is $100$. Three observed claims have mean $\$50$. Find the posterior predictive mean and variance for one future claim. (Answer: mean $\$44.2857$, variance $114.2857$ dollars squared.)
A hypothetical portfolio has prior $\theta\sim N(500,400)$, known process variance $900$, and two observed costs with sample mean $\$560$. Find the posterior predictive mean and variance. (Answer: mean $\$528.2353$, variance $1{,}111.7647$ dollars squared.)
Suppose the prior is $\theta\sim N(100,400)$, the known process variance is $100$, and one observed cost is $\$140$. Find the posterior predictive distribution and the probability that the next cost exceeds $\$150$. (Answer: $N(132,180)$; probability $0.0898$.)
After observing data, suppose the posterior distribution of the mean is $N(300,64)$, while the known process variance is $225$. Find the predictive variance and the probability a future cost lies between $\$283$ and $\$317$. (Answer: variance $289$ dollars squared; probability $0.6827$.)
A hypothetical model has prior $\theta\sim N(1{,}000,40{,}000)$, known process variance $90{,}000$, and three observed costs with mean $\$1{,}300$. Find the posterior predictive mean and variance for the next cost. (Answer: mean $\$1{,}171.4286$, variance $107{,}142.8571$ dollars squared.)

What You'll Learn