Bayesian Premiums as Posterior Expectations

Why a Bayesian Premium Is a Predictive Expectation

A risk’s observed experience is informative about its future losses, but it does not usually reveal the risk’s underlying characteristics with certainty. Bayesian credibility represents that uncertainty with a latent parameter or risk type. It begins with a prior distribution, updates that distribution using the observed data, and averages the possible future outcomes using the updated probabilities.

The premium of interest is therefore not automatically the prior average or an estimate of the latent parameter. It is the expected value of the specified future loss, conditional on the experience observed so far. The treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $\Theta$ denote an unknown risk parameter or risk type, with prior distribution $\pi$. Let $D$ denote observed experience and $X_{\text{new}}$ a future loss. The Bayesian premium for that future loss is its posterior predictive mean, $E[X_{\text{new}}\mid D]$. If the future loss has conditional mean $m(\Theta)=E[X_{\text{new}}\mid\Theta]$, and the future outcome is conditionally independent of $D$ given $\Theta$, then $E[X_{\text{new}}\mid D]=E[m(\Theta)\mid D]$.

Three quantities must be kept distinct. The prior mean, $E[m(\Theta)]$, averages the conditional expected loss before the new experience is observed. The posterior mean, $E[m(\Theta)\mid D]$, averages it after observing $D$. The posterior predictive mean, $E[X_{\text{new}}\mid D]$, is the expected future outcome conditional on that experience. Under the stated conditional-independence assumption, the last two quantities are equal when they refer to the same future loss. The posterior mean of $\Theta$ itself is a different quantity unless $m(\Theta)=\Theta$.

Updating Risk-Type Probabilities

For a finite set of risk types, updating follows directly from Bayes’ rule. Write the possible types as $\theta_1,\ldots,\theta_k$, with prior probabilities $P(\Theta=\theta_j)$. For observed data $D$, the posterior probability of type $\theta_j$ is proportional to its prior probability times the likelihood of the data under that type:

P(\Theta=\theta_j\mid D) = \frac{P(D\mid\Theta=\theta_j)P(\Theta=\theta_j)} {\sum_{r=1}^k P(D\mid\Theta=\theta_r)P(\Theta=\theta_r)}.

Once these posterior probabilities are known, the premium is a weighted average of conditional expected future losses. The weights are posterior probabilities, not prior probabilities. The conditioning event $D$ must also describe only the experience actually observed: for example, zero claims over a stated exposure is not the same evidence as one claim.

Specify the risk model.
State the possible risk types, their prior probabilities, and the distribution of the observed and future outcomes conditional on each type.

Update using the data.
Calculate each type’s likelihood for the observed experience and apply Bayes’ rule to obtain posterior probabilities.

Average the future conditional mean.
Multiply each type’s expected future loss by its posterior probability and sum. Check that the result refers to the requested future exposure and loss measure.

Frequency Premiums

For a frequency model, the future outcome may be the number of claims in a future exposure period. Conditional on a risk type, calculate the expected count for that exposure. The Bayesian premium is the posterior average of those conditional expected counts.

Worked Example: Updating a Claim-Rate Premium After No Claims

A hypothetical portfolio segment has one of two possible claim rates per exposure unit: $\Theta=1$ or $\Theta=2$, each with prior probability $1/2$. Conditional on $\Theta=\lambda$, the claim count in exposure $t$ is Poisson with mean $\lambda t$. The observed experience is zero claims over exposure $t=\ln 2$. Calculate the expected claim count for a future exposure of 1 unit.

Method. First update the two risk-type probabilities using the likelihood of observing zero claims. For a Poisson count with mean $\lambda t$, the probability of zero is $e^{-\lambda t}$.

For $\lambda=1$, the likelihood is $e^{-\ln 2}=1/2$. For $\lambda=2$, it is $e^{-2\ln 2}=1/4$. Multiplying by the prior probabilities gives unnormalized posterior weights $(1/2)(1/2)=1/4$ and $(1/2)(1/4)=1/8$. Their sum is $3/8$, so the posterior probabilities are $ (1/4)/(3/8)=2/3$ for rate 1 and $(1/8)/(3/8)=1/3$ for rate 2.

The conditional expected future counts in exposure 1 are 1 and 2. Thus the predictive mean is $(2/3)(1)+(1/3)(2)=4/3$ claims. The prior mean rate was $(1/2)(1)+(1/2)(2)=3/2$ claims per unit, so the observed no-claim experience has reduced the expected rate for the future period from $3/2$ to $4/3$.

Check. The posterior probabilities sum to 1, and $4/3$ lies between the two possible conditional future means, 1 and 2. A common mistake is to keep the prior weights and report $3/2$; that ignores the observed zero count. Another is to use the observation exposure $\ln 2$ as the future exposure, even though the question asks for a future exposure of 1 unit.

Severity Premiums

A severity model may assign different loss distributions to different risk types. After observing a loss, the posterior probabilities reflect how compatible that observation is with each type. The predictive severity mean is then the posterior average of each type’s conditional expected severity.

Worked Example: Updating Expected Severity After a Large Loss

A hypothetical insurer considers two severity types, A and B, with prior probabilities $1/2$ each. For type A, a claim is \$100 with probability 0.75 and \$300 with probability 0.25. For type B, those probabilities are reversed. One observed claim is \$300. Find the expected severity of the next claim, assuming the claims are independent conditional on type.

Method. The likelihood of the observed \$300 claim is 0.25 under A and 0.75 under B. The unnormalized posterior weights are $(1/2)(0.25)=0.125$ and $(1/2)(0.75)=0.375$. Their sum is 0.5, giving posterior probabilities $0.125/0.5=0.25$ for A and $0.375/0.5=0.75$ for B.

The conditional expected severity for type A is $(0.75)(\$100)+(0.25)(\$300)=\$150$. For type B it is $(0.25)(\$100)+(0.75)(\$300)=\$250$. Therefore the predictive mean is $(0.25)(\$150)+(0.75)(\$250)=\$225$.

As a direct check, the predictive probability of a \$100 next claim is $(0.25)(0.75)+(0.75)(0.25)=0.375$, while the probability of a \$300 next claim is $0.625$. The resulting mean is $(0.375)(\$100)+(0.625)(\$300)=\$225$. The prior expected severity was $(1/2)(\$150)+(1/2)(\$250)=\$200$. The observed large claim shifts posterior probability toward type B, raising the predictive mean to \$225. The common mistake avoided here is treating the observed loss as if it were the next loss; the premium averages the next claim’s distribution, not the already-observed amount.

Aggregate Premiums and Shared Risk Types

For an aggregate future loss, frequency and severity may both depend on the same latent risk type. In that case, calculate the conditional expected aggregate loss for each type first, then average those amounts using posterior probabilities. In general, the posterior expectation of a product is not the product of posterior expectations.

Justification: Conditional on $\Theta$, suppose a future claim count $N$ has mean $\lambda(\Theta)t$, and each claim has conditional mean severity $m(\Theta)$. If claim count and severities follow the stated conditional frequency-severity model, then the conditional expected aggregate loss is $\lambda(\Theta)t\,m(\Theta)$. Taking conditional expectation given the observed data gives $$ E[\text{future aggregate loss}\mid D] = E[\lambda(\Theta)t\,m(\Theta)\mid D]. $$ This follows by first averaging the aggregate loss for a fixed type, then averaging those conditional means over the posterior distribution of the type.

Worked Example: A Posterior Aggregate Premium with Dependent Frequency and Severity

A hypothetical risk has two equally likely prior types. Type A has a Poisson claim rate of 1 per exposure unit and a fixed claim severity of \$100. Type B has a Poisson claim rate of 3 per exposure unit and a fixed severity of \$300. The observed experience is zero claims over exposure $\ln 2$. Calculate the expected aggregate loss for future exposure 1.

Method. The zero-claim likelihoods are $e^{-\ln 2}=1/2$ for A and $e^{-3\ln 2}=1/8$ for B. Multiplying by the prior probabilities gives weights $1/4$ and $1/16$, with total $5/16$. Thus the posterior probabilities are $(1/4)/(5/16)=4/5$ for A and $(1/16)/(5/16)=1/5$ for B.

For future exposure 1, type A’s expected aggregate loss is $(1)(\$100)=\$100$. Type B’s is $(3)(\$300)=\$900$. The posterior predictive premium is $(4/5)(\$100)+(1/5)(\$900)=\$80+\$180=\$260$.

The posterior mean rate is $(4/5)(1)+(1/5)(3)=1.4$, and the posterior mean severity is $(4/5)(\$100)+(1/5)(\$300)=\$140$. Multiplying these separate means gives $1.4(\$140)=\$196$, which is not the correct premium. It fails because the high-frequency type is also the high-severity type. The correct posterior mean of the product is $(4/5)(1\cdot \$100)+(1/5)(3\cdot \$300)=\$260$. A common mistake is to average frequency and severity separately and multiply, as though their latent risk-type dependence did not matter.

Derivation and Interpretation

The general posterior predictive formula follows from conditional expectation. Assume the future outcome is conditionally independent of observed data $D$ given $\Theta$. Then, conditional on $D$, average the future outcome first for each possible $\Theta$, and then over the posterior distribution of $\Theta$:

E[X_{\text{new}}\mid D] = E\!\left[E[X_{\text{new}}\mid\Theta,D]\mid D\right] = E\!\left[E[X_{\text{new}}\mid\Theta]\mid D\right].

The first equality is the law of iterated expectations. The second uses conditional independence: once the risk type is known, the observed experience gives no further information about the future outcome. If that assumption does not hold, the conditional mean must remain $E[X_{\text{new}}\mid\Theta,D]$; it is not valid to replace it by $E[X_{\text{new}}\mid\Theta]$.

A Bayesian premium is a mean, not a complete description of predictive uncertainty. Two risks can have the same posterior predictive mean but different predictive distributions. The premium also depends on the quantity being priced: a future claim count, one claim’s severity, or aggregate loss over a specified exposure are different targets. State that target before calculating.

Common Exam Traps

Confusing the parameter with the premium. The posterior mean $E[\Theta\mid D]$ equals a premium only when the target’s conditional expected value is $\Theta$, or when a specified transformation makes it so.
Using prior weights after observing data. Recalculate the posterior weights from the likelihood and prior; do not average future means using the original prior unless the data leave the probabilities unchanged.
Using the wrong exposure. A frequency rate must be multiplied by the future exposure being priced, not by the historical exposure.
Multiplying averages when the latent type affects both frequency and severity. Calculate the conditional aggregate mean by type before taking the posterior average.
Dropping the conditioning assumption. The simple predictive-mean identity requires that the future outcome be independent of the observed data given the latent risk type.

Key takeaway: Update the risk-type distribution with observed experience, calculate the conditional expected future loss for each type, and average those values with posterior probabilities. That posterior predictive expectation is the Bayesian premium for the stated future outcome.

Check Your Understanding

For each problem, identify the posterior probabilities before calculating the requested future expectation.

A risk has claim rate 1 or 2 per exposure unit, with equal prior probabilities. Conditional on its rate, counts are Poisson. It has zero claims over exposure $\ln 2$. What is the expected count over future exposure 1.5? (Answer: 2 claims)
Severity type A has prior probability 0.3 and pays \$100 with probability 0.8 or \$400 with probability 0.2. Type B has prior probability 0.7 and pays \$100 with probability 0.2 or \$400 with probability 0.8. An observed claim is \$400. What is the expected severity of the next claim, assuming conditional independence given type? (Answer: \$322.58)
Two equally likely types have, respectively, claim rate 1 and fixed severity \$50, or claim rate 3 and fixed severity \$250. There are zero claims over historical exposure $\ln 2$. What is the expected aggregate loss over future exposure 2? (Answer: \$380)
A Bernoulli claim probability $\Theta$ is either 0.25 or 0.75, each with prior probability 0.5. Two independent exposure units both have claims. Find the posterior mean of $\Theta$ and the expected count over four future exposure units. (Answer: Posterior mean $0.70$; expected future count $2.8$)

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