Sensitivity of Bayesian Premiums to Prior Choice

Why Prior Sensitivity Matters

A Bayesian premium combines observed experience with assumptions about the population before that experience was observed. When experience is sparse, the prior can have appreciable influence; when experience is extensive, the data generally receive more weight. A premium calculation is therefore not fully understood until its response to plausible prior choices has been examined. This tutorial is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

A sensitivity comparison changes one or more prior specifications while keeping the observed data and likelihood model fixed. This isolates the effect of the prior from the effect of the experience. The comparison should identify what changed, calculate the resulting posterior premiums, and explain whether the difference matters for the decision at hand. There is no universal dollar or percentage difference that is material in every application.

Definition: A prior-sensitivity analysis compares posterior quantities under two or more specified prior distributions while holding the likelihood model and observed data fixed. For premium analysis, the quantity compared is usually a posterior expected loss or cost for a stated future period. A premium difference measures sensitivity to the prior; it is not, by itself, evidence that one prior is correct.

Normal-Normal Premiums: Separating Mean and Strength

Suppose $X_1,\ldots,X_n$, conditional on an unknown mean $\theta$, are independent with $X_j\mid\theta\sim N(\theta,\sigma^2)$. The variance $\sigma^2$ is known. Suppose also that $\theta\sim N(m_0,\tau^2)$, independently of the observations conditional on $\theta$. The posterior expected cost of a future observation with conditional mean $\theta$ is the posterior mean of $\theta$.

Result: Define the prior strength in observation units as $k=\sigma^2/\tau^2$. If the observed sample mean is $\bar{x}$, the posterior premium for one future observation is $$ E[\theta\mid X_1,\ldots,X_n] =\frac{n\bar{x}+km_0}{n+k} =Z\bar{x}+(1-Z)m_0,\qquad Z=\frac{n}{n+k}. $$ Here $Z$ is the weight on the data and $1-Z$ is the weight on the prior mean. This form requires the stated normal model, known common variance, and normal prior. It is not a general formula for arbitrary likelihoods or priors.

The strength $k$ is measured in units of observations because $k=\sigma^2/\tau^2$. A smaller prior variance means a larger $k$, hence more prior weight. Holding $k$ fixed while changing $m_0$ gives an exact sensitivity calculation:

\Delta E[\theta\mid X] =\frac{k}{n+k}\Delta m_0.

Thus a change in prior mean is attenuated by the prior weight. Holding $m_0$ fixed instead, increasing $k$ moves the posterior mean toward $m_0$. In differential form,

\frac{\partial E[\theta\mid X]}{\partial k} =\frac{n(m_0-\bar{x})}{(n+k)^2}.

The sign is informative: increasing prior strength raises the premium if the prior mean exceeds the observed mean, lowers it if the prior mean is below the observed mean, and has no effect if the two means agree.

Why the Weighted Average Appears

The weighting formula follows from combining normal likelihood and prior precision. The likelihood for the sample mean is proportional, as a function of $\theta$, to a normal kernel with mean $\bar{x}$ and variance $\sigma^2/n$. Multiplying that kernel by the prior density adds their precision terms. In the exponent, the terms involving $\theta$ are

-\frac{1}{2}\left\{\frac{n}{\sigma^2}(\theta-\bar{x})^2+ \frac{1}{\tau^2}(\theta-m_0)^2\right\}.

Collecting the coefficients of $\theta^2$ gives posterior precision $n/\sigma^2+1/\tau^2$. The coefficient determining the posterior mean is $n\bar{x}/\sigma^2+m_0/\tau^2$. Dividing the latter by the former and multiplying numerator and denominator by $\sigma^2$ gives

\frac{n\bar{x}+(\sigma^2/\tau^2)m_0}{n+\sigma^2/\tau^2} =\frac{n\bar{x}+km_0}{n+k}.

This derivation explains why the more precise source of information receives more weight: precision, rather than variance, adds when independent normal information is combined.

Worked Example: Changing the Prior Mean at Fixed Strength

A hypothetical insurer models individual claim costs $X_j$, conditional on an unknown mean $\theta$, as independent $N(\theta,200^2)$ amounts. Four claims have sample mean $\bar{x}=\$900$. Consider two normal priors with the same variance $\tau^2=100^2$, but prior means $\$800$ and $\$1{,}000$. Calculate both posterior premiums for one future claim and their difference.

Method and assumptions. Use the normal-normal posterior mean. The individual claim variance is $\sigma^2=200^2=\$40{,}000$, and $n=4$. The common prior strength is $k=\sigma^2/\tau^2=40{,}000/10{,}000=4$.

For prior mean $m_0=\$800$, the posterior premium is

\frac{4(900)+4(800)}{4+4} =\frac{3{,}600+3{,}200}{8} =\frac{6{,}800}{8} =\$850.00.

For prior mean $m_0=\$1{,}000$, it is

\frac{4(900)+4(1{,}000)}{8} =\frac{3{,}600+4{,}000}{8} =\frac{7{,}600}{8} =\$950.00.

The premium difference is $\$950.00-\$850.00=\$100.00$. The prior means differ by $\$200$, and the prior weight is $k/(n+k)=4/8=0.5$, so the sensitivity formula independently gives $0.5(\$200)=\$100$. This agreement checks the calculations.

Common mistake avoided: reversing the weights would assign $k/(n+k)$ to the data and $n/(n+k)$ to the prior. With the $\$800$ prior, that incorrect calculation gives $\{4(900)+4(800)\}/8=\$850$ only because $n=k$, so it fails to reveal the error here. The weights happen to be equal; in a case where they differ, the data weight must be $n/(n+k)$. The next example makes that distinction consequential.

Changing Prior Strength

A prior mean and a prior strength answer different questions. The mean specifies the prior’s central level; the strength specifies how concentrated the prior information is around that level. To isolate strength sensitivity, keep the prior mean, observed data, and likelihood model fixed, and compare different strengths. As the strength increases, the posterior premium moves toward the prior mean. The movement is small when the prior mean is already close to the observed rate.

The same comparison has a direct interpretation in a Gamma-Poisson frequency model. Let $C$ be the count observed over exposure $e$, with $C\mid\lambda\sim\operatorname{Poisson}(e\lambda)$. Let $\lambda\sim\operatorname{Gamma}(\alpha,\beta)$, using shape $\alpha$ and rate $\beta$. The prior mean is $m=\alpha/\beta$, and $\beta$ is the prior strength in exposure units. Since the posterior is $\operatorname{Gamma}(\alpha+C,\beta+e)$, its mean is

E[\lambda\mid C]=\frac{\alpha+C}{\beta+e} =\frac{\beta m+e(C/e)}{\beta+e}.

This is a weighted average of the prior frequency $m$ and observed frequency $C/e$, with weights proportional to prior exposure $\beta$ and actual exposure $e$. It gives the expected claim frequency per exposure unit, not the probability of at least one claim. The formula assumes a constant Poisson rate over the exposure and a Gamma prior using the stated rate parameterization.

Worked Example: Same Prior Mean, Different Strengths

A hypothetical portfolio has 30 claims over 3 policy-years, and claim counts are modeled as Poisson conditional on a frequency $\lambda$ per policy-year. Compare posterior expected annual frequencies under two Gamma priors, both with mean 4 claims per policy-year but with strengths $\beta=1$ and $\beta=9$ policy-years.

Method and assumptions. The observed frequency is $C/e=30/3=10$ claims per policy-year. For a Gamma prior with mean $m=4$ and rate $\beta$, its shape is $\alpha=\beta m$. Thus the first prior has $\alpha=1(4)=4$; the second has $\alpha=9(4)=36$.

With $\beta=1$, the posterior shape is $4+30=34$, and the posterior rate is $1+3=4$. Therefore,

E[\lambda\mid C]=\frac{34}{4}=8.5 \quad\text{claims per policy-year}.

With $\beta=9$, the posterior shape is $36+30=66$, and the posterior rate is $9+3=12$. Therefore,

E[\lambda\mid C]=\frac{66}{12}=5.5 \quad\text{claims per policy-year}.

The stronger prior moves the posterior frequency closer to its mean of 4. The difference between the posterior means is $8.5-5.5=3.0$ claims per policy-year. As a check, the weighted-average form gives $\{1(4)+3(10)\}/4=8.5$ and $\{9(4)+3(10)\}/12=5.5$.

Common mistake avoided: the observed rate $10$ is weighted by actual exposure $e=3$, not by the number of claims $C=30$. Claims contribute to the posterior shape, while exposure contributes to the posterior rate; confusing these quantities produces an invalid weighting calculation.

Comparing Competing Prior Specifications

In practice, analysts may consider priors that differ in both mean and strength. The comparison should report the posterior premium under each complete specification; changing both parameters at once does not identify the separate contribution of either change. When useful, add controlled comparisons: change only the mean at a fixed strength, then change only the strength at a fixed mean. These comparisons make the source of the difference transparent.

The absolute dollar difference is often more useful than a percentage alone. A difference should be interpreted in context: for example, relative to a pricing tolerance, an expected-loss component, or the financial consequence of the decision. Such a benchmark must be stated rather than treated as a universal definition of materiality.

Worked Example: Assessing the Difference Against a Decision Threshold

A hypothetical insurer observes $n=8$ claim costs with sample mean $\bar{x}=\$1{,}200$. Conditional on an unknown mean $\theta$, costs are independent normal observations with known standard deviation $\sigma=\$300$. Compare two candidate normal priors: prior A has $m_A=\$1{,}000$ and strength $k_A=2$; prior B has $m_B=\$1{,}400$ and strength $k_B=6$. A pricing review treats a change exceeding $\$100$ per claim as material. Determine whether the prior choice crosses that threshold.

Method and assumptions. For each prior, calculate the posterior mean using the same normal likelihood and observed sample. The corresponding prior variances are $\tau_A^2=\sigma^2/k_A=90{,}000/2=45{,}000$ and $\tau_B^2=90{,}000/6=15{,}000$, so both strengths are consistent with the specified likelihood variance.

Under prior A,

E[\theta\mid X,A] =\frac{8(1{,}200)+2(1{,}000)}{8+2} =\frac{9{,}600+2{,}000}{10} =\$1{,}160.00.

Under prior B,

E[\theta\mid X,B] =\frac{8(1{,}200)+6(1{,}400)}{8+6} =\frac{9{,}600+8{,}400}{14} =\$1{,}285.71.

The premium difference is $\$1{,}285.71-\$1{,}160.00=\$125.71$, rounded to the nearest cent. It exceeds the stated $\$100$ threshold by $\$25.71$, so the prior choice is material under this review’s criterion. The directions of the changes also make sense: prior A pulls the observed mean down, while prior B pulls it up. This is a decision-specific conclusion, not a claim that prior B is preferable.

Common mistake avoided: the difference is not the distance between the prior means, $\$400$, nor the distance between either prior mean and the data. It is the difference between the two posterior premiums after each prior has been combined with the same eight observations.

A Practical Sensitivity Workflow

Fix the data model.
State the likelihood, observed experience, exposure units, and the future quantity being priced. Keep these fixed across the prior comparisons.

Specify each prior completely.
Report its mean and strength, or its distribution parameters and parameterization. For a Gamma prior, state explicitly whether the second parameter is a rate or a scale.

Calculate posterior premiums consistently.
Use the same posterior quantity for every prior, such as the posterior expected claim cost or frequency for the same future period.

Quantify and interpret the difference.
Report the absolute difference and, when helpful, a relative difference. Compare it with an explicit actuarial decision benchmark and explain which prior feature caused the movement.

Common Exam Traps

Reversing the weights. In the normal-normal model, $n/(n+k)$ weights the data and $k/(n+k)$ weights the prior. The source with greater precision receives greater weight.
Treating strength as the prior variance. Here $k=\sigma^2/\tau^2$, so larger $k$ means a more concentrated and stronger prior, not a more diffuse one.
Mixing parameterizations. For a Gamma prior expressed with shape and rate, the mean is $\alpha/\beta$. If the second parameter is instead a scale, the mean formula and update must be adjusted.
Comparing unlike quantities. An annual frequency should be compared with another annual frequency, and a per-claim expected cost with another per-claim expected cost. Exposure units and future periods must match.
Calling every difference material. A numerical change is not automatically important. State the threshold or business consequence used to assess materiality.

Key takeaway: With experience held fixed, prior sensitivity is controlled by both the distance between the prior mean and the observed experience and the prior’s relative strength. Calculate posterior premiums under each stated prior, quantify their difference, and interpret it against an explicit decision context.

Check Your Understanding

For each problem, hold the specified likelihood and observed experience fixed when comparing prior choices.

For a normal-normal model, $n=5$, $\bar{x}=\$620$, $\sigma=\$100$, and prior mean $m_0=\$500$ with strength $k=5$, calculate the posterior premium for one future observation. (Answer: $\$560.00$)
In a Gamma-Poisson model, a prior has mean 3 claims per exposure unit and strength $\beta=2$ exposure units. Observed experience is 20 claims over 4 exposure units. Find the posterior expected frequency. (Answer: $4.3333$ claims per exposure unit)
In a normal-normal model, $n=6$ and $k=3$. With the observed data and strength fixed, the prior mean increases by $\$300$. By how much does the posterior premium change? (Answer: $\$100.00$)
A Gamma-Poisson prior has mean 7 claims per year and rate $\beta=5$ years. The data are 20 claims over 5 years. Calculate the posterior expected annual frequency. (Answer: $5.5000$ claims per year)
In a normal-normal model, $\bar{x}=\$1{,}000$, $n=10$, and $\sigma^2=\$2{,}500$. Compare prior A $(m_A=\$900,k_A=5)$ with prior B $(m_B=\$1{,}100,k_B=10)$. What is the absolute difference between the posterior premiums? (Answer: $\$83.33$)

What You'll Learn