Credibility Estimation of Exponential Severity Means

Introduction

When severity observations are modeled as exponential with an unknown mean, a Bayesian estimate combines the observed losses with prior information about that mean. The inverse-gamma distribution is conjugate for this parameterization: after observing independent severities, the posterior distribution remains inverse-gamma. Its expected value can often be written as a credibility-weighted average of the sample mean and prior mean.

The distinction between an inverse-gamma prior on the mean and a gamma prior on the exponential rate is important. This tutorial uses $\theta$ for the unknown exponential mean, so the conditional exponential rate is $1/\theta$. The prior’s shape also determines whether its expected value exists; a posterior expectation may be finite even when the prior expectation is not.

Definition: Let $X_1,\ldots,X_n$, conditional on an unknown severity mean $\theta>0$, be independent and identically distributed exponential random variables with mean $\theta$. Use the inverse-gamma parameterization $$ \pi(\theta)=\frac{\beta^\alpha}{\Gamma(\alpha)}\theta^{-(\alpha+1)}\exp\left(-\frac{\beta}{\theta}\right),\qquad \theta>0, $$ where $\alpha>0$ is the shape and $\beta>0$ is the scale parameter in this inverse-gamma parameterization. The conditional density is $f(x\mid\theta)=\theta^{-1}\exp(-x/\theta)$ for $x>0$. The prior expectation is $E[\theta]=\beta/(\alpha-1)$ if $\alpha>1$; it is infinite if $0<\alpha\leq 1$.

Posterior Distribution and Expected Severity

The model assumes that the severities are positive, fully observed, and conditionally independent given one common mean $\theta$. It does not account for censoring, truncation, or different severity means across risks. Under these assumptions, the sample affects the posterior through its count $n$ and total $T=\sum_{j=1}^n X_j$.

Theorem: With the inverse-gamma prior above and $n$ independent exponential observations with total $T$, the posterior distribution of $\theta$ is inverse-gamma with shape $\alpha+n$ and scale $\beta+T$. Its expectation is $$ E[\theta\mid X_1,\ldots,X_n]=\frac{\beta+T}{\alpha+n-1}, $$ provided $\alpha+n>1$. If $\alpha+n\leq1$, the posterior expectation is infinite.

To see why the update holds, the likelihood for the observed sample is proportional, as a function of $\theta$, to $\theta^{-n}\exp(-T/\theta)$. Multiplying this likelihood by the prior density gives the posterior kernel:

\pi(\theta\mid X_1,\ldots,X_n)\ \propto\ \theta^{-(\alpha+n+1)}\exp\left(-\frac{\beta+T}{\theta}\right).

This is the kernel of an inverse-gamma density with shape $\alpha+n$ and scale $\beta+T$. For an inverse-gamma variable with shape $a>1$ and scale $b$, its expectation follows by substituting $u=b/\theta$ in the defining integral:

E[\theta]=\frac{b^a}{\Gamma(a)}\int_0^\infty \theta^{-a}e^{-b/\theta}\,d\theta =\frac{b}{\Gamma(a)}\int_0^\infty u^{a-2}e^{-u}\,du =\frac{b\Gamma(a-1)}{\Gamma(a)} =\frac{b}{a-1}.

The integral is finite only for $a>1$. Taking $a=\alpha+n$ and $b=\beta+T$ gives the stated posterior expectation and its existence condition.

Credibility-Weighted Average

If $\alpha>1$, the prior expected severity is finite. Write it as $\mu_0=\beta/(\alpha-1)$, and write the sample mean as $\bar X=T/n$ for $n>0$. Since $\beta=(\alpha-1)\mu_0$ and $T=n\bar X$, the posterior expectation becomes

E[\theta\mid X_1,\ldots,X_n] =\frac{(\alpha-1)\mu_0+n\bar X}{(\alpha-1)+n} =Z\bar X+(1-Z)\mu_0, \qquad Z=\frac{n}{n+\alpha-1}.

Thus $Z$ is the weight on the observed sample mean, while $1-Z$ is the weight on the prior mean. Both weights are nonnegative and sum to one when $\alpha>1$. More observations increase $Z$; a larger $\alpha-1$, with the prior mean held fixed, gives the prior greater weight.

The prior contribution in this expression is $\alpha-1$, not $\alpha$. The subtraction arises because the expected value of an inverse-gamma distribution with shape $\alpha$ is $\beta/(\alpha-1)$, rather than $\beta/\alpha$. If $\alpha\leq1$, there is no finite prior mean to use in this weighted-average representation. The posterior expectation can nevertheless exist once the data make $\alpha+n>1$.

Worked Example: A Finite Prior Mean and Three Observed Severities

A hypothetical portfolio’s individual claim severities are modeled as conditionally independent exponential amounts with common unknown mean $\theta$, measured in dollars. Before observing data, the actuary uses an inverse-gamma prior with shape $\alpha=5$ and scale $\beta=\$1{,}200$. Three fully observed severities are $\$250$, $\$400$, and $\$550$. Find the posterior expected severity using both the posterior formula and the weighted-average form.

Method and data summary. The observations follow the assumed exponential model. Their count and total are $n=3$ and $T=250+400+550=\$1{,}200$, so the sample mean is $\bar X=1{,}200/3=\$400$.

Posterior update. The posterior shape is $\alpha+n=5+3=8$, and its scale is $\beta+T=1{,}200+1{,}200=\$2{,}400$. Since $8>1$, the posterior expectation exists:

E[\theta\mid X_1,X_2,X_3]=\frac{2{,}400}{8-1} =\frac{2{,}400}{7} =\$342.857\ldots

Weighted-average check. The prior mean is $1{,}200/(5-1)=\$300$. The data weight is $Z=3/(3+5-1)=3/7$, and the prior weight is $4/7$. Therefore,

\frac{3}{7}(400)+\frac{4}{7}(300) =\frac{1{,}200+1{,}200}{7} =\$342.857\ldots

Both methods agree. Rounded only at the final step to the nearest cent, the posterior expected severity is \$342.86. The common mistake avoided here is using $\alpha=5$, instead of $\alpha-1=4$, as the prior weight.

Worked Example: Updating a Prior With a Smaller Mean

For a hypothetical equipment policy, suppose fully observed claim severities are conditionally independent exponential amounts with common mean $\theta$, in dollars. The prior for $\theta$ is inverse-gamma with $\alpha=3$ and $\beta=\$1{,}800$. Two severities are $\$800$ and $\$1{,}200$. Calculate the posterior expected severity and verify its credibility interpretation.

Calculate the observed statistics. There are $n=2$ observations, their total is $T=800+1{,}200=\$2{,}000$, and their mean is $\bar X=2{,}000/2=\$1{,}000$.

Update and take the expectation. The posterior shape is $3+2=5$, and the posterior scale is $1{,}800+2{,}000=\$3{,}800$. The posterior mean exists because $5>1$:

E[\theta\mid X_1,X_2]=\frac{3{,}800}{5-1}=\$950.00.

Check using credibility weights. The prior expectation is $1{,}800/(3-1)=\$900$. The sample weight is $Z=2/(2+3-1)=1/2$, and the prior weight is $1/2$. Hence $E[\theta\mid X_1,X_2]=(1/2)(1{,}000)+(1/2)(900)=\$950$. The equality checks both the update and the weighting calculation.

Worked Example: A Finite Posterior Mean Without a Finite Prior Mean

Consider a hypothetical product with independent, fully observed exponential severities conditional on their common mean $\theta$. The prior is inverse-gamma with $\alpha=0.5$ and $\beta=\$1{,}000$. Two severities are $\$800$ and $\$1{,}200$. Determine whether the prior and posterior expected severities exist, and calculate the posterior expectation.

Check the prior. The prior shape is $0.5\leq1$, so its expected value does not exist as a finite number; in fact, the expectation is infinite. A finite-prior-mean weighted-average interpretation is therefore unavailable.

Update from the observations. Here $n=2$ and $T=800+1{,}200=\$2{,}000$. The posterior shape is $0.5+2=2.5$, and the posterior scale is $1{,}000+2{,}000=\$3{,}000$. Since $2.5>1$, the posterior expected severity is finite:

E[\theta\mid X_1,X_2]=\frac{3{,}000}{2.5-1} =\frac{3{,}000}{1.5} =\$2{,}000.00.

This result is not a weighted average of the sample mean and a finite prior mean: the prior mean is not finite. It is still a well-defined posterior expectation because the observed data raise the posterior shape above one. This distinction matters whenever checking existence conditions, not just computing a ratio.

Common Exam Trap

For a finite prior mean, an incorrect shortcut is to treat the shape $\alpha$ as the prior’s pseudo-count. In the first example, that would incorrectly calculate

\frac{\alpha\mu_0+n\bar X}{\alpha+n} =\frac{5(300)+3(400)}{5+3} =\frac{2{,}700}{8} =\$337.50.

The correction is to use the prior strength $\alpha-1$, giving $((5-1)(300)+3(400))/((5-1)+3)=\$342.86$, consistent with the posterior formula. The incorrect result also fails the direct posterior check $2{,}400/(8-1)=\$342.86$. A reliable sequence is to update shape and scale first, check whether the posterior shape exceeds one, and only then compute its expectation or use the weighted-average form.

This conjugate update is useful when an estimate of mean severity must reflect both prior information and a small amount of claim data. It also provides a quick internal check: when the prior mean is finite, the posterior estimate must lie between the prior mean and sample mean because it is a convex combination of them. When the prior mean is not finite, do not force that interpretation; calculate the posterior expectation directly and check its existence condition.

Check Your Understanding

For each problem, assume the stated severities are fully observed and conditionally independent exponential amounts with a common unknown mean. Report dollar amounts to the nearest cent.

A prior is inverse-gamma with $\alpha=4$ and $\beta=\$900$. Two observed severities are $\$300$ and $\$500$. Find the posterior expected severity. (Answer: \$340.00)
A prior is inverse-gamma with $\alpha=2.5$ and $\beta=\$600$. Three severities are $\$100$, $\$200$, and $\$300$. State the prior mean and calculate the posterior expected severity. (Answer: Prior mean \$400.00; posterior expected severity \$266.67)
A prior is inverse-gamma with $\alpha=0.8$ and $\beta=\$500$. One observed severity is $\$1{,}000$. Does the prior mean exist, and what is the posterior expected severity? (Answer: The prior mean is infinite; the posterior expected severity is \$1,875.00)
A prior is inverse-gamma with $\alpha=1$ and $\beta=\$700$. Three observed severities have total $\$2{,}100$. Does the prior mean exist, and calculate the posterior expected severity. (Answer: The prior mean is infinite; the posterior expected severity is \$933.33)
A prior is inverse-gamma with $\alpha=6$ and $\beta=\$2{,}500$. Four observed severities have sample mean $\$600$. Calculate the posterior expected severity and its weight on the data. (Answer: \$544.44; data weight $4/9$)

What You'll Learn