Predictive Severity With an Uncertain Mean

Why Predictive Severity Includes Parameter Uncertainty

A forecast for one future claim severity depends both on the severity process and on what is known about its mean. If the mean is uncertain, substituting a single estimate for it omits the uncertainty that remains after observing past claims. A posterior predictive distribution accounts for both sources of variation by averaging the conditional severity distribution over the posterior distribution of the mean.

This tutorial develops that calculation for a specific positive-severity model: claim severities are conditionally exponential, with an unknown mean, and the prior for that mean is inverse gamma. The resulting predictive distribution is a Lomax distribution. This is a model-based result, not a claim that all insurance severities are exponential. It is appropriate only when the conditional exponential assumption and the stated prior are reasonable for the application.

Model and assumptions: Let $\Theta$ be the unknown mean severity, measured in dollars. Conditional on $\Theta=\theta$, each observed severity $X_j$ and one future severity $X_{\mathrm{new}}$ are independent and identically distributed with density $f(x\mid\theta)=\theta^{-1}\exp(-x/\theta)$ for $x>0$. Assume the observed severities are complete, uncensored, and drawn from this same conditional distribution. Use the inverse-gamma prior $\Theta\sim\operatorname{InvGamma}(\alpha,\beta)$, with density $f(\theta)=\beta^\alpha\theta^{-(\alpha+1)}\exp(-\beta/\theta)/\Gamma(\alpha)$ for $\theta>0$, where $\alpha,\beta>0$. Here $\beta$ has dollar units; this is not a Gamma shape-rate prior on $\Theta$.

Updating the Mean After Observed Severities

Suppose $n$ positive severities $x_1,\ldots,x_n$ have been observed, and let $T=\sum_{j=1}^n x_j$. Under the conditional exponential model, their likelihood as a function of $\theta$ is proportional to $\theta^{-n}\exp(-T/\theta)$. Multiplication by the inverse-gamma prior gives a posterior with the same form:

\Theta\mid x_1,\ldots,x_n\sim\operatorname{InvGamma}(a,b), \qquad a=\alpha+n,\qquad b=\beta+T.

The update adds one to the shape for each observed claim and adds each observed severity to the prior scale parameter. The posterior mean of $\Theta$ is $b/(a-1)$ when $a>1$; its variance is $b^2/[(a-1)^2(a-2)]$ when $a>2$. These are moments of the unknown conditional mean, not yet the predictive moments of a future severity.

Worked Example: Updating and Forecasting One Severity

A hypothetical portfolio uses the model above. Its prior for the mean severity is $\operatorname{InvGamma}(4,2000)$, with $\beta$ measured in dollars. Three observed severities are $400, $600, and $1,000. Find the posterior predictive mean and variance for one future severity, and calculate the probability that it is between $2,000 and $6,000.

Method: update the inverse-gamma parameters, then use the posterior predictive distribution. The number of observations is $n=3$, and their total is $T=400+600+1{,}000=2{,}000$. Thus $a=4+3=7$ and $b=2{,}000+2{,}000=4{,}000$.

The predictive mean, which exists because $a=7>1$, is $b/(a-1)=4{,}000/6=\$666.67$, rounded to the nearest cent. The predictive variance, which exists because $a>2$, is $b^2a/[(a-1)^2(a-2)]=4{,}000^2(7)/(6^2\cdot5)=112{,}000{,}000/180=\$622{,}222.22$ squared, rounded to the nearest cent.

For the interval probability, the predictive survival probabilities at the endpoints are $P(X_{\mathrm{new}}>2{,}000\mid x)=(4{,}000/6{,}000)^7=(2/3)^7=128/2{,}187\approx0.0585$ and $P(X_{\mathrm{new}}>6{,}000\mid x)=(4{,}000/10{,}000)^7=0.4^7=0.0016384$. Therefore, $P(2{,}000<X_{\mathrm{new}}\le6{,}000\mid x)=0.0585277-0.0016384=0.0569$, to four decimal places. The checks are that the posterior shape and scale both exceed their prior values, and the interval probability is positive and smaller than the probability of exceeding $2,000.

Deriving the Posterior Predictive Distribution

The future severity is uncertain even if $\Theta$ were known. When $\Theta$ is unknown, its posterior uncertainty must also be included. The predictive density is obtained by integrating the conditional density over the posterior distribution of $\Theta$:

f(x\mid x_1,\ldots,x_n) =\int_0^\infty f(x\mid\theta)f(\theta\mid x_1,\ldots,x_n)\,d\theta.

Write $a=\alpha+n$ and $b=\beta+T$, so the posterior density is inverse gamma with these parameters. For $x>0$, substitution into the integral gives

\begin{aligned} f(x\mid x_1,\ldots,x_n) &=\frac{b^a}{\Gamma(a)} \int_0^\infty \theta^{-(a+2)} \exp\left(-\frac{b+x}{\theta}\right)\,d\theta\\ &=\frac{b^a}{\Gamma(a)} \frac{\Gamma(a+1)}{(b+x)^{a+1}} =\frac{a b^a}{(b+x)^{a+1}}. \end{aligned}

For the second equality, set $z=(b+x)/\theta$. Then the integral becomes $\Gamma(a+1)/(b+x)^{a+1}$, and $\Gamma(a+1)=a\Gamma(a)$. Integrating this density above $x$ gives the survival function. The cumulative distribution function follows by subtracting the survival probability from one.

Theorem: Under the stated conditional exponential model and inverse-gamma prior, the posterior predictive density and survival function for one future severity are $f(x\mid x_1,\ldots,x_n)=ab^a/(b+x)^{a+1}$ for $x>0$, and $P(X_{\mathrm{new}}>x\mid x_1,\ldots,x_n)=[b/(b+x)]^a$ for $x\ge0$, where $a=\alpha+n$ and $b=\beta+\sum_jx_j$. This is a Lomax distribution with shape $a$ and scale $b$. For $0\le l<u$, its interval probability is $[b/(b+l)]^a-[b/(b+u)]^a$.

Predictive Mean and Variance

The Lomax survival function also gives the predictive moments. For a nonnegative severity, the mean can be calculated by integrating its survival function:

E[X_{\mathrm{new}}\mid x_1,\ldots,x_n] =\int_0^\infty \left(\frac{b}{b+x}\right)^a dx =\frac{b}{a-1},\qquad a>1.

The second raw moment is twice the integral of $x$ times the survival function. It is finite only for $a>2$, and equals $2b^2/[(a-1)(a-2)]$. Subtracting the square of the mean gives the predictive variance:

\operatorname{Var}(X_{\mathrm{new}}\mid x_1,\ldots,x_n) =\frac{2b^2}{(a-1)(a-2)}-\left(\frac{b}{a-1}\right)^2 =\frac{ab^2}{(a-1)^2(a-2)},\qquad a>2.

The existence conditions matter. If $a\le1$, the predictive mean is infinite. If $1<a\le2$, the mean is finite but the variance is infinite. It is not valid to report a finite variance merely because the posterior distribution for $\Theta$ has a finite mean.

Worked Example: Calculating an Interval Probability

For a hypothetical book of risks, the prior for mean severity is $\operatorname{InvGamma}(3,3000)$. Two observed severities are $500 and $1,500. Calculate the posterior predictive mean and variance for one future claim, and the probability its severity is more than $2,500 but no more than $7,500.

Method: update first, then subtract predictive tail probabilities. Here $n=2$ and $T=500+1{,}500=2{,}000$. The posterior parameters are $a=3+2=5$ and $b=3{,}000+2{,}000=5{,}000$.

The predictive mean is $5{,}000/(5-1)=\$1{,}250.00$. The predictive variance is $5{,}000^2(5)/[(5-1)^2(5-2)]=125{,}000{,}000/48=\$2{,}604{,}166.67$ squared, rounded to the nearest cent.

At $2,500, the survival probability is $(5{,}000/7{,}500)^5=(2/3)^5=32/243\approx0.1316872$. At $7,500, it is $(5{,}000/12{,}500)^5=(2/5)^5=0.01024$. Subtracting gives $0.1316872-0.01024=0.1214472$, or $0.1214$ to four decimal places. The interval probability is less than the probability of exceeding $2,500, as required.

Separating Severity Variation From Mean Uncertainty

The law of total variance explains why the predictive variance exceeds the conditional variance at a fixed mean. Given $\Theta$, an exponential severity has conditional mean $\Theta$ and conditional variance $\Theta^2$. Therefore,

\operatorname{Var}(X_{\mathrm{new}}\mid x) =E[\Theta^2\mid x]+\operatorname{Var}(\Theta\mid x).

The first term is average claim-level variation conditional on the mean; the second is variation in the mean itself. This identity gives the same predictive variance as the Lomax formula when $a>2$. It also identifies the contribution that a plug-in forecast leaves out.

Worked Example: Quantifying the Added Parameter Uncertainty

A hypothetical portfolio has prior $\operatorname{InvGamma}(4,3000)$ for its mean severity and observes claims of $1,000 and $2,000. Calculate the predictive mean and variance for one future severity, and the probability that it exceeds $6,000.

Method: update the posterior and apply the variance decomposition. The observed total is $T=1{,}000+2{,}000=3{,}000$, so $a=4+2=6$ and $b=3{,}000+3{,}000=6{,}000$. The predictive mean is $6{,}000/(6-1)=\$1{,}200.00$.

The posterior second moment of the mean is $E[\Theta^2\mid x]=b^2/[(a-1)(a-2)]=36{,}000{,}000/(5\cdot4)=1{,}800{,}000$ dollars squared. Since $E[\Theta\mid x]^2=1{,}200^2=1{,}440{,}000$, the posterior variance of the mean is $1{,}800{,}000-1{,}440{,}000=360{,}000$ dollars squared. Thus the predictive variance is $1{,}800{,}000+360{,}000=2{,}160{,}000$ dollars squared. Direct substitution checks this result: $6(6{,}000^2)/(5^2\cdot4)=2{,}160{,}000$.

The probability of exceeding $6,000 is $[6{,}000/(6{,}000+6{,}000)]^6=(1/2)^6=0.015625$, or $0.0156$ to four decimal places. The predictive variance exceeds the average conditional variance by exactly the posterior variance of $\Theta$, as the decomposition requires.

Common Mistake: Treating the Posterior Mean as Known

A plug-in calculation replaces $\Theta$ with its posterior mean and then uses an exponential distribution as though that value were known exactly. In the second worked example, this would use a fixed mean of $1,250 and report variance $1{,}250^2=\$1{,}562{,}500$ squared. The correct predictive variance is $\$2{,}604{,}166.67$ squared because it includes uncertainty in $\Theta$. The plug-in exponential tail probability above $2,500 would be $\exp(-2)=0.1353$, whereas the posterior predictive probability is $(2/3)^5=0.1317$. These forecasts answer different questions.

Other checks are equally important: update $b$ using the sum of severities, keep its units consistent with the losses, and use the posterior predictive survival function rather than the conditional exponential survival function when $\Theta$ remains uncertain. If observations are censored, truncated, or not conditionally identically distributed, the likelihood and update need to reflect those facts. A different severity family or prior may also produce a different posterior predictive distribution.

Key takeaway: With conditional exponential severities and an inverse-gamma prior on the unknown mean, update to $a=\alpha+n$ and $b=\beta+\sum_jx_j$. The future severity then has a Lomax predictive distribution. Its mean is $b/(a-1)$ for $a>1$, and its variance is $ab^2/[(a-1)^2(a-2)]$ for $a>2$. Both the model assumptions and the moment conditions must be checked before using these formulas.

Check Your Understanding

For each problem, use the conditional exponential and inverse-gamma model defined above.

A prior is $\operatorname{InvGamma}(2,1000)$, and one severity of $500 is observed. Find the posterior predictive mean and the probability that the next severity exceeds $1,500. (Answer: mean $750.00; probability 0.1250)
A prior is $\operatorname{InvGamma}(5,4000)$, and observed severities are $1,000 and $3,000. Find the probability that the next severity is between $4,000 and $12,000. (Answer: 0.0569)
The posterior parameters are $a=4$ and $b=3000$. Calculate the posterior predictive variance. (Answer: 2,000,000 dollars squared)
The posterior parameters are $a=2$ and $b=9000$. State whether the predictive mean and variance are finite, and give the mean if it is finite. (Answer: mean is finite and equals $9,000.00; variance is infinite)
The posterior parameters are $a=5$ and $b=2500$. Find the probability that the next severity is positive and no more than $2,500. (Answer: 0.9688)

What You'll Learn