Predictive Quantiles and Tail Probabilities for Claims

Why Predictive Tails Need Care

A forecast of future claims is uncertain for two reasons: future outcomes vary even when a risk parameter is known, and the parameter itself is uncertain after observing experience. A posterior predictive distribution accounts for both. Its upper tail is often the part most relevant to a decision such as setting a reserve threshold or assessing the chance that a claim count exceeds a specified level.

A posterior predictive quantile is not generally obtained by inserting a posterior mean into a model and taking that model’s quantile. Instead, the forecast averages over plausible parameter values, weighted by their posterior probabilities. This distinction can materially affect high quantiles and exceedance probabilities.

Definition: For a future random quantity $Y$, its posterior predictive distribution is the distribution of $Y$ after averaging its conditional distribution over the posterior distribution of the unknown parameter. For a discrete count $N$, the $p$-quantile is the smallest integer $n$ for which $P(N\le n\mid\text{data})\ge p$. Its exceedance probability at threshold $k$ is $P(N\ge k\mid\text{data})$.

The discreteness matters: the probability mass at a particular count can make the cumulative probability jump over the requested percentile. Also, “at least $k$” includes $k$, so its complement is the probability of a count no greater than $k-1$, not $k$.

Poisson-Gamma Predictive Counts

Suppose the annual claim count rate $\lambda$ has posterior distribution $\operatorname{Gamma}(\alpha,\beta)$, using the shape-rate convention. Thus its density is proportional to $\lambda^{\alpha-1}e^{-\beta\lambda}$, its posterior mean is $\alpha/\beta$, and $\beta$ has units of time when $\lambda$ is measured per unit time. Conditional on $\lambda$, let the count $N$ over future exposure $t$ have a Poisson distribution with mean $t\lambda$. The exposure must use the same time unit as the rate.

Integrating out $\lambda$ gives the posterior predictive probability mass function:

P(N=n\mid\text{data})= \frac{\Gamma(n+\alpha)}{\Gamma(\alpha)n!} \left(\frac{\beta}{\beta+t}\right)^\alpha \left(\frac{t}{\beta+t}\right)^n,\qquad n=0,1,2,\ldots

This is a negative binomial distribution under the parameterization displayed in the formula. The first fraction is the rising-factorial coefficient; the two remaining terms depend on the posterior rate and the future exposure. One useful check is that the probabilities sum to one, since this is the negative binomial probability mass function with success probability $\beta/(\beta+t)$.

Derivation: The predictive probability is the conditional Poisson probability averaged over the posterior density:

\begin{aligned} P(N=n\mid\text{data}) &=\int_0^\infty \frac{(t\lambda)^n e^{-t\lambda}}{n!} \frac{\beta^\alpha}{\Gamma(\alpha)}\lambda^{\alpha-1}e^{-\beta\lambda}\,d\lambda\\ &=\frac{t^n\beta^\alpha}{n!\Gamma(\alpha)} \int_0^\infty \lambda^{n+\alpha-1}e^{-(\beta+t)\lambda}\,d\lambda\\ &=\frac{t^n\beta^\alpha}{n!\Gamma(\alpha)} \frac{\Gamma(n+\alpha)}{(\beta+t)^{n+\alpha}}. \end{aligned}

The final equality uses the Gamma integral: for $a>0$ and $c>0$, $\int_0^\infty x^{a-1}e^{-cx}\,dx=\Gamma(a)/c^a$. Rearranging the result yields the displayed predictive probability mass function. The added uncertainty is visible in its variance:

E[N\mid\text{data}]=\frac{t\alpha}{\beta}, \qquad \operatorname{Var}(N\mid\text{data})= \frac{t\alpha}{\beta}+\frac{t^2\alpha}{\beta^2}.

The first variance term is the expected conditional Poisson variance. The second is the variance in the conditional mean $t\lambda$ as $\lambda$ varies over its posterior distribution. A Poisson forecast with the posterior mean rate retains the first source of variation but omits the second.

Worked Example: A 90th-Percentile Claim Count

In a hypothetical portfolio, the posterior annual claim rate is $\lambda\mid\text{data}\sim\operatorname{Gamma}(3,2)$, using shape 3 and rate 2 per year. Find the 90th percentile of the count during one future year, and calculate the probability of at least four claims. Use the posterior predictive distribution.

Method and parameters. Here $\alpha=3$, $\beta=2$ per year, and $t=1$ year. Therefore $\beta/(\beta+t)=2/3$ and $t/(\beta+t)=1/3$. For integer $n$, the coefficient is $\Gamma(n+3)/(\Gamma(3)n!)=(n+1)(n+2)/2$.

The probabilities from zero through four are:

\begin{aligned} P(N=0)&=(2/3)^3=8/27=0.296296,\\ P(N=1)&=3(2/3)^3(1/3)=8/27=0.296296,\\ P(N=2)&=6(2/3)^3(1/3)^2=16/81=0.197531,\\ P(N=3)&=10(2/3)^3(1/3)^3=80/729=0.109739,\\ P(N=4)&=15(2/3)^3(1/3)^4=40/729=0.054870. \end{aligned}

The cumulative probabilities are $F(0)=0.296296$, $F(1)=0.592593$, $F(2)=0.790123$, and $F(3)=656/729=0.899863$. Since $F(3)<0.90$, the 90th percentile is not 3. Adding $P(N=4)=40/729$ gives $F(4)=696/729=0.954733$, so the smallest count reaching 0.90 is 4 claims.

For at least four claims, use the complement of counts zero through three: $P(N\ge4)=1-F(3)=1-656/729=73/729=0.1001$, to four decimal places. As a check, the percentile condition is satisfied because $F(3)=0.899863<0.90\le0.954733=F(4)$. The common mistake avoided here is treating the 90th percentile as the count for which the cumulative probability is merely closest to 0.90; the definition requires the first count whose cumulative probability reaches or exceeds 0.90.

Working from a Tail Threshold

If the question supplies a threshold rather than a percentile, calculate the cumulative probability just below that threshold and take its complement. This avoids summing a long upper tail. For a count threshold $k$, the identity is $P(N\ge k)=1-F(k-1)$. For example, $P(N>k)=1-F(k)$; confusing these statements creates a one-count error.

Worked Example: Exceeding a Count Threshold Over Two Years

A hypothetical insured group has a posterior annual claim rate $\lambda\mid\text{data}\sim\operatorname{Gamma}(4,4)$, with shape 4 and rate 4 per year. Find the probability of at least three claims during the next two years and the 95th-percentile count.

Method and parameters. Set $\alpha=4$, $\beta=4$ per year, and $t=2$ years. Then $\beta/(\beta+t)=2/3$ and $t/(\beta+t)=1/3$. For $n=0,1,2,\ldots$, use the coefficient $\Gamma(n+4)/(\Gamma(4)n!$).

The probabilities needed for the threshold and percentile are:

\begin{aligned} P(N=0)&=(2/3)^4=16/81=0.197531,\\ P(N=1)&=4(2/3)^4(1/3)=64/243=0.263374,\\ P(N=2)&=10(2/3)^4(1/3)^2=160/729=0.219479,\\ P(N=3)&=20(2/3)^4(1/3)^3=320/2187=0.146319,\\ P(N=4)&=35(2/3)^4(1/3)^4=560/6561=0.085353,\\ P(N=5)&=56(2/3)^4(1/3)^5=896/19683=0.045521. \end{aligned}

For the exceedance probability, add the probabilities below three: $F(2)=16/81+64/243+160/729=(144+192+160)/729=496/729=0.680384$. Therefore $P(N\ge3)=1-496/729=233/729=0.3196$, to four decimal places.

For the 95th percentile, the cumulative probabilities at 4 and 5 are $F(4)=5984/6561=0.912056$ and $F(5)=0.957577$. Since the first is below 0.95 and the second is at least 0.95, the 95th percentile is 5 claims. The check is $F(4)<0.95\le F(5)$. The method uses two years of exposure directly; using $t=1$ would forecast the wrong period.

Posterior Uncertainty in Individual Severity Tails

The same averaging principle applies when forecasting a claim amount. Consider a hypothetical individual claim whose severity $X$, conditional on its rate parameter $\theta$, is exponentially distributed with rate $\theta$. Suppose the posterior distribution of $\theta$ is $\operatorname{Gamma}(\alpha,\beta)$ in the shape-rate convention. Then $P(X>x\mid\theta)=e^{-\theta x}$, for $x\ge0$. Averaging this survival probability over the posterior gives:

P(X>x\mid\text{data}) =E[e^{-\theta x}\mid\text{data}] =\left(\frac{\beta}{\beta+x}\right)^\alpha.

The corresponding predictive cumulative distribution is $F_X(x)=1-(\beta/(\beta+x))^\alpha$. For $0<p<1$, solving $F_X(x_p)=p$ gives the predictive quantile $x_p=\beta\big((1-p)^{-1/\alpha}-1\big)$. Here $x$ and $\beta$ must use the same monetary unit: if $x$ is in dollars, $\beta$ is in dollars as well. This predictive distribution is not exponential; mixing exponential distributions over an uncertain rate creates a heavier tail.

Worked Example: A Predictive Severity Quantile

For a hypothetical line of coverage, suppose the posterior claim severity rate is $\theta\mid\text{data}\sim\operatorname{Gamma}(2,10000)$, with shape 2 and rate 10,000 dollars. Conditional on $\theta$, an individual claim severity is exponential with rate $\theta$. Find the probability that severity exceeds $10,000 and the 90th-percentile predictive severity.

Method and parameters. The predictive survival function is $(\beta/(\beta+x))^\alpha$, with $\alpha=2$ and $\beta=10{,}000$ dollars. At $x=\$10{,}000$:

P(X>\$10{,}000\mid\text{data}) =\left(\frac{10{,}000}{10{,}000+10{,}000}\right)^2 =\left(\frac12\right)^2=0.25.

For the 90th percentile, set the predictive survival probability to $1-0.90=0.10$: \[ \left(\frac{10{,}000}{10{,}000+x_{0.90}}\right)^2=0.10. \] Taking positive square roots gives $10{,}000/(10{,}000+x_{0.90})=\sqrt{0.10}$. Hence $x_{0.90}=10{,}000(1/\sqrt{0.10}-1)=10{,}000(\sqrt{10}-1)=\$21{,}622.78$, rounded to the nearest cent. Substitution gives survival probability 0.10, checking that the cumulative probability is 0.90. This method avoids a plug-in forecast: using the posterior mean rate $2/10{,}000=0.0002$ in a single exponential model would instead give exceedance probability $e^{-2}=0.1353$ at $10,000, below the posterior predictive value of 0.25.

Interpreting and Checking Upper-Tail Results

The posterior predictive tail includes both process variation and uncertainty about the underlying risk parameter. As the count example’s variance formula shows, ignoring posterior uncertainty removes a nonnegative variance component. For the severity example, averaging $e^{-\theta x}$ over uncertain rates gives a larger tail at the illustrated threshold than evaluating the survival function only at the posterior mean rate. This does not mean every posterior predictive quantile must be larger than every plug-in quantile under every model and percentile; it means the plug-in distribution is not the full predictive distribution and should not be substituted without justification.

In practical calculations, identify the event precisely before using a cumulative probability. For a discrete variable, $N\ge k$ and $N>k$ differ by the mass at $k$. For a continuous severity distribution, the probability at a single exact value is zero under the model, so $P(X>x)$ and $P(X\ge x)$ are equal. Also check whether the Gamma parameter is a rate or a scale. In the shape-rate convention used here, the posterior mean is $\alpha/\beta$, and the predictive count uses $\beta/(\beta+t)$; replacing rate by scale reverses the role of $\beta$.

Exam trap: A common wrong calculation for “at least four claims” is $1-F(4)$. That is $P(N>4)$, which excludes exactly four claims. The correction is $P(N\ge4)=1-F(3)$. Always translate the event into the complementary counts before calculating.

These calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus. A reliable solution states the posterior parameterization, matches exposure units, identifies whether the target is a quantile or a tail probability, and checks the final result against a cumulative probability or an equivalent survival probability.

Check Your Understanding

For each problem, use the stated posterior predictive model and take care with the event boundary.

A hypothetical portfolio has $\lambda\mid\text{data}\sim\operatorname{Gamma}(2,3)$, with rate 3 per year, and the forecast covers one year. Find the probability of at least two claims. (Answer: $5/32=0.1563$)
A claim count rate has posterior $\operatorname{Gamma}(1,2)$, with rate 2 per year, and one future year is exposed. Find the 90th-percentile count. (Answer: 2 claims)
A hypothetical individual severity model has $X\mid\theta$ exponential with rate $\theta$, and $\theta\mid\text{data}\sim\operatorname{Gamma}(3,6000)$, with rate parameter in dollars. Find $P(X>\$6{,}000\mid\text{data})$. (Answer: 0.1250)
A hypothetical count rate has posterior $\operatorname{Gamma}(3,3)$, with rate 3 per year. Forecast two years and find the probability of zero claims. (Answer: $27/125=0.2160$)
A hypothetical portfolio has $\lambda\mid\text{data}\sim\operatorname{Gamma}(2,2)$, with rate 2 per year, and one year of future exposure. Find the probability of at least three claims. (Answer: $1/9=0.1111$)

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