Why CDFs and Quantiles Matter
A severity model is useful not only for describing a loss distribution but also for answering practical questions about probabilities and thresholds. An insurer might need the probability that a hypothetical claim exceeds a specified amount, the chance that a loss falls in a particular range, or the loss amount exceeded by only 10% of claims. These are CDF, survival-probability, and quantile calculations.
The formulas depend on the model’s parameterization. In particular, the exponential rate is not the same as its scale, and the two common Pareto models use different supports and quantile formulas. Begin by recording the distribution, its parameters, and the units of the loss.
For a continuous distribution, endpoints do not affect an interval probability: \(P(a<X\le b)=P(a\le X\le b)\). A percentile is not an upper-tail probability: the 90th percentile has 90% of the distribution at or below it and 10% above it. For distributions with jumps, a quantile is conventionally defined as the smallest \(x\) for which \(F_X(x)\ge p\); the continuous models considered here have no jumps.
Model Formulas and How to Invert Them
The following parameterizations are used throughout. All loss amounts and scale parameters must use the same units. For example, a rate stated per dollar must be paired with a loss amount in dollars.
| Model and parameters | CDF \(F_X(x)\) | Quantile \(q_p\) |
|---|---|---|
| Exponential, rate \(\lambda>0\), \(x\ge0\) | \(1-e^{-\lambda x}\) | \(-\ln(1-p)/\lambda\) |
| Gamma, shape \(\alpha>0\), scale \(\theta>0\), \(x\ge0\) | \(\gamma(\alpha,x/\theta)/\Gamma(\alpha)\) | Inverse of the gamma CDF |
| Pareto Type I, minimum \(\theta>0\), shape \(\alpha>0\), \(x\ge\theta\) | \(1-(\theta/x)^\alpha\) | \(\theta(1-p)^{-1/\alpha}\) |
| Pareto Type II (Lomax), scale \(\theta>0\), shape \(\alpha>0\), \(x\ge0\) | \(1-(\theta/(\theta+x))^\alpha\) | \(\theta((1-p)^{-1/\alpha}-1)\) |
| Lognormal, \(\ln X\sim N(\mu,\sigma^2)\), \(\sigma>0\) | \(\Phi((\ln x-\mu)/\sigma)\), \(x>0\) | \(\exp(\mu+\sigma z_p)\) |
| Weibull, shape \(k>0\), scale \(\lambda>0\), \(x\ge0\) | \(1-e^{-(x/\lambda)^k}\) | \(\lambda[-\ln(1-p)]^{1/k}\) |
For the gamma model, \(\Gamma(\alpha)\) is the gamma function and \(\gamma(\alpha,t)=\int_0^t u^{\alpha-1}e^{-u}\,du\) is the lower incomplete gamma function. Except in special cases, gamma CDF values and quantiles are obtained from a numerical routine or table. Here \(z_p\) is the \(p\)-quantile of the standard normal distribution, so \(\Phi(z_p)=p\). The notation \(X\sim N(\mu,\sigma^2)\) uses \(\sigma^2\) as the variance, and \(\sigma\) as the standard deviation.
Exponential and Gamma Probabilities
An exponential loss with rate \(\lambda\) has \(S_X(x)=e^{-\lambda x}\) for \(x\ge0\). Its upper-tail calculation is often direct, and its interval probability is the difference between two CDF values. A gamma loss with shape \(\alpha\) and scale \(\theta\) has density \(f_X(x)=x^{\alpha-1}e^{-x/\theta}/(\Gamma(\alpha)\theta^\alpha)\) for \(x>0\). When \(\alpha\) is a positive integer, it is an Erlang model and some probabilities have simple finite-sum formulas.
Worked Example: Exponential Interval Probability and Percentile
Suppose a hypothetical claim amount \(X\), measured in dollars, follows an exponential distribution with rate \(\lambda=0.002\) per dollar. Find (a) \(P(250<X\le800)\) and (b) the 90th percentile. Give the probability to four decimals and the percentile to the nearest cent.
Method and model. The loss is nonnegative and has \(F_X(x)=1-e^{-0.002x}\). The rate is per dollar, so the dollar amounts can be substituted directly.
(a) Interval probability. First calculate each CDF value:
Subtract the lower CDF value from the upper one:
(b) 90th percentile. Set the CDF equal to \(0.90\) and solve:
Check and trap avoided. At this percentile, the survival probability is \(e^{-0.002(1151.2926)}=0.10\), as required. A common error is to calculate the 90th percentile using \(-\ln(0.90)/\lambda\); that instead sets the survival probability to 0.90 and gives the 10th percentile.
Worked Example: Gamma Probability With an Erlang Model
A hypothetical severity \(X\), in dollars, follows a gamma distribution with shape \(\alpha=2\) and scale \(\theta=1000\). Find \(P(500<X\le2000)\), to four decimals.
Method and model. Since the shape is the integer 2, this is an Erlang model. Its CDF simplifies to \(F_X(x)=1-e^{-x/\theta}(1+x/\theta)\) for \(x\ge0\). Evaluate the CDF at both endpoints, then subtract.
At \(x=2000\), \(x/\theta=2\), so:
At \(x=500\), \(x/\theta=0.5\), so:
Therefore:
Check and trap avoided. The answer is between 0 and 1, and it is less than \(F_X(2000)\), as an interval probability must be. Subtracting the endpoint values in the reverse order would produce a negative number; using only \(F_X(2000)\) would incorrectly include losses at or below 500.
Pareto and Lognormal Calculations
Pareto Type I and Pareto Type II have different lower endpoints, so their quantile formulas must not be interchanged. Type I has no probability below its minimum \(\theta\); Type II (also called Lomax) begins at zero. Both have survival functions that make upper-tail probabilities especially convenient.
Worked Example: Distinguishing the Two Pareto Quantiles
Consider two hypothetical claim models. Model I is Pareto Type I with minimum \(\theta=500\) dollars and shape \(\alpha=2.5\). Model II is Pareto Type II with scale \(\theta=1000\) dollars and shape \(\alpha=2\). For Model I, find \(P(X>1000)\) and its 90th percentile. For Model II, find its 90th percentile. Give probabilities to four decimals and percentiles to the nearest cent.
Model I probability. Its survival function is \(S_X(x)=(\theta/x)^\alpha\) for \(x\ge\theta\). Since 1000 is above the 500-dollar minimum:
Model I percentile. Invert \(1-(\theta/q_p)^\alpha=p\), which gives \(q_p=\theta(1-p)^{-1/\alpha}\):
Model II percentile. For Type II, solving \(1-(\theta/(\theta+q_p))^\alpha=p\) gives \(q_p=\theta((1-p)^{-1/\alpha}-1)\). Therefore:
Check and trap avoided. Model I’s percentile exceeds its minimum of 500 dollars, and Model II’s percentile is nonnegative. Omitting the “\(-1\)” in the Type II formula would give the wrong result: it would treat the Type II model as though it had Type I’s positive minimum.
For a lognormal loss, the logarithm—not the loss itself—is normal. Thus a probability about \(X\) is converted to a probability about \(\ln X\), then standardized using \((\ln x-\mu)/\sigma\). This is also why a lognormal quantile is exponentiated after finding the corresponding normal quantile.
Worked Example: A Lognormal Interval and Median
A hypothetical loss \(X\), measured in dollars, satisfies \(\ln X\sim N(6,0.16)\). Find \(P(e^{5.6}<X\le e^{6.4})\) and the median loss. Give the probability to four decimals and the median to the nearest cent.
Method and model. The normal mean is \(\mu=6\), and the standard deviation is \(\sigma=\sqrt{0.16}=0.4\). Transform each boundary by taking its natural logarithm, then standardize:
Therefore, using standard normal probabilities:
(The value before rounding using more precise normal probabilities is approximately \(0.6827\).) The median corresponds to \(p=0.50\), for which \(z_{0.50}=0\):
Check and trap avoided. The standardized endpoints are symmetric about zero, so the interval probability should be about 68%. Using \(0.16\) as the standard deviation instead of taking its square root would produce incorrect standardized values; the second parameter in the stated normal distribution is the variance.
Weibull Probabilities and Quantiles
For a Weibull loss with shape \(k\) and scale \(\lambda\), the survival function is \(S_X(x)=e^{-(x/\lambda)^k}\). The scale \(\lambda\) is a loss amount, while \(k\) controls the shape. The quantile formula follows by setting this survival function equal to \(1-p\) and solving, as shown earlier.
Worked Example: Weibull Interval Probability and Upper Percentile
Suppose a hypothetical claim amount \(X\), in dollars, follows a Weibull distribution with shape \(k=2\) and scale \(\lambda=1000\) dollars. Find \(P(500<X\le1500)\) and the 90th percentile. Give the probability to four decimals and the percentile to the nearest cent.
Method and model. Use \(S_X(x)=e^{-(x/1000)^2}\). The interval probability is \(S_X(500)-S_X(1500)\), since it is the probability of exceeding 500 but not exceeding 1500.
For the 90th percentile, substitute \(p=0.90\) into the inverted CDF:
Check and trap avoided. At this percentile, \(S_X(q_{0.90})=e^{-2.3025851}=0.10\), confirming that 10% of losses exceed it. The interval calculation uses survival at the lower endpoint minus survival at the upper endpoint; reversing that subtraction gives a negative probability.
A Reliable Calculation Process
A short, consistent workflow prevents most errors in parametric probability and percentile problems.
Record its parameterization, support, and parameters. For example, distinguish a gamma scale from a rate and Pareto Type I from Type II.
Use \(F_X(b)-F_X(a)\) for an interval, \(F_X(x)\) for “at most,” and \(1-F_X(x)\) or \(S_X(x)\) for “exceeds.”
For a quantile, set \(F_X(q_p)=p\). For a lognormal model, take logarithms of loss thresholds before standardizing.
A probability must be between 0 and 1; a quantile must lie in the model’s support. Check that the survival probability at a \(p\)-quantile is \(1-p\).
The main exam traps are confusing a CDF with a survival probability, using a rate where a scale is required, and applying a formula from the wrong Pareto parameterization. In a lognormal calculation, using the variance as the standard deviation is another frequent source of error. Writing down the model and the event before substituting values makes these errors easier to detect. These probability and quantile techniques are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each hypothetical loss, identify the appropriate model formula before calculating. Round probabilities to four decimals and dollar amounts to the nearest cent where applicable.
- A loss \(X\) is exponential with rate \(0.004\) per dollar. Find \(P(X>300)\). (Answer: \(0.3010\))
- A loss \(X\) is Pareto Type I with minimum \(\theta=2\) and shape \(\alpha=3\). Find its 80th percentile. (Answer: \(2\sqrt[3]{5}\approx3.42\))
- A loss \(X\) is Weibull with shape \(k=1.5\) and scale \(\lambda=4\). Find \(P(X\le2)\). (Answer: \(1-e^{-0.5^{1.5}}\approx0.2978\))
- A loss satisfies \(\ln X\sim N(5,0.16)\). Find \(P(e^{4.6}<X\le e^{5.4})\). (Answer: \(\Phi(1)-\Phi(-1)\approx0.6827\))
- A loss is gamma with shape \(\alpha=3\) and scale \(\theta=2\). Using its Erlang survival formula, find \(P(X>4)\). (Answer: \(5e^{-2}\approx0.6767\))