Weibull Quantiles And Survival Calculations

From a Weibull Model to a Probability

A severity model is useful when it turns a monetary threshold into a probability or a requested probability into a monetary threshold. For a Weibull model, both calculations follow from the survival function, but the shape parameter affects how the threshold is transformed. A reliable calculation therefore begins by identifying the parameterization and distinguishing a cumulative probability from a survival probability.

Let $X$ be a nonnegative loss amount. In this tutorial, the Weibull model has scale parameter $\theta>0$ and shape parameter $\tau>0$. The scale $\theta$ has the same units as $X$, such as dollars; the shape $\tau$ is dimensionless. The model assumes a continuous severity distribution on $x\geq 0$.

Definition: If $X$ has a Weibull distribution with scale $\theta>0$ and shape $\tau>0$, then for $x\geq 0$, $$ S_X(x)=P(X>x)=\exp\left[-\left(\frac{x}{\theta}\right)^\tau\right], \qquad F_X(x)=1-S_X(x). $$ For $x<0$, $F_X(x)=0$ and $S_X(x)=1$.

The parameterization matters. The quantity inside the power is the dimensionless ratio $x/\theta$, not $x\theta$. At $x=\theta$, the survival probability is $e^{-1}$, whatever the positive value of $\tau$. Thus $\theta$ is a useful reference point on the loss scale, but it is not generally the mean or the median.

Quantiles: Inverting the Survival Function

For $0<p<1$, the $p$-quantile $q_p$ is the loss amount satisfying $F_X(q_p)=p$. Since $F_X(q_p)=1-S_X(q_p)$, the corresponding survival probability is $1-p$. This distinction is important: a 90th percentile has 10% of the distribution above it.

Result: Under the Weibull model above, the $p$-quantile is $$ q_p=\theta\left[-\ln(1-p)\right]^{1/\tau}, \qquad 0<p<1. $$ Equivalently, the loss amount with exceedance probability $r$, where $0<r<1$, is $$ x_r=\theta\left[-\ln(r)\right]^{1/\tau}. $$

The inverse exponent is $1/\tau$. The expression $-\ln(1-p)$ comes from the survival probability $1-p$, not from $p$ itself. These two features account for many incorrect answers on quantile calculations.

Identify the target probability.
For a $p$-quantile, the cumulative probability is $p$ and the survival probability is $1-p$.

Isolate the power.
Set the Weibull survival function equal to $1-p$, take the natural logarithm, and solve for $(q_p/\theta)^\tau$.

Apply the inverse power and scale.
Raise the result to $1/\tau$, then multiply by $\theta$ to restore the loss units.

Worked Example: A High Quantile and a Threshold Probability

A hypothetical equipment-replacement severity $X$, in dollars, follows a Weibull model with $\theta=\$1{,}000$ and $\tau=2$. Find the 95th percentile and the probability that a loss exceeds $\$1{,}500$.

Method. Use the inverse survival function for the percentile and the survival function for the threshold. The model is continuous, so $P(X>1{,}500)=S_X(1{,}500)$.

95th percentile. The exceedance probability at this quantile is $1-0.95=0.05$. Therefore, $$ q_{0.95} =1{,}000[-\ln(0.05)]^{1/2} =1{,}000(2.995732\ldots)^{1/2} =1{,}000(1.730818\ldots) =\$1{,}730.82. $$

Exceedance probability at $\$1{,}500$. The scale ratio is $1{,}500/1{,}000=1.5$, so $$ P(X>1{,}500) =\exp[-(1.5)^2] =\exp(-2.25) =0.105399\ldots =0.1054. $$

Check. The 95th percentile is above $\theta$, as expected because $S_X(\theta)=e^{-1}\approx0.3679$, well above 0.05. Also, $\$1{,}500$ is below the 95th percentile, so its exceedance probability should exceed 0.05; $0.1054$ does. Both final numerical answers are rounded to four decimal places for the probability and to the nearest cent for the amount. The common mistake avoided is using $0.95$, rather than $0.05$, as the survival probability.

Why the Quantile Formula Holds

The formula follows by solving the model equation, rather than by memorizing a separate rule. For $0<p<1$, continuity and the definition of a quantile give $F_X(q_p)=p$, and hence $S_X(q_p)=1-p$. Substitute $q_p$ into the Weibull survival function:

\begin{aligned} \exp\left[-\left(\frac{q_p}{\theta}\right)^\tau\right]&=1-p,\\ -\left(\frac{q_p}{\theta}\right)^\tau&=\ln(1-p),\\ \left(\frac{q_p}{\theta}\right)^\tau&=-\ln(1-p),\\ \frac{q_p}{\theta}&=\left[-\ln(1-p)\right]^{1/\tau},\\ q_p&=\theta\left[-\ln(1-p)\right]^{1/\tau}. \end{aligned}

The logarithm is defined because $0<1-p<1$; its value is negative, so $-\ln(1-p)>0$, as required for a nonnegative loss quantile. Multiplication by $\theta$ gives $q_p$ the same monetary units as the severity. This derivation also shows why the power after taking the logarithm must be $1/\tau$.

Shape Parameters Below and Above One

The shape parameter changes the way $x/\theta$ enters the exponent. If $\tau<1$, raising a ratio above one to the power $\tau$ makes it smaller; if $\tau>1$, it makes the ratio larger. Consequently, for thresholds above $\theta$, the survival probabilities can differ substantially across shape choices, even when the scale is unchanged. These are direct comparisons of the survival formula, not a substitute for evaluating the requested probability.

For thresholds below $\theta$, the direction of the comparison reverses: if $0<x/\theta<1$, a power below one makes the powered ratio larger, while a power above one makes it smaller. At $x=\theta$, all positive shapes give $S_X(\theta)=e^{-1}$. Keeping these cases separate prevents a misleading general claim about which shape has the larger survival probability at every threshold.

Worked Example: A Shape Below One

A hypothetical small-property loss severity has Weibull parameters $\theta=\$600$ and $\tau=0.5$. Find its median and the probability that a loss exceeds $\$900$.

Median. The median is the 50th percentile, so its survival probability is $1-0.50=0.50$. With $1/\tau=2$, $$ q_{0.50} =600[-\ln(0.50)]^2 =600(0.693147\ldots)^2 =600(0.480453\ldots) =\$288.27. $$

Exceedance probability. At $\$900$, the scale ratio is $900/600=1.5$, and $(1.5)^{0.5}=\sqrt{1.5}=1.224745\ldots$. Thus, $$ P(X>900) =\exp[-(1.5)^{0.5}] =\exp(-1.224745\ldots) =0.2936. $$

Check. Since $\$900>\theta$, the survival probability must be below $S_X(600)=e^{-1}\approx0.3679$; $0.2936$ meets that check. The median is below $\theta$, also consistent with the fact that $S_X(\theta)\approx0.3679<0.50$. The answers are rounded only at the final step: the amount to the nearest cent and the probability to four decimal places. The common mistake avoided is treating $1/\tau$ as $\tau$ in the quantile formula.

Worked Example: A Shape Above One

A hypothetical service-interruption severity $X$, measured in dollars, follows a Weibull model with $\theta=\$1{,}500$ and $\tau=3$. Find the 90th percentile and the probability that a loss exceeds $\$2{,}500$.

90th percentile. Its survival probability is $1-0.90=0.10$. Therefore, $$ q_{0.90} =1{,}500[-\ln(0.10)]^{1/3} =1{,}500(2.302585\ldots)^{1/3} =1{,}500(1.32050\ldots) =\$1{,}980.75. $$

Exceedance probability. The scale ratio is $2{,}500/1{,}500=5/3$, and its cube is $125/27=4.629629\ldots$. Hence, $$ P(X>2{,}500) =\exp\left[-\left(\frac{5}{3}\right)^3\right] =\exp\left(-\frac{125}{27}\right) =0.009770\ldots =0.0098. $$

Check. The 90th percentile is greater than $\theta$, consistent with $S_X(\theta)\approx0.3679>0.10$. The threshold $\$2{,}500$ is above the 90th percentile, so its exceedance probability should be below 0.10; $0.0098$ is. The amount is rounded to the nearest cent and the probability to four decimal places. The common mistake avoided is rounding the percentile calculation too early or using a survival probability of 0.90 for a 90th percentile.

Threshold Probabilities and Exam Checks

A threshold exceedance question usually requires less algebra than a quantile question: substitute the threshold directly into $S_X(x)$. If instead a problem gives an exceedance probability and asks for the threshold, use $x_r=\theta[-\ln(r)]^{1/\tau}$. In both cases, confirm which probability is being requested. For a continuous Weibull severity, $P(X>x)=P(X\geq x)$, because the probability of a loss at exactly one point is zero.

Check the units. The ratio $x/\theta$ must be dimensionless, and the resulting quantile must have the units of $\theta$.
Check the tail. A high percentile has a small survival probability: a 90th percentile uses 0.10 inside the logarithm.
Check the inverse. The quantile expression uses the exponent $1/\tau$, not $\tau$.
Check the direction. Since $S_X(x)$ decreases as $x$ increases, a threshold above a known quantile must have a smaller exceedance probability.

This tutorial treats $X$ as an unmodified continuous severity with known Weibull parameters. If observations are truncated or censored, or if coverage terms change the payment variable, the relevant probability calculation must reflect those features rather than automatically applying the unmodified loss formula.

Key takeaway: For a Weibull severity, calculate threshold exceedance with $S_X(x)=\exp[-(x/\theta)^\tau]$. Find the $p$-quantile using the survival probability $1-p$: $q_p=\theta[-\ln(1-p)]^{1/\tau}$. Check the probability tail, inverse exponent, and monetary units before reporting a result.

Check Your Understanding

For each hypothetical severity, use the stated Weibull scale and shape. Answers use four decimal places for probabilities and the nearest cent for amounts.

A component-repair severity has $\theta=\$800$ and $\tau=2$. Calculate $P(X>\$1{,}200)$. (Answer: 0.1054)
A small-property severity has $\theta=\$500$ and $\tau=0.5$. Find its median using the survival probability at the median. (Answer: $240.23)
A travel-disruption severity has $\theta=\$2{,}000$ and $\tau=1.5$. Calculate the probability that a loss exceeds $\$1{,}000$. (Answer: 0.7022)
A hypothetical equipment severity has $\theta=\$1{,}200$ and $\tau=3$. Find the 90th percentile, using the correct upper-tail probability in the inversion. (Answer: $1,584.60)
Two hypothetical severities both have $\theta=\$1{,}000$. Model A has $\tau=0.5$ and Model B has $\tau=2$. At a threshold of $\$2{,}000$, calculate each exceedance probability and identify which is larger. (Answer: Model A: 0.2431; Model B: 0.0183; Model A is larger.)

What You'll Learn