Ground-Up Severity as a Random Variable
A claim amount is uncertain before it is observed, so it can be represented by a random variable. In rating work, it is important to specify which claim amount the variable describes. Ground-up severity is the loss amount before applying a deductible, policy limit, or other contract adjustment. It describes the size of a loss, not how often losses occur.
For this tutorial, \(X\) denotes ground-up severity for a hypothetical covered loss, measured in dollars unless otherwise stated. It is nonnegative: \(P(X\ge 0)=1\). A model may allow \(X=0\), for example if the modeled variable includes outcomes with no positive loss, or it may assign probability only to positive amounts. The choice must agree with how the loss population is defined. No frequency assumption is needed to describe the distribution of \(X\).
The CDF answers a question about the amount at or below a threshold; the survival function answers a question about the amount strictly above it. Both describe the same probability distribution, but their endpoint conventions matter. In particular, \(P(X\ge x)\) is not always equal to \(S_X(x)\). If there is positive probability at exactly \(x\), then \(P(X\ge x)=P(X>x)+P(X=x)\).
A CDF is nondecreasing, takes values between 0 and 1, and is right-continuous. Since severity is nonnegative, \(F_X(x)=0\) for \(x<0\). As \(x\) grows without bound, \(F_X(x)\) approaches 1 and \(S_X(x)\) approaches 0. The CDF can have jumps: the size of its jump at \(x\) is the probability \(P(X=x)\). A continuous severity model has no such point masses, so \(P(X=x)=0\) at any single amount.
Reading Cumulative, Tail, and Interval Probabilities
A threshold probability can be obtained directly from the appropriate function. For an amount \(x\), \(P(X\le x)=F_X(x)\) and \(P(X>x)=S_X(x)\). For a strict lower-tail event, \(P(X<x)\), the value is the left-hand limit \(F_X(x-)\), which can differ from \(F_X(x)\) when there is an atom at \(x\). Consequently, endpoint notation should be translated into words before selecting a formula.
For two thresholds \(a<b\), the interval probability \(P(a<X\le b)\) is obtained by subtracting the cumulative probability through \(a\) from the cumulative probability through \(b\). Other endpoint combinations can differ when either boundary has positive probability. If \(X\) is continuous, boundary points have probability zero, so changing a strict endpoint to an inclusive one does not change the interval probability.
The survival-function identity follows from \(F_X(x)=1-S_X(x)\): subtracting the CDF values gives \([1-S_X(b)]-[1-S_X(a)]=S_X(a)-S_X(b)\). This is useful when a model gives the tail probability more directly than the CDF. It also supplies a check: the two methods must produce the same interval probability.
A severity quantile is another way to express a threshold. Under the usual lower-quantile convention, the \(p\)-quantile is the smallest \(x\) for which \(F_X(x)\ge p\). For a continuous, strictly increasing CDF, it satisfies \(F_X(x_p)=p\), equivalently \(S_X(x_p)=1-p\). With a discrete distribution, a quantile may lie at a jump, so equality to \(p\) need not hold.
Worked Examples
Worked Example: A Discrete Severity Distribution and Its Atoms
A hypothetical small-asset coverage model assigns ground-up loss \(X\), in thousands of dollars, the following distribution: \(P(X=0)=0.05\), \(P(X=0.5)=0.25\), \(P(X=2)=0.40\), and \(P(X=10)=0.30\). Find \(F_X(2)\), \(S_X(2)\), \(P(X\ge2)\), and \(P(0.5<X\le10)\). State why the two probabilities involving the value 2 are different.
Method. Add the probabilities at or below a threshold to obtain the CDF. Use the strict-tail definition for \(S_X\), and include the mass at the boundary when the event is \(X\ge2\). The probabilities sum to \(0.05+0.25+0.40+0.30=1.00\), as required for a distribution.
At 2, the outcomes at or below the threshold are 0, 0.5, and 2. Thus \(F_X(2)=0.05+0.25+0.40=0.70\). The only outcome strictly above 2 is 10, so \(S_X(2)=P(X>2)=0.30\).
For the inclusive event \(X\ge2\), include both the mass at 2 and the mass at 10: \(P(X\ge2)=0.40+0.30=0.70\). Equivalently, \(P(X\ge2)=1-F_X(2-)=1-(0.05+0.25)=0.70\).
For the interval \(0.5<X\le10\), the possible outcomes are 2 and 10. Its probability is \(0.40+0.30=0.70\). The CDF calculation checks this: \(F_X(10)-F_X(0.5)=1.00-(0.05+0.25)=0.70\).
Check. \(S_X(2)=0.30\), while \(P(X\ge2)=0.70\), because \(P(X=2)=0.40\). The common mistake is to use \(1-F_X(2)\) for \(P(X\ge2)\); that expression is \(P(X>2)\), not \(P(X\ge2)\), when there is an atom at 2.
Worked Example: Exponential Severity Tail and Interval Probability
For a hypothetical equipment-loss portfolio, suppose ground-up severity \(X\), measured in thousands of dollars, has an exponential distribution with mean 25 thousand dollars. Find \(P(X>25)\) and \(P(5<X\le25)\), reporting probabilities to four decimal places.
Model and assumptions. The exponential model has support \(x\ge0\), mean \(\theta=25\) thousand dollars, and rate \(\lambda=1/\theta=1/25=0.04\) per thousand dollars. Its survival function is \(S_X(x)=e^{-\lambda x}\) for \(x\ge0\), and its CDF is \(F_X(x)=1-e^{-\lambda x}\). It is continuous, so boundary points have zero probability.
First, \(P(X>25)=S_X(25)=e^{-(0.04)(25)}=e^{-1}=0.367879\ldots\). Rounded only at the final step to four decimal places, this is \(0.3679\).
For the interval, subtract the tail at the upper endpoint from the tail at the lower endpoint: \(P(5<X\le25)=S_X(5)-S_X(25)\). The two tails are \(S_X(5)=e^{-(0.04)(5)}=e^{-0.2}=0.818731\ldots\) and \(S_X(25)=e^{-1}=0.367879\ldots\). Therefore the interval probability is \(0.818730753-0.367879441=0.450851312\), or \(0.4509\) to four decimal places.
Check. The interval is a subset of \(X>5\), whose probability is about 0.8187, and the calculated probability 0.4509 is smaller. The common mistake is to subtract in the wrong order, which would give a negative probability; the larger tail at the lower threshold must come first.
Worked Example: Uniform Severity, a Tail Threshold, and a Conditional Tail
Suppose a hypothetical specialty-property model has ground-up severity \(X\) uniformly distributed from 0 to 80 thousand dollars. Find \(P(12<X\le52)\), the threshold \(q\) with \(P(X>q)=0.10\), and \(P(X>60\mid X>20)\). Report the conditional probability to four decimal places.
Model and assumptions. \(X\sim\operatorname{Uniform}(0,80)\), where amounts are in thousands of dollars. The CDF is \(F_X(x)=0\) for \(x<0\), \(x/80\) for \(0\le x\le80\), and 1 for \(x>80\). Within the support, \(S_X(x)=1-x/80\). The distribution is continuous.
The interval probability is \(F_X(52)-F_X(12)=52/80-12/80=40/80=0.50\).
For the threshold, solve \(S_X(q)=0.10\): \(1-q/80=0.10\), so \(q/80=0.90\) and \(q=72\). Thus 72 thousand dollars is the 90th percentile, and 10% of modeled losses strictly exceed it.
For the conditional probability, use the definition \(P(A\mid B)=P(A\cap B)/P(B)\), provided \(P(B)>0\). Here \(A=\{X>60\}\) is contained in \(B=\{X>20\}\), so the numerator is \(P(X>60)=1-60/80=0.25\), and the denominator is \(P(X>20)=1-20/80=0.75\). Therefore \(P(X>60\mid X>20)=0.25/0.75=1/3=0.3333\) to four decimal places.
Check. The interval spans 40 of the model's 80 thousand-dollar range, giving probability 0.50. The conditional probability is smaller than 1 because the event \(X>60\) is only part of the conditioning event \(X>20\). The common mistake is to divide by \(P(X\le20)\), rather than by the probability of the stated conditioning event.
Using the Functions Correctly
A useful solution process is to translate the wording into an event before doing arithmetic. This prevents a formula intended for one endpoint convention from being applied to another.
Confirm that \(X\) is ground-up severity in the stated units, not a count of claims or an amount after a contractual adjustment.
For example, “exceeds \(x\)” means \(X>x\), whereas “at least \(x\)” means \(X\ge x\). These differ when \(P(X=x)>0\).
Use \(F_X\) for probabilities at or below a threshold and \(S_X\) for strict upper tails. For an interval, subtract cumulative probabilities or subtract the upper-threshold tail from the lower-threshold tail.
Confirm that thresholds use the model's units and that the result is between 0 and 1. For a nested interval or tail event, its probability cannot exceed that of the containing event.
For continuous models, a density \(f_X\), when it exists, describes probability per unit of severity rather than probability at one exact amount. An interval probability is the area under the density on that interval. A point has zero area, so \(P(X=x)=0\). This density interpretation does not apply to a distribution with atoms; such a distribution can have positive probability concentrated at an exact amount.
These distinctions matter in rating because severity models summarize the distribution of gross loss amounts, and probability statements about thresholds are used to describe the loss profile. The model's support and units must match the question. For example, a threshold of 25 means 25 dollars if \(X\) is in dollars, but 25 thousand dollars if the model explicitly measures \(X\) in thousands. A unit conversion error changes the probability being calculated, not just its presentation.
Common Mistakes and Exam Traps
- Replacing “greater than” with “at least.” \(S_X(x)=P(X>x)\). If there is an atom at \(x\), \(P(X\ge x)=S_X(x)+P(X=x)\).
- Subtracting interval probabilities in the wrong order. For \(a<b\), \(F_X(b)-F_X(a)\) and \(S_X(a)-S_X(b)\) are nonnegative; reversing either subtraction gives the wrong sign.
- Assuming every severity distribution has a density. A discrete or mixed distribution can have atoms, so its CDF may jump and endpoint probabilities may matter.
- Mixing severity units. Apply a model parameter and threshold in the same units before evaluating an exponential, uniform, or other distribution formula.
- Treating a severity probability as a claim frequency. A statement such as \(P(X>25)\) is conditional on the modeled severity variable; it does not say how many claims occur.
Check Your Understanding
For each hypothetical model, identify the relevant event and calculate before checking the answer.
- A severity model assigns probabilities 0.10, 0.20, 0.40, and 0.30 to losses of 0, 1, 4, and 9 thousand dollars, respectively. Find \(F_X(4)\), \(S_X(4)\), and \(P(X\ge4)\). (Answer: \(F_X(4)=0.7000\), \(S_X(4)=0.3000\), and \(P(X\ge4)=0.7000\).)
- Ground-up severity is exponential with mean 40 thousand dollars. Find \(P(X>30)\) and \(P(10<X\le30)\), rounding probabilities to four decimals. (Answer: \(P(X>30)=0.4724\); \(P(10<X\le30)=0.3064\).)
- Severity is uniform from 0 to 100 thousand dollars. Find \(P(X\le35)\), \(P(20<X\le75)\), and the 80th percentile. (Answer: \(0.3500\), \(0.5500\), and 80 thousand dollars.)
- A continuous hypothetical severity distribution has \(S_X(15)=0.72\) and \(S_X(50)=0.18\), with amounts in thousands of dollars. Find \(P(15<X\le50)\), and explain why no further distributional assumption is needed. (Answer: \(0.72-0.18=0.5400\); the survival-function interval identity applies directly.)
- Ground-up severity is exponential with mean 50 thousand dollars. Calculate \(P(X>80\mid X>30)\), rounding to four decimals. (Answer: \(e^{-(80-30)/50}=e^{-1}=0.3679\).)