Why the Payment Distribution Has Boundary Atoms
An ordinary deductible and a payment limit transform a ground-up loss into an insurer payment. This transformation can concentrate probability at particular payment amounts, even when the loss itself varies across a range. In particular, all losses at or below the deductible produce a zero payment, while all sufficiently large losses produce the same maximum payment. The resulting payment distribution can therefore have point masses at both ends of its range.
The distinction between a loss being equal to a threshold and being strictly greater than it matters. A loss exactly equal to the deductible produces no payment; a loss exactly equal to the deductible plus the limit already produces the maximum payment. Those boundary cases determine the point masses.
The payment is zero when \(X\leq d\), rises with \(X\) between \(d\) and \(d+u\), and is capped at \(u\) when \(X\geq d+u\). Thus \(Y\) can take values only in \([0,u]\). For distributions with atoms, use the left limit \(F_X(t-)=P(X<t)\) to distinguish losses strictly below \(t\) from losses at or above it.
Deriving the CDF and Point Masses
For a payment level \(y\) strictly between zero and the limit, the event \(Y\leq y\) is exactly the event that the loss does not exceed \(d+y\). This gives the central CDF relation. The endpoint cases must be stated separately because the payment is bounded below by zero and above by \(u\).
Because \(Y\geq 0\), no payment can be at or below a negative value, so \(F_Y(y)=0\) for \(y<0\).
For \(0\leq y<u\), a payment is at most \(y\) precisely when \(X\leq d+y\). Hence \(F_Y(y)=F_X(d+y)\).
Since \(Y\leq u\) for every loss, \(F_Y(y)=1\) whenever \(y\geq u\).
The middle step follows directly by checking the payment transformation. If \(X\leq d\), then \(Y=0\leq y\). If \(d<X<d+u\), then \(Y=X-d\), so \(Y\leq y\) is equivalent to \(X\leq d+y\). If \(X\geq d+u\), then \(Y=u>y\), so this case is excluded for \(y<u\). Together these cases establish the stated event equivalence, including equality at \(X=d+y\).
The zero-payment formula includes losses equal to the deductible, because \(X=d\) gives \((X-d)^+=0\). The maximum-payment formula includes losses equal to \(d+u\), because that loss produces payment \(u\) exactly. The left limit in the maximum-payment formula is essential: \(1-F_X((d+u)-)\) counts all losses at or above the threshold, including an atom at the threshold.
For a continuous loss distribution, there is no probability at an individual loss value. In that case, the endpoint probabilities simplify to \(P(Y=0)=F_X(d)\) and \(P(Y=u)=1-F_X(d+u)\). This simplified upper formula should not be used without checking continuity at \(d+u\).
Worked Examples
Worked Example: A Discrete Loss With Both Endpoint Atoms
A hypothetical equipment policy has deductible \(d=1\) and payment limit \(u=2\). Its ground-up loss \(X\), in thousands of dollars, has the following distribution. Find the payment distribution, \(F_Y(1.5)\), and the probability of a maximum payment.
| \(x\) | 0 | 2 | 3 | 5 |
|---|---|---|---|---|
| \(P(X=x)\) | 0.15 | 0.25 | 0.35 | 0.25 |
Method. Apply the payment transformation to each possible loss. The maximum-payment threshold is \(d+u=1+2=3\).
For \(X=0\), \(Y=\min((0-1)^+,2)=0\), with probability \(0.15\). For \(X=2\), \(Y=\min(1,2)=1\), with probability \(0.25\). For \(X=3\), \(Y=\min(2,2)=2\), with probability \(0.35\). For \(X=5\), \(Y=\min(4,2)=2\), with probability \(0.25\). The two losses at or above 3 combine at the cap.
Therefore \(P(Y=0)=0.15\), \(P(Y=1)=0.25\), and \(P(Y=2)=0.35+0.25=0.60\). These probabilities sum to \(0.15+0.25+0.60=1.00\).
Since \(0\leq1.5<2\), \(F_Y(1.5)=F_X(1+1.5)=F_X(2.5)=0.15+0.25=0.40\). The maximum-payment probability is \(P(X\geq3)=0.35+0.25=0.60\), which agrees with the mass at \(Y=2\).
Check and trap avoided. The CDF at 1.5 includes payments of both 0 and 1, giving \(0.40\). It does not include the payment of 2. Treating the cap threshold as strictly greater than 3 would incorrectly omit the \(X=3\) loss from the maximum-payment probability.
Worked Example: A Mixed Loss Distribution
For a hypothetical coverage, \(P(X=0)=0.20\). With probability \(0.80\), \(X\) follows an exponential distribution with rate \(0.5\) per thousand dollars, conditional on being in that continuous component. The deductible is \(d=1\) thousand dollars and the limit is \(u=3\) thousand dollars. Find the probability of zero payment, the probability of maximum payment, and \(F_Y(2)\).
Method and model. For the exponential component, \(P(X\leq t)=1-e^{-0.5t}\) and \(P(X\geq t)=e^{-0.5t}\) for \(t\geq0\). The total distribution combines the atom at zero with this component, weighted by its probability \(0.80\). The cap threshold is \(d+u=4\).
Zero payment occurs for \(X\leq1\). The atom at zero contributes \(0.20\), and the exponential component contributes \(0.80(1-e^{-0.5(1)})\). Thus \(P(Y=0)=0.20+0.80(1-e^{-0.5})\). Using \(e^{-0.5}=0.6065307\), this is \(0.20+0.80(0.3934693)=0.20+0.3147754=0.5147754\), or 0.5148 to four decimal places.
A maximum payment occurs for \(X\geq4\). The atom at zero contributes nothing to this event, so \(P(Y=3)=0.80e^{-0.5(4)}=0.80e^{-2}\). Using \(e^{-2}=0.1353353\), the probability is \(0.80(0.1353353)=0.1082682\), or 0.1083 to four decimal places.
Because \(2<3\), the CDF relation gives \(F_Y(2)=F_X(1+2)=F_X(3)\). Therefore \(F_Y(2)=0.20+0.80(1-e^{-0.5(3)})\). Using \(e^{-1.5}=0.2231302\), this is \(0.20+0.80(0.7768698)=0.20+0.6214958=0.8214958\), or 0.8215 to four decimal places.
Check and trap avoided. The probabilities of zero payment and maximum payment are different events, separated by the interior range. In particular, the atom at zero belongs in the zero-payment probability but not in the exponential tail used for the cap probability.
Worked Example: An Atom Exactly at the Cap Threshold
A hypothetical repair-loss model assigns probabilities \(0.20, 0.25, 0.15,\) and \(0.40\) to losses \(X=2,4,5,\) and \(7\), respectively. The deductible is \(d=2\) and the payment limit is \(u=3\). Find \(P(Y=0)\), \(P(Y=2)\), \(P(Y=3)\), and \(F_Y(2)\).
Method. The cap threshold is \(d+u=2+3=5\). Apply the transformation at each loss, keeping the equality at the cap threshold.
For \(X=2\), the payment is \(\min(0,3)=0\), with probability \(0.20\). For \(X=4\), it is \(\min(2,3)=2\), with probability \(0.25\). For \(X=5\), it is \(\min(3,3)=3\), with probability \(0.15\). For \(X=7\), it is \(\min(5,3)=3\), with probability \(0.40\). Hence \(P(Y=0)=0.20\), \(P(Y=2)=0.25\), and \(P(Y=3)=0.15+0.40=0.55\).
Since \(2<u=3\), the CDF formula gives \(F_Y(2)=F_X(d+2)=F_X(4)=0.20+0.25=0.45\). This also equals \(P(Y=0)+P(Y=2)=0.20+0.25=0.45\). The payment probabilities sum to \(0.20+0.25+0.55=1.00\).
Check and trap avoided. Here \(F_X(5)=0.60\), so using \(1-F_X(5)\) would give \(0.40\), not the correct cap probability. The correct value is \(P(X\geq5)=0.15+0.40=0.55\), equivalently \(1-F_X(5-)\). The incorrect calculation excludes the \(0.15\) probability at \(X=5\), even though that loss produces the maximum payment.
Reading the Distribution Correctly
A deductible-limit payment distribution has a simple structure, but its masses depend on the ground-up distribution at specific points. The zero payment gathers all losses through \(d\). Payments strictly between zero and \(u\) correspond one-to-one with losses strictly between \(d\) and \(d+u\), shifted down by \(d\). The maximum payment gathers the entire upper tail beginning at \(d+u\).
| Payment event | Corresponding loss event | Probability |
|---|---|---|
| \(Y=0\) | \(X\leq d\) | \(F_X(d)\) |
| \(Y=y,\ 0<y<u\) | \(X=d+y\) | \(F_X(d+y)-F_X((d+y)-)\) |
| \(Y=u\) | \(X\geq d+u\) | \(1-F_X((d+u)-)\) |
The table also clarifies why a continuous ground-up model can still produce a payment distribution with atoms: the many losses at or below the deductible all map to zero, and the many losses at or above the cap threshold all map to \(u\). A continuous ground-up distribution has no interior point masses, but it generally still has these two endpoint masses.
When using the CDF, check which interval contains the requested payment level before substituting. For \(y<u\), \(F_Y(y)=F_X(d+y)\); at \(y=u\) and above, the CDF is already one. The CDF value at the cap is not itself the cap atom: the atom is the jump at the cap, or \(F_Y(u)-F_Y(u-)\). This distinction is useful whenever a question asks for a point probability rather than a cumulative probability.
These distributional calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The most reliable approach is to mark the deductible \(d\), the cap threshold \(d+u\), and any atoms of \(X\) at those boundaries before calculating.
Check Your Understanding
For each hypothetical loss model, determine the requested payment probabilities or CDF values. Keep equality at the deductible and cap threshold in the appropriate event.
- A loss \(X\) takes values \(1,3,6\) with probabilities \(0.20,0.50,0.30\). For \(d=2\) and \(u=2\), find \(P(Y=0)\), \(P(Y=2)\), and \(F_Y(1.5)\). (Answer: \(0.2000,\ 0.3000,\ 0.7000\).)
- A loss is uniform on \([0,10]\). For \(d=2\) and \(u=3\), find \(P(Y=0)\), \(P(Y=3)\), and \(F_Y(2)\). (Answer: \(0.2000,\ 0.5000,\ 0.4000\).)
- A loss \(X\) takes values \(2,5,8\) with probabilities \(0.10,0.30,0.60\). For \(d=3\) and \(u=2\), find \(P(Y=0)\), \(P(Y=2)\), and \(F_Y(1)\). (Answer: \(0.1000,\ 0.9000,\ 0.1000\).)
- A mixed loss has \(P(X=0)=0.25\); with probability \(0.75\), \(X\) is exponential with rate \(0.25\) per thousand dollars. For \(d=2\) and \(u=2.5\), find \(P(Y=0)\), \(P(Y=2.5)\), and \(F_Y(1)\). (Answer: \(0.5451,\ 0.2435,\ 0.6457\).)
- A loss \(X\), in thousands of dollars, takes values \(0,2,4,6,9\) with probabilities \(0.10,0.20,0.25,0.15,0.30\). For \(d=2\) and \(u=4\), find \(P(Y=0)\), \(P(Y=4)\), and \(F_Y(3)\). (Answer: \(0.3000,\ 0.4500,\ 0.5500\).)