Atoms at the Attachment Point

Why an Atom at the Attachment Matters

A stop-loss cover pays the amount by which aggregate loss exceeds an attachment. If loss equals the attachment exactly, the payment is zero. That simple fact matters when the aggregate-loss distribution has a point mass there: the probability of exceeding the attachment is then different from the probability of reaching or exceeding it, even though the mass at the attachment contributes nothing to the stop-loss payment.

This distinction is especially important for discrete and mixed distributions. A formula may use a strict tail probability, such as $P(X>d)$, or an inclusive one, such as $P(X\ge d)$. Either can be used in a correctly matched calculation, but mixing the two can add or subtract the atom at $d$ incorrectly.

Definition: Let $X$ be an aggregate loss with finite expectation, and let $d$ be a fixed attachment. The unlimited stop-loss payment is $(X-d)^+=\max(X-d,0)$. Its expected payment, or stop-loss premium on an expected-value basis, is $\pi(d)=E[(X-d)^+]$. When $X=d$, the payment is exactly zero.

Strict and Inclusive Tails

Use $P(X>d)$ for the strict tail: the probability that loss is greater than the attachment. Use $P(X\ge d)$ for the inclusive tail: the probability that loss is at least the attachment. If $F_X(x)=P(X\le x)$, then $P(X>d)=1-F_X(d)$, while $P(X\ge d)=1-F_X(d-)$, where $F_X(d-)=P(X<d)$. Their difference is the probability mass at the attachment:

P(X\ge d)-P(X>d)=P(X=d).

The strict and inclusive tails are therefore equal when there is no atom at $d$, as in a continuous distribution. They differ when $P(X=d)>0$. The payment remains zero at $X=d$, so the presence of that mass must not be mistaken for a positive stop-loss payment.

Key result: If $X$ has finite expectation, then $$ \pi(d)=E[X\mathbf{1}_{\{X>d\}}]-dP(X>d) =E[X\mathbf{1}_{\{X\ge d\}}]-dP(X\ge d). $$ Either expression is correct when the same strict or inclusive event is used in both terms. In the inclusive expression, the contribution from $X=d$ is $dP(X=d)-dP(X=d)=0$.

For a discrete distribution with probability masses $p_j=P(X=x_j)$, the direct calculation is

\pi(d)=\sum_{x_j>d}p_j(x_j-d).

A term at $x_j=d$ may also be included in the sum because it equals $p_j(d-d)=0$. The strict form makes the zero-payment condition explicit. For a mixed distribution with atoms and a density $f_X$, the corresponding calculation adds the positive contributions from both parts:

\pi(d)=\sum_{x_j>d}p_j(x_j-d)+\int_d^\infty (x-d)f_X(x)\,dx.

An atom at $d$ contributes zero and is omitted from the sum. The integral starts at $d$; including or excluding its single endpoint does not change the value of an ordinary density integral.

Why the Tail-Integral Formula Works

The tail-integral formula is useful when a distribution is described by survival probabilities rather than by a full probability table or density. For any fixed loss value $x$, the positive excess $(x-d)^+$ is the length of the interval from $d$ up to $x$, if $x>d$; otherwise that length is zero. Equivalently, it is the integral of an indicator over all thresholds above $d$:

(x-d)^+=\int_d^\infty \mathbf{1}_{\{x>t\}}\,dt.

Taking expectations and interchanging the expectation and integral gives the following identity. The interchange is valid because the integrand is nonnegative; finite $E[X]$ ensures the resulting stop-loss expectation is finite for nonnegative aggregate loss.

\begin{aligned} \pi(d) &=E\!\left[\int_d^\infty \mathbf{1}_{\{X>t\}}\,dt\right]\\ &=\int_d^\infty E[\mathbf{1}_{\{X>t\}}]\,dt\\ &=\int_d^\infty P(X>t)\,dt. \end{aligned}

The strict tail appears inside the integral. Replacing it with $P(X\ge t)$ gives the same integral: the two functions can differ at points where $X$ has atoms, but those individual points have zero length in the integration. In contrast, at a particular attachment $d$, $P(X>d)$ and $P(X\ge d)$ need not be equal. A discrete calculation or a tail-moment formula must handle that difference explicitly.

Worked Example: A Discrete Loss With Mass at the Attachment

A hypothetical group stop-loss cover attaches at $d=\$100{,}000$. Aggregate loss $X$, in thousands of dollars, equals 0 with probability 0.20, 100 with probability 0.50, or 250 with probability 0.30. Calculate the expected payment and both tail probabilities at the attachment.

Method. Apply the discrete stop-loss sum, then check it using the tail-moment formula. The common mistake to avoid is counting the probability at 100 as a positive-payment event.

The strict tail includes only the outcome above 100, so $P(X>100)=0.30$. The inclusive tail includes both 100 and 250, so $P(X\ge100)=0.50+0.30=0.80$.

In the direct payment calculation, the outcomes 0 and 100 produce no payment. At 250, the payment is $250-100=150$. Thus

\pi(100)=0.20(0)+0.50(100-100)+0.30(250-100) =0+0+45=45

The answer is 45 thousand dollars, or $\$45{,}000$. Check it with the strict tail-moment formula: $E[X\mathbf{1}_{\{X>100\}}]=0.30(250)=75$, and $100P(X>100)=100(0.30)=30$, so $75-30=45$. With the inclusive form, $E[X\mathbf{1}_{\{X\ge100\}}]=0.50(100)+0.30(250)=50+75=125$, and $100P(X\ge100)=100(0.80)=80$, so $125-80=45$. Both checks agree with the direct payment calculation.

Worked Example: A Mixed Distribution at the Attachment

A hypothetical aggregate loss, measured in thousands of dollars, has probability 0.25 of being exactly 100. With probability 0.75, it is uniformly distributed from 40 to 160. A cover attaches at $d=100$. Find the expected payment and the strict and inclusive tail probabilities.

Method. Calculate the atom and continuous contributions separately. The atom at 100 has zero excess; the uniform component has weight 0.75. The common mistake to avoid is either omitting that weight or treating the atom at the attachment as a positive payment.

The strict tail probability from the atom is zero. For the uniform component, the interval above 100 runs from 100 to 160, a length of 60 out of the full length $160-40=120$. Its conditional probability of exceeding 100 is $60/120=0.50$. Therefore

P(X>100)=0.75(0.50)=0.375, \qquad P(X\ge100)=0.375+0.25=0.625.

The continuous part has density $1/120$ on $[40,160]$, conditional on being in that part of the mixture. Its expected excess contribution is

0.75\int_{100}^{160}(x-100)\frac{1}{120}\,dx =0.75\left[\frac{(x-100)^2}{240}\right]_{100}^{160} =0.75\left(\frac{3600}{240}\right) =0.75(15)=11.25.

The atom contributes $0.25(100-100)=0$. Therefore $\pi(100)=11.25$ thousand dollars, or $\$11{,}250$. As a check, the continuous component's conditional excess when it is above 100 is the average of values from 0 to 60, namely 30; its unconditional contribution is $0.75(0.50)(30)=11.25$. The calculation agrees.

One-Sided Changes in the Stop-Loss Premium

An atom also affects how the expected payment changes as the attachment moves. Write $\pi(d)=E[(X-d)^+]$. For a small increase in $d$, only losses strictly above $d$ continue to generate payment; a loss exactly equal to $d$ already pays zero. For a small decrease in $d$, losses equal to $d$ now lie above the lower attachment and contribute to the change. The resulting one-sided derivatives, when $E[X]$ is finite, are

\pi'_+(d)=-P(X>d), \qquad \pi'_-(d)=-P(X\ge d).

Here $\pi'_+(d)$ is the derivative from attachments above $d$, and $\pi'_-(d)$ is the derivative from attachments below $d$. A point mass at $d$ makes the left-hand slope more negative than the right-hand slope. This gives another way to see why the two tail probabilities must not be confused.

Worked Example: Checking the One-Sided Slopes at an Atom

A hypothetical aggregate loss, in thousands of dollars, equals 60 with probability 0.20, 100 with probability 0.30, or 160 with probability 0.50. At $d=100$, calculate the stop-loss premium and the one-sided derivatives. Check the slopes by moving the attachment to 90 and 110.

Method. First calculate payments from outcomes strictly above each attachment; then compare the resulting changes per unit of attachment. The common mistake to avoid is using the strict tail on both sides of an atom.

At $d=100$, only the outcome 160 pays, and its excess is $160-100=60$. Thus $\pi(100)=0.50(60)=30$ thousand dollars. The strict tail is $P(X>100)=0.50$, while the inclusive tail is $P(X\ge100)=0.30+0.50=0.80$. Therefore

\pi'_+(100)=-0.50, \qquad \pi'_-(100)=-0.80.

For an attachment of 110, only the loss of 160 pays, by $160-110=50$, so $\pi(110)=0.50(50)=25$. The change from 100 to 110 is $(25-30)/(110-100)=-5/10=-0.50$, matching the right-hand derivative. For an attachment of 90, the losses of 100 and 160 pay excesses 10 and 70. Hence $\pi(90)=0.30(10)+0.50(70)=3+35=38$. The change from 90 to 100 is $(30-38)/(100-90)=-8/10=-0.80$, matching the left-hand derivative. Both checks confirm that the atom makes the two slopes different.

Common Tail-Probability Trap

The tail-moment formula requires the same event in both terms. In the second worked example, the strict-event moment is $E[X\mathbf{1}_{\{X>100\}}]=0.375(130)=48.75$ thousand dollars: conditional on the uniform part exceeding 100, the average loss is 130. If one incorrectly subtracts the inclusive tail term, the result is $48.75-100(0.625)=-13.75$, which cannot be a stop-loss payment. The calculation has subtracted $100$ times the atom's probability without adding that atom to the first term.

Correct the mismatch in either of two ways. With strict tails, subtract $100(0.375)$, giving $48.75-37.50=11.25$. With inclusive tails, add the atom's moment $100(0.25)=25$ to obtain $E[X\mathbf{1}_{\{X\ge100\}}]=73.75$, then subtract $100(0.625)=62.50$, again giving $11.25$. The direct payment formula is often the safest check: an outcome equal to the attachment contributes zero.

Key takeaway: At an atom, $P(X>d)$ and $P(X\ge d)$ are different. The stop-loss payment at $X=d$ is still zero, so use a consistent strict or inclusive event in the tail-moment formula, or calculate the excess directly. Tail integrals are unchanged by the value of the tail at an individual threshold, but pointwise tail probabilities are not.

Check Your Understanding

For each problem, distinguish the probability at the attachment from the probability of a positive stop-loss payment.

A hypothetical aggregate loss $X$, in thousands of dollars, is 0 with probability 0.40, 80 with probability 0.25, and 120 with probability 0.35. A cover attaches at 80. Find $\pi(80)$, $P(X>80)$, and $P(X\ge80)$. (Answer: $\pi(80)=14$ thousand dollars; $P(X>80)=0.3500$; $P(X\ge80)=0.6000$.)
A hypothetical loss is exactly 30 with probability 0.20; otherwise it is uniform on $[0,100]$. The attachment is 30. Find the expected payment and both tail probabilities. (Answer: $\pi(30)=19.60$ dollars; $P(X>30)=0.5600$; $P(X\ge30)=0.7600$.)
A hypothetical discrete aggregate loss, in thousands of dollars, takes values 50, 100, 150, and 250 with respective probabilities 0.10, 0.20, 0.30, and 0.40. At attachment 100, calculate the payment expectation using the direct sum and report both tail probabilities. (Answer: $\pi(100)=75$ thousand dollars; $P(X>100)=0.7000$; $P(X\ge100)=0.9000$.)
A hypothetical loss has probability 0.15 of being exactly 200. Otherwise, it is uniform on $[100,300]$. At attachment 200, calculate the expected payment and both tail probabilities. (Answer: $\pi(200)=21.25$ dollars; $P(X>200)=0.4250$; $P(X\ge200)=0.5750$.)
A hypothetical aggregate loss, in thousands of dollars, is 0 with probability 0.20, 100 with probability 0.30, and 180 with probability 0.50. At attachment 100, find $\pi(100)$, $\pi'_-(100)$, and $\pi'_+(100)$. (Answer: $\pi(100)=40$ thousand dollars; $\pi'_-(100)=-0.8000$; $\pi'_+(100)=-0.5000$.)

The treatment of atoms at an attachment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

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