Common Errors in Stop-Loss Threshold Calculations

Why Threshold Language Matters

Stop-loss calculations can be short, but their answers depend on reading each threshold precisely. A loss equal to a deductible produces no payment under an ordinary stop-loss contract. Yet that same loss is included in the event “loss is at least the deductible.” If the loss distribution has probability mass at the threshold, confusing these events changes a conditional expectation even though the payment at the threshold remains zero.

A related source of error is confusing a layer’s exhaustion point with its limit amount. If a layer attaches at $d$ and exhausts at $u$, its maximum payment is $u-d$, not $u$. The calculations below use hypothetical aggregate losses and contracts, with no expenses or premium loading. They are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For aggregate loss $X$ and threshold $d$, the ordinary stop-loss payment is $(X-d)_+=\max(X-d,0)$. Its expected payment, or stop-loss premium on a pure expected-value basis, is $E[(X-d)_+]$. A layer with attachment $d$ and exhaustion point $u>d$ pays $\min((X-d)_+,u-d)$. It pays nothing at or below $d$, increases dollar for dollar between $d$ and $u$, and is capped at $u-d$ at or above $u$.

Strict Exceedance, Inclusive Exceedance, and Conditional Excess

The events $X>d$ and $X\ge d$ differ only when $P(X=d)>0$. In contrast, the payment $(X-d)_+$ is zero whenever $X=d$. Thus, including the threshold in the event does not add a payment to the numerator of a conditional excess calculation, but it can add probability to the denominator.

Provided $P(X>d)>0$, the expected stop-loss payment and the conditional excess above $d$ are related by

E[(X-d)_+]=P(X>d)E[X-d\mid X>d].

To see why, split the expectation into outcomes above and not above $d$. On $X>d$, the payment is $X-d$; on $X\le d$, it is zero. Therefore

\begin{aligned} E[(X-d)_+] &=E[(X-d)\mathbf{1}_{\{X>d\}}]\\ &=P(X>d)E[X-d\mid X>d]. \end{aligned}

The indicator $\mathbf{1}_{\{X>d\}}$ equals one on the stated event and zero otherwise. If the question instead conditions on $X\ge d$, the probability in the denominator must be $P(X\ge d)$. That conditional expectation is a different quantity when there is an atom at $d$.

Worked Example: An Atom at the Deductible

A hypothetical insurer models an annual aggregate loss $X$ in dollars by the following discrete distribution. There is no discounting (interest rate $0\%$), no mortality model, and no loading or expenses.

Loss $x$	0	10	20	30
Probability $P(X=x)$	0.20	0.30	0.30	0.20

For deductible $d=\$20$, calculate the expected stop-loss payment and the conditional excess given (a) $X>20$ and (b) $X\ge20$. The probabilities sum to $0.20+0.30+0.30+0.20=1.00$.

Step 1: Calculate the stop-loss payment at each possible loss. The payments are $0,0,0,10$, respectively. The loss equal to the deductible produces zero payment.

Step 2: Calculate the unconditional expected payment.

$$ E[(X-20)_+]=0.20(0)+0.30(0)+0.30(0)+0.20(10)=2. $$

The expected stop-loss payment is \$2.00.

Step 3: Condition on strict exceedance. Only the loss of 30 is strictly greater than 20, so $P(X>20)=0.20$. Its excess is 10. Hence $E[X-20\mid X>20]=10$, and the check $0.20(10)=2$ agrees with the unconditional expectation.

Step 4: Condition on inclusive exceedance. Here $P(X\ge20)=0.30+0.20=0.50$. The loss of 20 contributes zero excess, while the loss of 30 contributes expected excess $0.20(10)=2$. Therefore $E[X-20\mid X\ge20]=2/0.50=4$.

Method and trap avoided. Compute the numerator from payments first, then use the probability of the exact conditioning event. Using $P(X\ge20)=0.50$ to answer the question conditioned on $X>20$ would give $2/0.50=4$, which is wrong for that event; the correct answer is 10.

Attachment and Exhaustion in a Layer

For a layer attaching at $d$ and exhausting at $u$, the amount of covered loss is the part of $X$ above $d$, capped at the layer width $u-d$. Its expected payment can be calculated directly from the layer payment or as the difference between two ordinary stop-loss expectations:

\min((X-d)_+,u-d)=(X-d)_+-(X-u)_+,\qquad u>d.

This identity can be checked for every possible loss value. If $X\le d$, both sides are zero. If $d<X\le u$, the right-hand side is $X-d$, the full amount above attachment. If $X>u$, the right-hand side is $(X-d)-(X-u)=u-d$, the layer’s maximum payment. Taking expectations gives

E[\min((X-d)_+,u-d)]=E[(X-d)_+]-E[(X-u)_+].

The point $u$ in this formula is the exhaustion point, where the layer has paid its full width. It is not the dollar limit paid above the attachment. If a contract describes a limit as an amount of coverage rather than an exhaustion point, translate that wording into the corresponding upper point before applying the formula.

Worked Example: Reading a Layer’s Exhaustion Point

A hypothetical aggregate loss has the following distribution, in dollars. The contract covers the layer attaching at $d=\$40$ and exhausting at $u=\$100$. There is no discounting (interest rate $0\%$), no mortality model, and no loading or expenses.

Loss $x$	0	40	80	120
Probability $P(X=x)$	0.25	0.25	0.30	0.20

The probabilities sum to 1.00. Find the expected layer payment.

Step 1: Identify the width. The layer width is $u-d=100-40=60$, so the maximum payment is \$60.

Step 2: Calculate payments outcome by outcome. At $X=0$ and $X=40$, the payment is zero. At $X=80$, it is $80-40=40$. At $X=120$, the payment is capped at 60. The expected payment is

$$ 0.25(0)+0.25(0)+0.30(40)+0.20(60)=12+12=24. $$

Step 3: Check using stop-loss differences.

$$ E[(X-40)_+]=0.30(40)+0.20(80)=12+16=28, $$

and

$$ E[(X-100)_+]=0.20(20)=4. $$

Thus the layer expectation is $28-4=24$, agreeing with the direct calculation. The expected payment is \$24.00.

Method and trap avoided. The layer is capped at \$60, not \$100. Treating the exhaustion point of \$100 as a \$100 limit above the attachment would assign a payment of \$80 at $X=120$, producing $0.30(40)+0.20(80)=28$, not the correct 24.

Discrete Masses at Layer Boundaries

At an attachment point $d$, a loss exactly equal to $d$ has zero layer payment. At an exhaustion point $u$, a loss exactly equal to $u$ has already generated the full payment $u-d$. These statements follow from the payment function itself; they do not require a continuous distribution. In a discrete model, the probability at either boundary must be included in the appropriate event probability, but its payment must still be calculated from the function.

Worked Example: Boundary Atoms and Two Different Conditional Averages

A hypothetical aggregate loss $X$, in dollars, has this discrete distribution. A layer attaches at $d=\$100$ and exhausts at $u=\$150$. Assume interest rate $0\%$, no mortality model, and no loading or expenses.

Loss $x$	0	50	100	150	200
Probability $P(X=x)$	0.10	0.20	0.40	0.20	0.10

Find the expected layer payment, the conditional uncapped excess $E[X-100\mid X>100]$, and the conditional layer payment given $X\ge100$. The probabilities sum to 1.00.

Step 1: Calculate the layer payments. The layer width is $150-100=50$. The payments at losses $0,50,100,150,200$ are $0,0,0,50,50$. Therefore

E[\min((X-100)_+,50)]=0.20(50)+0.10(50)=15.

The expected layer payment is \$15.00.

Step 2: Calculate the uncapped stop-loss expectation. Above 100, the excesses at losses 150 and 200 are 50 and 100. Thus

$$ E[(X-100)_+]=0.20(50)+0.10(100)=20. $$

Since $P(X>100)=0.20+0.10=0.30$, the conditional uncapped excess is $20/0.30=66.666\ldots$, or \$66.67 to the nearest cent.

Step 3: Condition the capped payment on $X\ge100$. This event has probability $0.40+0.20+0.10=0.70$. The payment is zero at $X=100$ and \$50 at $X=150$ and $X=200$. Its unconditional expected contribution on this event is still $15$, so

E[\min((X-100)_+,50)\mid X\ge100]=\frac{15}{0.70}=21.428\ldots.

The conditional layer payment is \$21.43 to the nearest cent. The uncapped excess conditional on strict exceedance and the capped payment conditional on inclusive exceedance answer different questions; they should not be substituted for one another.

Method and trap avoided. Keep the payment type and the conditioning event explicit. The expected layer payment is \$15, not the uncapped stop-loss expectation of \$20, and the atom at $X=100$ belongs in the denominator for $X\ge100$ even though it contributes zero payment.

A Reliable Threshold-Calculation Check

Before accepting a result, write down the payment function, the event used for any conditioning, and whether a stated upper threshold is an exhaustion point or a limit amount. For a discrete distribution, evaluate the payment at every support point, including points equal to the boundaries. A direct calculation and a stop-loss-difference calculation should agree when both are available.

Translate the wording.
Write the strict or inclusive event exactly as stated, and identify the attachment $d$ and exhaustion point $u$ if a layer is involved.

Evaluate payments at boundaries.
At $X=d$, an ordinary stop-loss payment is zero. At $X=u$, a layer has reached its maximum payment $u-d$.

Separate numerator and denominator.
For a conditional expectation, first total the relevant probability-weighted payments, then divide by the probability of the exact conditioning event.

Check the result another way.
For a layer, compare its direct expected payment with $E[(X-d)_+]-E[(X-u)_+]$.

A common mistake is to assume that changing $X>d$ to $X\ge d$ never matters because the payment at $X=d$ is zero. It does not change the stop-loss numerator, but it can change a conditional mean by changing the conditioning probability. Another is to confuse an unconditional expectation, which averages payments over all outcomes, with an average conditional on exceedance, which divides by a tail probability.

Key takeaway: A threshold determines both a payment rule and, sometimes, a conditioning event. Apply the payment rule at equality, use the exact strict or inclusive event in the denominator, and treat an exhaustion point as the upper boundary of a layer—not as the layer’s width.

Check Your Understanding

For each hypothetical model, assume no discounting (interest rate $0\%$), no mortality model, and no loading or expenses. State each requested amount in dollars to the nearest cent when needed.

$X$ takes values $0,20,40$ with probabilities $0.50,0.30,0.20$, respectively. For $d=20$, find $E[(X-d)_+]$, $P(X>d)$, and $E[X-d\mid X>d]$. (Answer: \$4.00; 0.2000; \$20.00)
$X$ takes values $0,30,60,90$ with probabilities $0.20,0.30,0.30,0.20$, respectively. For $d=50$, find $E[(X-d)_+]$ and $E[X-d\mid X\ge50]$. (Answer: \$11.00; \$22.00)
$X$ takes values $0,25,50,75,100$ with probabilities $0.20,0.20,0.30,0.20,0.10$, respectively. A layer attaches at 25 and exhausts at 75. Find its expected payment directly and by stop-loss differences. (Answer: \$22.50 by either method)
$X$ takes values $10,40,70,100$ with probabilities $0.10,0.40,0.30,0.20$, respectively. For $d=70$, find the stop-loss expectation, $E[X-d\mid X>d]$, and $E[X-d\mid X\ge d]$. (Answer: \$6.00; \$30.00; \$12.00)
$X$ takes values $20,40,70,90,120$ with probabilities $0.20,0.20,0.30,0.20,0.10$, respectively. A layer attaches at 40 and exhausts at 90. Find its expected payment and the uncapped stop-loss expectation at 40. (Answer: \$24.00; \$27.00)

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