One-Sided Derivatives at Discrete Attachments

Why One-Sided Derivatives Matter

A stop-loss premium changes as the attachment increases: raising the attachment reduces the amount paid. When aggregate loss has a probability mass exactly at the attachment, however, the premium curve has different marginal changes immediately to the left and right. The distinction is important because a loss equal to the attachment begins to generate payment on one side, but not the other.

The one-sided derivatives express these marginal changes as tail probabilities. Their difference also gives a direct way to identify the probability of an atom at a specified loss value. This tutorial develops those results for an integrable aggregate loss $X$, including the boundary issue that arises if the attachment domain is restricted to nonnegative values.

Definition: For an integrable loss $X$, define its stop-loss transform at any real attachment $a$ by $$ \pi(a)=E[(X-a)_+], \qquad (z)_+=\max(z,0). $$ Here, $a$ is the attachment and $(X-a)_+$ is the payment per loss under an unlimited stop-loss cover. Defining $\pi$ for all real $a$ makes both left and right derivatives meaningful at every real attachment. In a loss-insurance application, $X$ is usually nonnegative and the contractual attachment is usually nonnegative, but the mathematical extension to negative $a$ is well-defined.

Integrability means $E[|X|]<\infty$; it ensures that the stop-loss transform is finite for each real $a$. We use the standard tail notation $P(X>a)$, $P(X\geq a)$, and $P(X=a)$. These probabilities are not interchangeable when $X$ has positive probability at $a$.

The One-Sided Derivative Result

Theorem: If $E[|X|]<\infty$ and $\pi(a)=E[(X-a)_+]$ is defined for all real $a$, then at every real $a$, $$ \pi'_+(a)=-P(X>a), \qquad \pi'_-(a)=-P(X\geq a). $$ Consequently, $$ \pi'_+(a)-\pi'_-(a)=P(X=a). $$ The right derivative is the limit from larger attachment values; the left derivative is the limit from smaller attachment values.

The signs are negative because increasing the attachment cannot increase the stop-loss payment. If there is no atom at $a$, the two tail probabilities are equal, so the one-sided derivatives agree. If $P(X=a)>0$, the left derivative is more negative: just below $a$, losses equal to $a$ are in the payment layer, while just above $a$, they are not.

Why the Derivative Formulas Hold

The key is to examine the change in payment for a small change in attachment. For $h>0$, consider the right difference quotient $$ \frac{\pi(a+h)-\pi(a)}{h} = E\left[\frac{(X-a-h)_+-(X-a)_+}{h}\right]. $$ For each fixed loss value $X$, the expression inside the expectation is between $-1$ and $0$. As $h$ decreases to zero, it tends to $-1$ if $X>a$, and to $0$ if $X\leq a$. In particular, when $X=a$, a small increase in the attachment still produces no payment, so the limiting contribution is zero. The bounded convergence theorem therefore gives $$ \pi'_+(a)=E[-\mathbf{1}_{\{X>a\}}]=-P(X>a). $$

For the left derivative, take $h<0$ and use the same quotient. As $h$ increases to zero from below, the quotient tends to $-1$ if $X\geq a$, and to $0$ if $X<a$. At $X=a$, lowering the attachment by $|h|$ creates a payment of $|h|$; dividing the change in payment by the negative attachment change gives $-1$. The quotient remains bounded between $-1$ and $0$, so bounded convergence gives $$ \pi'_-(a)=E[-\mathbf{1}_{\{X\geq a\}}]=-P(X\geq a). $$ Subtracting the left derivative from the right derivative yields $$ \pi'_+(a)-\pi'_-(a) =-P(X>a)+P(X\geq a) =P(X=a). $$

This derivation explains why the strict and non-strict inequalities differ: the outcome $X=a$ affects the limiting change from below, but not from above. The result requires the stop-loss expectation to be finite; the bounded difference quotients then justify taking the limit inside the expectation.

Worked Examples

Worked Example: A Discrete Loss With an Atom at the Attachment

A hypothetical aggregate claim $X$, measured in thousands of dollars, has the distribution $P(X=0)=0.20$, $P(X=5)=0.30$, and $P(X=9)=0.50$. Find the stop-loss premium at attachment $a=5$, both one-sided derivatives there, and the probability mass identified by their difference.

Method. Calculate the expected payment directly from the probability table, then apply the one-sided derivative formulas. This avoids treating the mass at the attachment as part of both tails.

Premium. At $a=5$, the payment is $0$ when $X=0$, $0$ when $X=5$, and $9-5=4$ when $X=9$. Thus $$ \pi(5)=0.20(0)+0.30(0)+0.50(4)=2.00 $$ thousand dollars.

Right derivative. The event $X>5$ consists only of $X=9$, so $$ \pi'_+(5)=-P(X>5)=-0.50. $$

Left derivative. The event $X\geq5$ consists of $X=5$ and $X=9$, so $$ \pi'_-(5)=-P(X\geq5)=-(0.30+0.50)=-0.80. $$

Check. The derivative gap is $-0.50-(-0.80)=0.30$, equal to $P(X=5)=0.30$. The premium is in thousands of dollars, while the derivatives are changes in thousands of dollars of premium per thousand-dollar change in attachment.

Worked Example: A Mixed Distribution at Its Atom

A hypothetical aggregate loss has a $0.40$ probability of being exactly $6$ thousand dollars. With the remaining probability $0.60$, it is uniformly distributed from $0$ to $10$ thousand dollars. Find $\pi(6)$, the two one-sided derivatives at $6$, and the derivative gap.

Method. Separate the point mass from the continuous component. The atom at $6$ pays zero at attachment $6$, but it contributes to the left derivative.

Premium from the atom. When $X=6$, $(X-6)_+=0$, so the atom contributes $0.40(0)=0$.

Premium from the uniform component. Conditional on being in the uniform component, the expected excess above $6$ is $$ E[(X-6)_+\mid\text{uniform component}] =\frac{1}{10}\int_6^{10}(x-6)\,dx =\frac{1}{10}\left[\frac{(x-6)^2}{2}\right]_6^{10} =\frac{8}{10}=0.8. $$ Weighting this conditional expectation by the component probability gives $0.60(0.8)=0.48$ thousand dollars. Therefore $\pi(6)=0+0.48=0.48$ thousand dollars.

Right derivative. The atom at $6$ is not in the event $X>6$. In the uniform component, the conditional probability of exceeding $6$ is $(10-6)/10=0.40$. Thus $$ P(X>6)=0.60(0.40)=0.24, \qquad \pi'_+(6)=-0.24. $$

Left derivative. The event $X\geq6$ includes the entire atom and the uniform losses from $6$ to $10$. The uniform component has no atom at $6$, so including or excluding its endpoint does not change its probability: $$ P(X\geq6)=0.40+0.60(0.40)=0.64, \qquad \pi'_-(6)=-0.64. $$

Check. The gap is $-0.24-(-0.64)=0.40$, exactly the point mass at $6$. The continuous component contributes equally to the two tail probabilities; only the atom creates the difference.

Worked Example: Checking Derivatives From a Probability Table

A hypothetical aggregate loss $X$, in thousands of dollars, takes values $1,3,4,8$ with respective probabilities $0.10,0.20,0.30,0.40$. At attachment $a=3$, calculate the premium and one-sided derivatives. Then verify each derivative using a finite change of $0.5$ thousand dollars.

Method. First compute the stop-loss premium at each attachment from the table. The local finite differences provide an independent check on the tail-probability formulas.

At $a=3$, the positive payments are $4-3=1$ when $X=4$, and $8-3=5$ when $X=8$. Hence $$ \pi(3)=0.10(0)+0.20(0)+0.30(1)+0.40(5)=0.30+2.00=2.30 $$ thousand dollars.

For the right derivative, $P(X>3)=P(X=4)+P(X=8)=0.30+0.40=0.70$, so $\pi'_+(3)=-0.70$. For the left derivative, $P(X\geq3)=0.20+0.30+0.40=0.90$, so $\pi'_-(3)=-0.90$. Their difference, $0.20$, equals $P(X=3)$.

Right finite difference. At $a=3.5$, the payments are $0$ for $X=1,3$, $0.5$ for $X=4$, and $4.5$ for $X=8$. Thus $$ \pi(3.5)=0.30(0.5)+0.40(4.5)=0.15+1.80=1.95, \qquad \frac{\pi(3.5)-\pi(3)}{0.5}=\frac{1.95-2.30}{0.5}=-0.70. $$

Left finite difference. At $a=2.5$, the payments are $0$ for $X=1$, $0.5$ for $X=3$, $1.5$ for $X=4$, and $5.5$ for $X=8$. Therefore $$ \pi(2.5)=0.20(0.5)+0.30(1.5)+0.40(5.5)=0.10+0.45+2.20=2.75, \qquad \frac{\pi(2.5)-\pi(3)}{2.5-3}=\frac{2.75-2.30}{-0.5}=-0.90. $$

Check. Both finite differences match the respective one-sided derivatives. The probability at $X=3$ accounts for their $0.20$ difference.

Boundary Attachments and Common Traps

In loss applications, one often first defines the stop-loss premium only for $a\geq0$. Under that restricted definition, a left derivative at $a=0$ is not defined: it would require evaluating $\pi(0+h)$ for negative $h$. The theorem above avoids this problem by explicitly extending $\pi(a)=E[(X-a)_+]$ to every real $a$. If instead the contract and transform are defined only for nonnegative attachments, state the left-derivative result only for $a>0$.

At the boundary $a=0$, for nonnegative $X$, the right derivative remains well-defined on the nonnegative domain and is $-P(X>0)$. Under the all-real extension, the left derivative at zero is $-P(X\geq0)=-1$, since $X$ is nonnegative. These are different derivatives when there is positive probability at zero.

Common exam trap: Using $-P(X\geq a)$ for both derivatives incorrectly includes the atom in the right derivative. For the first example at $a=5$, that mistake gives $-0.80$ for the right derivative. The correct right derivative is $-P(X>5)=-0.50$; $-0.80$ is the left derivative. The difference, $0.30$, is the mass at the attachment.

A related error is to say that an atom makes the premium discontinuous. It does not: $(X-a)_+$ changes continuously with $a$, and its expectation is continuous. The atom changes the slope, not the premium value itself. The derivative gap records the size of that slope change.

Using the Result in Stop-Loss Calculations

The formulas translate a local premium sensitivity directly into tail probabilities. If the right derivative at $a$ is known, then $P(X>a)=-\pi'_+(a)$. If the left derivative is known, then $P(X\geq a)=-\pi'_-(a)$. Subtracting the two tail probabilities gives $P(X=a)$; equivalently, subtracting the left derivative from the right derivative gives the atom.

This interpretation is useful when a stop-loss premium curve is available numerically or from a table. A sharp change in slope at a particular attachment signals a discrete probability mass there. Conversely, if the distribution is continuous at an attachment, the left and right slopes coincide. The derivatives describe local changes; they do not by themselves provide the premium level, which still depends on the expected excess over the attachment.

Check the domain.
Confirm whether the transform is defined for all real attachments or only for nonnegative attachments. A left derivative at zero requires an extension to negative attachments.

Use the correct tail event.
For a right derivative use the strict event $X>a$; for a left derivative use the non-strict event $X\geq a$.

Check the atom.
Compute $\pi'_+(a)-\pi'_-(a)$. The result must equal $P(X=a)$, so it must be nonnegative and no greater than one.

Key takeaway: For the stop-loss transform $\pi(a)=E[(X-a)_+]$ defined on all real $a$, the right slope is $-P(X>a)$, the left slope is $-P(X\geq a)$, and their difference is exactly the probability mass at $a$.

Check Your Understanding

For each problem, use the strict tail for the right derivative and the non-strict tail for the left derivative. Attachments and losses are in the units stated.

A discrete loss takes values $0,2,6$ with probabilities $0.25,0.35,0.40$. At attachment $2$, find $\pi(2)$, $\pi'_+(2)$, and $\pi'_-(2)$. (Answer: $\pi(2)=1.60$, $\pi'_+(2)=-0.4000$, $\pi'_-(2)=-0.7500$.)
A loss has probability $0.15$ at $4$ and probability $0.85$ at $9$. At attachment $4$, what are the one-sided derivatives, and what probability mass does their difference identify? (Answer: $\pi'_+(4)=-0.8500$, $\pi'_-(4)=-1.0000$, and $P(X=4)=0.15$.)
A loss is $0$ with probability $0.20$, $5$ with probability $0.30$, and otherwise is uniformly distributed on $[0,10]$ with probability $0.50$. At attachment $5$, find both derivatives and the derivative gap. (Answer: $\pi'_+(5)=-0.2500$, $\pi'_-(5)=-0.5500$, and the gap is $0.3000=P(X=5)$.)
A nonnegative loss has $P(X=0)=0.25$, $P(X=3)=0.50$, and $P(X=7)=0.25$. If the stop-loss transform is extended to all real attachments, find the left and right derivatives at zero. Then state whether the left derivative would be defined if the transform were specified only for $a\geq0$. (Answer: With the all-real extension, $\pi'_-(0)=-1.0000$ and $\pi'_+(0)=-0.7500$; on the restricted domain $a\geq0$, the left derivative at zero is not defined.)
A finite loss distribution has probability $0.10$ at $1$, $0.25$ at $4$, and $0.65$ at $10$. At attachment $4$, calculate $\pi(4)$, both one-sided derivatives, and verify that their difference equals the attachment mass. (Answer: $\pi(4)=3.90$, $\pi'_+(4)=-0.6500$, $\pi'_-(4)=-0.9000$, and the difference is $0.2500=P(X=4)$.)

What You'll Learn