Stop-Loss Insurance

An insurer computes expected aggregate claims at $5 million but faces an$8 million tail scenario. Stop-loss reinsurance pays everything above a deductible $d$. The net stop-loss premium (the fair price) is a single expected value calculation. Get it wrong, and the reinsurance is mispriced. The FAM exam tests two computation paths (direct sum and limited expected value identity) and expects you to pick the faster one.

Stop-loss insurance pays the amount by which aggregate claims $S$ exceed a deductible $d$:

$$\text{Payment} = (S - d)_+ = \max(S - d, 0)$$

The net stop-loss premium is the expected value of this payment:

$$\pi(d) = E[(S - d)_+]$$

Method 1. Direct formula. Sum or integrate over the region where $S > d$:

$$\pi(d) = \sum_{s > d} (s - d) \cdot \Pr(S = s) \quad \text{(discrete)}$$

$$\pi(d) = \int_d^{\infty} (s - d)\, f_S(s)\, ds \quad \text{(continuous)}$$

Common mistakes

• Forgetting to subtract $d$ in the direct formula. The stop-loss payment when $S = s$ is $s - d$, not $s$. Writing $\sum_{s > d} s \cdot \Pr(S = s)$ computes the conditional expected aggregate, not the expected excess. Trap: a value significantly larger than the correct premium.
• Confusing $E[S \wedge d]$ with $E[S] \cdot \Pr(S \leq d)$. The limited expected value caps each outcome at $d$ before taking the expectation. It equals $E[S \mid S \leq d] \cdot \Pr(S \leq d) + d \cdot \Pr(S > d)$, not $E[S] \cdot \Pr(S \leq d)$. Trap: a smaller retained amount that inflates $\pi(d)$.
• Not accounting for the full tail. Truncating a direct sum underestimates the stop-loss premium. Example 2 still has about 6.17% probability above 8. Use the limited expected value identity to include the whole tail without summing each large outcome.

Bottom line

• Stop-loss payment: $(S - d)_+ = \max(S - d, 0)$; premium $\pi(d) = E[(S - d)_+]$
• Fundamental identity: $\pi(d) = E[S] - E[S \wedge d]$, total expected minus retained
• Boundary and shape: $\pi(0) = E[S]$; non-increasing and convex in $d$; $\pi(d) \geq \max(E[S]-d,0)$, with strictness only when losses can fall on both sides of the deductible.
• Discrete direct method: sum $(s - d)\Pr(S = s)$ over $s > d$, or apply the identity

Exam shortcut

When a problem gives a discrete aggregate distribution, use the identity $\pi(d)=E[S]-E[S\wedge d]$. Compute the mean from the frequency and severity moments, then cap the remaining outcomes at the deductible. The lower bound permits equality: losses 2 or 4 with equal probabilities and deductible 1 give premium 2, exactly the mean of 3 minus 1. Positive variance alone does not make the inequality strict.

The full lesson (about 2,351 words, 16 min read) adds 2 worked examples, all 5 common mistakes, a self-check, free in the app.

Learning objectives

• 2l

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