Stop-Loss Insurance

An insurer computes expected aggregate claims at $5millionbutfacesan$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 $S>d$:

$$\pi (d)=\sum _{s>d}(s-d)\cdot \mathrm{Pr}(S=s)\text{(discrete)}$$

$$\pi (d)={\int }_d^{\infty }(s-d)f_S(s)ds\text{(continuous)}$$

Common mistakes

• Forgetting to subtract $d$ in the direct formula. The stop-loss payment when $S=s$ is $s-d$, not $s$. Writing ${\sum }_{s>d}s\cdot \mathrm{Pr}(S=s)$ computes the conditional expected aggregate, not the expected excess. Trap: a value significantly larger than the correct premium.
• Confusing $E[S\wedge d]$ with $E[S]\cdot \mathrm{Pr}(S\le d)$. The limited expected value caps each outcome at $d$ before taking the expectation. It equals $E[S\mid S\le d]\cdot \mathrm{Pr}(S\le d)+d\cdot \mathrm{Pr}(S>d)$, not $E[S]\cdot \mathrm{Pr}(S\le d)$. Trap: a smaller retained amount that inflates $\pi (d)$.
• Not accounting for the full tail. In compound distributions, truncating the direct sum too early underestimates $\pi (d)$. With 1.6% probability beyond $s=8$, stopping there misses material mass. Trap: a premium slightly too small. Use the identity to avoid this entirely.

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, convex in $d$; $\pi (d)\ge \max (E[S]-d,0)$
• Discrete direct method: sum $(s-d)\mathrm{Pr}(S=s)$ over $s>d$, or apply the identity

Exam shortcut

When a problem gives a discrete aggregate distribution, use the identity $\pi (d)=E[S]-E[S\wedge d]$. Computing $E[S]$ from $E[N]\cdot E[X]$ is fast, and $E[S\wedge d]$ involves only a finite sum up to $d$. This avoids summing an infinite tail and gives two independent paths to check your answer. Remember: "SL = E minus LEV". Stop-Loss = Expected total minus Limited Expected Value. Write it at the top of your scratch paper.

The full lesson (about 2,387 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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