**This section of sample problems and solutions is a part of** **The Actuary’s Free Study Guide for Exam 4 / Exam C****, authored by Mr. Stolyarov. This is Section 34 of the Study Guide. See an index of all sections by following the link in this paragraph.**

**Stop-loss insurance** is “insurance on the aggregate losses, subject to a deductible” (Klugman, Panjer, and Willmot 2008, p. 208).

The **net stop-loss premium** is the expected cost of stop-loss insurance. It is equal to E((S-d)+), where S is the aggregate loss random variable, d is the deductible, and the ()+notation implies that the value S-d is only used if S-d > 0. Otherwise, 0 is used in the parentheses.

The following equation is true with respect to the net stop-loss premium for any aggregate distribution:

**Equation 34.1.** E((S-d)+) = d∞∫(1-FS(x))dx.

The following equation is true for any *continuous* aggregate distribution:

**Equation 34.2.** E((S-d)+) = d∞∫(x-d)fS(x)dx.

The following equation is true for any *discrete* aggregate distribution:

**Equation 34.3.** E((S-d)+) = x>dΣ((x-d)fS(x)).

The following theorems are also true:

**Theorem 34.4.** Suppose Pr(a

E((S-d)+) = ((b-d)/(b-a))E((S-a)+) + ((d-a)/(b-a))E((S-b)+).

**Theorem 34.5.** Assume Pr(S = kh) = fk ≥ 0 for some fixed h > 0, k is a nonnegative integer, and Pr(S = x) = 0 for all other x. Then, provided d = jh, with j a nonnegative integer,

E((S-d)+) = h*m=0∞Σ(1-FS((m+j)h)).

**Corollary 34.6 to Theorem 34.5.** Under the conditions of Theorem 9.5,

E((S-(j+1)h)+) = E((S-jh)+) – h(1- FS(jh)).

**Source: **

*Loss Models: From Data to Decisions,*(Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 9, pp. 208-210.

**Original Problems and Solutions from The Actuary’s Free Study Guide**

**Problem S4C34-1.** Aggregate losses covered under a policy of stop-loss insurance have a cumulative distribution function FS(x) = 1 – 1/x2 for x > 1, where S is the aggregate loss random variable. There is a deductible of 2 on the policy. Find E((S-d)+)), the net stop-loss premium for the policy.

**Solution S4C34-1.** We use the formula E((S-d)+) = d∞∫(1-FS(x))dx, where d = 2:

E((S-d)+) = 2∞∫(1 – (1 – 1/x2))dx = 2∞∫(1/x2)dx = (-1/x)│2∞ = **E((S-d)+) = ½ = 0.5**.

**Problem S4C34-2.** Aggregate losses covered under a policy of stop-loss insurance have a probability density function fS(x) of 1/40 for 200 ≤ x +)), the net stop-loss premium for the policy.

**Solution S4C34-2.** Since we have a continuous aggregate loss distribution, we use the formula E((S-d)+) = d∞∫(x-d)fS(x)dx =

210470∫(x-210)fS(x)dx = 210220∫(x-210)(1/40)dx + 220470∫(x-210)(1/500)dx =

(x-210)2/80│210220 + (x-210)2/1000│220470 =

1.25 + 67.6 – 0.1 = **E((S-d)+) = 68.75**.

**Problem S4C34-3.** The following is the distribution of aggregate losses covered under a policy of stop-loss insurance:

Pr(S = 230) = ¼

Pr(S = 340) = 1/5

Pr(S = 345) = 1/6

Pr(S = 370) = 3/60

Pr(S = 500) = 1/3.

A policy deductible of 300 applies. Find E((S-d)+)), the net stop-loss premium for the policy.

**Solution S4C34-3.** Since we have a discrete aggregate loss distribution, we use the formula

E((S-d)+) = x>dΣ((x-d)fS(x)), where d = 300.

E((S-d)+) = ((340-300)fS(340)) + ((345-300)fS(345)) + ((370-300)fS(370)) + ((500-300)fS(500)) =

40*(1/5) + 45*(1/6) + 70*(3/60) + 200*(1/3) = **E((S-d)+) = 85.6666666667**.

**Problem S4C34-4.** The following facts are known about a stop loss policy:

If the deductible were 100, the net stop-loss premium would be 500.

If the deductible were 50, the net stop-loss premium would be 530.

It is the case that Pr(50

Find the net stop-loss premium if the deductible were 85.

**Solution S4C34-4.** This is a case where Theorem 34.4 applies and

E((S-d)+) = ((b-d)/(b-a))E((S-a)+) + ((d-a)/(b-a))E((S-b)+), where d = 85, a = 50, and b = 100.

We also know that E((S-a)+) = 530 and E((S-b)+) = 500. Hence,

E((S-85)+) = ((100-85)/(100-50))530 + ((85-50)/(100-50))500) =

(15/50)530 + (35/50)500 = **E((S-85)+) = 509**.

**Problem S4C34-5.** For a policy of stop-loss insurance, it is known with respect to the aggregate loss random variable S that S occurs only in multiples of 300, and Pr(S = 300k) = (1/2)k for each k except 0. The policy has a deductible of 900. Find E((S-d)+)), the net stop-loss premium for the policy.

**Solution S4C34-5.** We apply Theorem 34.5:Assume Pr(S = kh) = fk ≥ 0 for some fixed h > 0, k is a nonnegative integer, and Pr(S = x) = 0 for all other x. Then, provided d = jh, with j a nonnegative integer, E((S-d)+) = h*m=0∞Σ(1-FS((m+j)h)).

This situation meets the criteria for Theorem 34.5 to apply.

Here, h = 300, and d = 900 = 3*h, so j = 3.

Thus, E((S-d)+) = 300*m=0∞Σ(1-FS((m+3)300)).

For every k, FS(300k) = (1/2)*(1-(1/2)k)/(1-(1/2)) = (1-(1/2)k).

Thus, 1 – FS(300k) = (1/2)k.

Hence, E((S-d)+) = 300*m=0∞Σ((1/2)m+3) = 300*(1/2)3/(1-(1/2)) = 300(1/2)2 = **E((S-d)+) = 75**.

**See other sections of** **The Actuary’s Free Study Guide for Exam 4 / Exam C****.**