**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 26 of the Study Guide. See an index of all sections by following the link in this paragraph.**

A **franchise deductible** differs from an ordinary deductible in that the loss is paid in full if the loss is in excess of the deductible. We can again apply franchise deductibles to per-payment and per-loss payment random variables.

The per-payment payment variable is as follows:

YP = undefined when X ≤ d. (This is so because the insurer does not need to make a payment.)

YP = X when X > d.

The per-loss payment variable is as follows:

YL = 0 when X ≤ d.

YL = X when X > d.

fY^P(y) = fX(y)/SX(d), y > d

SY^P(y) = 1 if 0 ≤ y ≤ d;

SY^P(y) = SX(y)/SX(d) if y > d.

FY^P(y) = 0 if 0 ≤ y ≤ d;

FY^P(y) = (FX(y) – FX(d))/(1 – FX(d)) if y > d.

hY^P(y) = 0 if 0

hY^P(y) = hX(y) if y > d.

fY^L(y) = FX(d), y = 0;

fY^L(y) = fX(y), y > d.

SY^L(y) = SX(d), 0 ≤ y ≤ d;

SY^L(y) = SX(y), y > d.

FY^L(y) = FX(d), 0 ≤ y ≤ d;

FY^L(y) = FX(y), y > d.

hY^L(y) = 0 if 0

hY^L(y) = hX(y) if y > d.

For an ordinary deductible, the expected cost per payment is (E(X) – E(X Λ d))/(1-F(d)).

For a franchise deductible, the expected cost per payment is (E(X) – E(X Λ d))/(1-F(d)) + d.

For an ordinary deductible, the expected cost per loss is E(X) – E(X Λ d).

For a franchise deductible, the expected cost per loss is E(X) – E(X Λ d) + d(1 – F(d)).

**Source: **

*Loss Models: From Data to Decisions,*(Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 8, pp. 182-183.

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

**Problem S4C26-1.** Losses from flying cows follow an exponential distribution with mean θ = 340. Flying Cow Insurance Company requires policyholders to have a franchise deductible of 200. Find fY^P(y), the pdf of the per-payment payment variable under this policy.

**Solution S4C26-1.** We use the formula fY^P(y) = fX(y)/SX(d), y > d.

For an exponential distribution, fX(x) = e-x/θ/θ, and SX(x) = e-x/θ. Here, d = 200.

Thus, fY^P(y) = fX(y)/SX(d) = (e-y/340/340)/(e-200/340) = **fY^P(y) =** **e(-y+200)/340/340, y > 200**.

**Problem S4C26-2.** Losses from flying cows follow an exponential distribution with mean θ = 340. Flying Cow Insurance Company requires policyholders to have a franchise deductible of 200. Find fY^L(y), the pdf of the per-loss payment variable under this policy.

**Solution S4C26-2.** We use the formula fY^L(y) = FX(d), y = 0; fY^L(y) = fX(y), y > d.

For an exponential distribution, fX(x) = e-x/θ/θ, and SX(x) = e-x/θ. Here, d = 200.

FX(200) = 1 – SX(200) = 1 – e-200/340 = 1 – e-10/17 = 0.444693627.

Thus,

**fY^L(y) =** **1 – e-10/17 = 0.444693627** **if y = 0;**

**fY^L(y) =** **e-y/340/340****, y > 200**.

**Problem S4C26-3.** Losses from house-smashing snakes (HSSs) are denoted by a random variable X that follows a Pareto distribution with α = 3 and θ = 1000. HSS Mutual Insurance Company requires policyholders to have an ordinary deductible of 500. Find the expected cost per loss under this policy.

**Relevant properties for Pareto distributions:** S(x) **=** θα/(x + θ)α

E(X) = θ/(α – 1)

E(X Λ K) =(θ/(α – 1))(1 – (θ/(K + θ))(α – 1))

**Solution S4C26-3.** The expected cost per loss for a policy with an ordinary deductible d is

E(X) – E(X Λ d) = E(X) – E(X Λ 500).

Here, E(X) = θ/(α – 1) = 1000/(3-1) = 500.

E(X Λ 500) = (θ/(α – 1))(1 – (θ/(K + θ))(α – 1)) = 500(1 – (1000/(500 + 1000))2) = 500(1 – (2/3)2) = 277.777777778.

Thus, E(X) – E(X Λ d) = 500 – 277.777777778 = **222.22222222222**.

**Problem S4C26-4.** Losses from house-smashing snakes (HSSs) are denoted by a random variable X that follows a Pareto distribution with α = 3 and θ = 1000. The House-Smashing Snake Special (HSSS) Mutual Insurance Company requires policyholders to have a *franchise* deductible of 500. Find the expected cost per loss under this policy.

**Relevant properties for Pareto distributions:** S(x) **=** θα/(x + θ)α

E(X) = θ/(α – 1)

E(X Λ K) =(θ/(α – 1))(1 – (θ/(K + θ))(α – 1))

**Solution S4C26-4.**

For a franchise deductible, the expected cost per loss is E(X) – E(X Λ d) + d(1 – F(d)).

From Solution S4C26-4, we already know that E(X) – E(X Λ d) = 222.22222222.

We need to find d(1 – F(d)). We know that d = 500. We need to find 1 – F(500) = S(500) =

10003/(500 + 1000)3 = 8/27. Thus, d(1 – F(d)) = 500*8/27 = 148.1481481.

Hence, E(X) – E(X Λ d) + d(1 – F(d)) = 222.22222222 + 148.1481481 = **370.3703704.**

**Problem S4C26-5.** Losses from flying cows follow an exponential distribution with mean θ = 340. Flying Cow Insurance Company requires policyholders to have a franchise deductible of 200. Find the expected cost per payment under this policy.

**Relevant properties for exponential distributions:** SX(x) = e-x/θ;

E(X) = θ; E(X Λ K) = θ(1 – e-K/θ).

**Solution S4C26-5.**

For a franchise deductible, the expected cost per payment is (E(X) – E(X Λ d))/(1-F(d)) + d.

We know that E(X) = θ = 340 and that d = 200.

We calculate E(X Λ 200) = 340(1 – e-200/340) = 151.1958332.

We find 1-F(d) = S(d) = S(200) = e-200/340 = 0.555306373.

Thus, the expected cost per payment is (340 – 151.1958332)/0.555306373 + 200 = **540**.

(Note that the memoryless property of an exponential distribution implies that the expected cost per paymentwith an *ordinary* deductible is the same as E(X), the expected cost per payment without any deductible at all.)

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