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

The existence of a deductible in an insurance policy will affect the frequency of payments under that policy. Let X be the loss amount random variable. Let d be a deductible, and let v be the probability that a loss results in a payment. That is, v = Pr(X > d). Let NL be the random variable denoting the *number* of losses, and let NP be the random variable denoting the *number* of payments. We can compare the probability generating functions (pgfs) of these two random variables as follows:

**PN^P(z) = PN^L(1 + v(z-1))**

The above is true generally. There is also an important special case in which PN^L(z) is a function of some parameter θ, such that PN^L(z; θ) = B(θ(z-1)), where the function B(z) is itself entirely independent of θ. In that case, the following is true: **PN^P(z) = PN^L(z; vθ) = B(vθ(z-1))**.

It is also possible to obtain the pgf of NL by knowing the pdf of NP. This can be done as follows:

**PN^L(z) = PN^P(1 – v-1 + zv-1).**

If PN^P(z) is a function of some parameter θ, then **PN^L(z) = PN^P(z; θ/v)**.

**Source: **

*Loss Models: From Data to Decisions,*(Third Edition), 2008, by Klugman, S.A., Panjer, H.H. and Willmot, G.E., Chapter 8, pp. 192-196.

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

**Problem S4C30-1.** Losses from falling cows follow an exponential distribution with mean θ = 340. The *number* of losses for an insured per year follows a Poisson distribution with mean λ = 0.5. The Simple Cow Insurance Company initially does not require any deductible from policyholders insured from this peril. However, the imposition of an ordinary deductible of 200 is being contemplated. What is v, the probability that a loss will exceed this deductible?

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

**Solution S4C30-1.** We find v = Pr(X > d), where d = 200. Essentially, v = SX(200) = e-200/340 = **v = 0.555306373**.

**Problem S4C30-2.** Losses from falling cows follow an exponential distribution with mean θ = 340. The *number* of losses for an insured per year follows a Poisson distribution with mean λ = 0.5. The Simple Cow Insurance Company initially does not require any deductible from policyholders insured from this peril. However, the imposition of an ordinary deductible of 200 is being contemplated. What would the pgf of the number of payments be if this deductible were imposed?

**Relevant properties for Poisson distributions:** P(z) = eλ(z-1).

**Solution S4C30-2.**We want to find PN^P(z), which we will do using the formula

PN^P(z) = PN^L(1 + v(z-1)).

Here, λ = 0.5, so PN^L(z) = e0.5(z-1).

Thus, PN^L(1 + v(z-1)) = e0.5(1+v(z-1)-1) = e0.5(v(z-1)), where v, from Solution S4C30-1, is 0.555306373.

Thus, PN^P(z) = e0.5(0.555306373(z-1)) = **PN^P(z) =** **e0.2776531865(z-1)**.

**Problem S4C30-3.** Losses from falling cows follow an exponential distribution with mean θ = 340. The *number* of losses for an insured per year follows a Poisson distribution with mean λ = 0.5. The Simple Cow Insurance Company initially does not require any deductible from policyholders insured from this peril. However, the imposition of an ordinary deductible of 200 is being contemplated. If this deductible were imposed, what fraction of policyholders would be expected to have at least one claim paid to them?

**Relevant properties for Poisson distributions:** P(z) = eλ(z-1).

pk**=** e-λ*λk/k!

**Solution S4C30-3.** We want to find Pr(NP ≥ 1). From Solution S4C30-2, we know that

PN^P(z) = e0.2776531865(z-1). This is the pgf of a Poisson distribution with a mean λ = 0.2776531865.

Thus, Pr(NP ≥ 1) = 1 – Pr(NP = 0) = 1 – e-0.2776531865 = **Pr(NP ≥ 1) = 0.2424404922**.

**Problem S4C30-4.** Losses from house-smashing snakes (HSSs) follow a Pareto distribution with α = 3 and θ = 1000. HSS Mutual Insurance Company requires policyholders to have an ordinary deductible of 500. The *number* of losses for an insured per year follows a geometric distribution with β = 0.4. Find PN^P(z) for this policy.

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

**Relevant properties for geometric distributions:** P(z) = (1 – β(z-1))-1.

**Solution S4C30-4.** First, we find v = Pr(X > d) = SX(500) = 10003/(500 + 1000)3 = v = 8/27.

Here, PN^L(z) = (1 – β(z-1))-1 = (1 – 0.4(z-1))-1

We note that this is a case where the formula PN^L(z; θ) = B(θ(z-1)) applies, with β serving the role of the parameter θ (not to be confused with the θ from the Pareto distribution above!).

Thus, PN^L(z; β) = PN^L(z; 0.4), so PN^P(z) = PN^L(z; vβ) = PN^L(z; (8/27)*0.4) = PN^L(z; 16/135) =

**PN^P(z) = (1 – 16(z-1)/135)-1**.

**Problem S4C30-5.** Losses from house-smashing snakes (HSSs) follow a Pareto distribution with α = 3 and θ = 1000. HSS Mutual Insurance Company requires policyholders to have an ordinary deductible of 500. The *number* of losses for an insured per year follows a geometric distribution with β = 0.4. What fraction of policyholders will incur losses but be unable to file claims under this policy, because the losses of these policyholders are not in excess of the deductible?

**Relevant properties for geometric distributions:** pk = βk/(1+β)k+1.

**Solution S4C30-5.** We want to find Pr(NP = 0 and NL ≠ 0).

We can think of this value in the following way:

Pr(NL ≠ 0) = Pr(NP = 0 and NL ≠ 0) + Pr(NP ≠ 0 and NL ≠ 0).

If NP ≠ 0 and NL ≠ 0, then this is the same as simply saying that NP ≠ 0, since a payment cannot occur without a loss. Thus,

Pr(NL ≠ 0) = Pr(NP = 0 and NL ≠ 0) + Pr(NP ≠ 0) →

Pr(NP = 0 and NL ≠ 0) = Pr(NL ≠ 0) – Pr(NP ≠ 0) →

Pr(NP = 0 and NL ≠ 0) = 1 – Pr(NL = 0) – (1 – Pr(NP = 0)) →

Pr(NP = 0 and NL ≠ 0) = Pr(NP = 0) – Pr(NL = 0).

We know from Solution S4C30-4 that PN^P(z) = (1 – 16(z-1)/135)-1. Thus, NP follows a geometric distribution with β = 16/135.

We know from Solution S4C30-4 that PN^L(z) = (1 – 0.4(z-1))-1. Thus, NL follows a geometric distribution with β = 0.4.

Therefore, Pr(NP = 0) = 1/(1+16/135) = 0.8940397351.

Pr(NL = 0) = 1/(1+0.4) = 0.7142857143.

Hence, our desired answer is 0.8940397351 – 0.7142857143 = **0.1797540208**.

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