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

This section provides additional exam-style practice with a variety of syllabus topics.

Some of the problems in this section were designed to be similar to problems from past versions of Exam 4/C, offered jointly by the Casualty Actuarial Society and the Society of Actuaries. They use original exam questions as their inspiration – and the specific inspiration is cited to give students an opportunity to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

**Source:** Exam C Sample Questions and Solutions from the Society of Actuaries.

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

**Problem S4C114-1. Similar to Question 275 of the** **Exam C Sample Questions** **from the Society of Actuaries.** An insurer applies a deductible of 300 to each claim. Then the insurer pays 50% of the amount of loss amount above 300, subject to a maximum payment of 400.

It is hypothesized the amount of each loss follows an exponential distribution with mean 500. Values of this distribution are simulated using an inversion (inverse transformation) method, where the random variable U has a uniform distribution on the interval (0, 1). Higher values of U correspond to higher values of the simulated loss amount.

Four values of U were generated: 0.12, 0.34, 0.56, 0.89.

Using this pseudorandom sample data, find the average amount that the *insurer* would have to pay.

**Solution S4C114-1.** First, we can find the raw loss amounts.

Let u be any of the four given simulated values of U. Let X denote the loss amount.

We know that u = F(x) = 1 – e-x/500.

Thus, e-x/500 = 1 – u,

and x = -500ln(1 – u).

Thus, our simulated loss amounts are as follows:

-500*ln(1-0.12) = 63.91668575.

-500*ln(1-0.34) = 207.757722.

-500*ln(1-0.56) = 410.490276.

-500*ln(1-0.89) = 1103.637457.

Loss amounts of 63.91668575 and 207.757722 are each below the deductible of 300, so they result in payments of 0. The loss amount of 410.490276 results in a payment of

(410.490276 – 300)*0.5 = 55.245138.

The insurer will pay 50% of losses up to a total payment of 400. The total payment of 400 will occur when losses exceed the deductible of 300 by 400/0.5 = 800, so any loss amount above 1100 will result in a payment of 400. Thus, a loss amount of 1103.637457 results in a payment of 400.

The average payment is thus (0 + 0 + 55.245138 + 400)/4 = **113.8112845.**

**Problem S4C114-2. Similar to Question 276 of the** **Exam C Sample Questions** **from the Society of Actuaries.** You know that number of peanuts eaten by elephants follows the distribution function F(x) = 1 – θ/x for θ

You also know the following:

8 elephants ate fewer than 60 peanuts.

7 elephants ate between 60 and 90 peanuts.

10 elephants ate more than 90 peanuts.

Find the maximum likelihood estimate of θ.

**Solution S4C114-2.** Each elephant that ate fewer than 60 peanuts contributes a factor of F(60) = 1 – θ/60 to the likelihood function. Each elephant that ate between 60 and 90 peanuts contributes a factor of F(90) – F(60) = (1 – θ/90) – (1 – θ/60) = θ/60 – θ/90 = θ/180 to the likelihood function.

Each elephant that ate more than 90 peanuts contributes a factor of S(90) = θ/90 to the likelihood function.

Thus, L(θ) = (1 – θ/60)8*(θ/180)7*(θ/90)10.

The loglikelihood function is l(θ) = ln((1 – θ/60)8*(θ/180)7*(θ/90)10) =

8*ln(1 – θ/60) + 7*ln(θ/180) + 10*ln(θ/90).

l'(θ) = (-8/60)/(1 – θ/60) + 7/θ + 10/θ = 0 at the maximum.

Thus, 17/θ = (2/15)/(1 – θ/60) →

2θ = 255(1 – θ/60) →

2θ = 255 – 255θ/60 →

6.25θ = 255 → **θ = 40.8**.

**Problem S4C114-3. Similar to Question 278 of the** **Exam C Sample Questions** **from the Society of Actuaries.** You know the following data:

There are n total losses.

6 of the losses are below 100.

x of the losses are between 100 and 300.

y of the losses are between 300 and 500.

30 of the losses are between 500 and 800.

There are no losses greater than 800.

Assume that losses are uniformly distributed within each of the intervals given above. From the ogive constructed from this data, you know that

Fn(200) = 0.18, and Fn(350) = 0.40. Use this information to find x.

**Solution S4C114-3.** We can find an expression for n: n = 6 + x + y + 30 = 36 + x + y.

Thus, Fn(200) = (6 + (200-100)x/(300-100))/(36 + x + y) = (6 + 0.5x)/(36 + x + y) = 0.18.

Fn(350) = (6 + x + (350-300)y/(500-300))/(36 + x + y) = (6 + x + 0.25y)/(36 + x + y) = 0.40.

Thus, (36 + x + y)*0.18 = 6 + 0.5x →

6.48 + 0.18x + 0.18y = 6 + 0.5x →

0.18y = -0.48 + 0.32x →

y = -8/3 + (16/9)x.

Thus, (6 + x + 0.25(-8/3 + (16/9)x))/(36 + x + (-8/3 + (16/9)x)) = 0.40 →

(6 + x – 2/3 + (4/9)x)/(100/3 + (25/9)x) = 0.40 →

(16/3 + (13/9)x) = 0.40(100/3 + (25/9)x) →

16/3 + (13/9)x = 40/3 + (10/9)x →

8 = x/3 → **x = 24**.

**Problem S4C114-4. Similar to Question 279 of the** **Exam C Sample Questions** **from the Society of Actuaries.** Loss amounts (X) follow a distribution function F(x) = (x/200)2 for 0 ≤ x ≤ 200. An insurance company pays 50% of losses in excess of an ordinary deductible of 50, up to a maximum payment amount of 50. Given that a payment has been made, find the expected value of that payment.

**Solution S4C114-4.** If a payment has been made, then we know that the loss is in excess of the deductible of 50, so the *per-payment* expected value of payments is the *per-loss* expected value of payments divided by S(50) = 1 – (50/200)2 = 0.9375.

Also, the company will only make additional payments for losses less than 50 + 50/0.5 = 150.

We can find the per-loss expected value of payments as follows:

The amount paid per loss is Y = 0.5(X – 50) Λ 150.

f(x) = 2x/2002 = x/20000.

E(Y) = 50150∫0.5(x-50)*(x/20000)dx + S(150)*50 =

50150∫(x2/40000 – x/800)dx + (1-(150/200)2)*50 =

(x3/120000 – x2/1600)│50150 + 21.875 = 14.5833333 + 21.875 = 36.4583333333.

The per-payment expected payment is 36.4583333333/0.9375 = **38.8888888889.**

**Problem S4C114-5. Similar to Question 280 of the** **Exam C Sample Questions** **from the Society of Actuaries.** Claim frequency (N) follows a Poisson distribution with mean λ = 8.

Claim severity (X) is either 7 with probability 0.3 or k with probability 0.7. It is known that k > 7. A policy of aggregate stop-loss insurance has a deductible of 7 on the loss amount. The expected value of a payment under this policy is 65.80234824. Find the value of k.

**Solution S4C114-5.** Let S be aggregate payments. We want to find an expression for

E(S) – E(S Λ 7).

E(S) = E(N)*E(X) = 8*(7*0.3 + k*0.7) = 16.8 + 5.6k.

E(S Λ 7) is 0*Pr(S = 0) + 7*(1 – Pr(S = 0)). Why is this the case? If there is one claim, then that claim is of size 7 or of size k > 7, so the next-largest value of S after 0 is 7.

Pr(S = 0) = Pr(N = 0), i.e., the probability of no claims. Pr(N = 0) = e-8, so

E(S Λ 7) = 7(1 – e-8).

Thus, 16.8 + 5.6k – 7(1 – e-8) = 65.80234824 → 56 = 5.6k → **k = 10**.

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