**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 105 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 S4C105-1. Similar to Question 204 of the** **Exam C Sample Questions** **from the Society of Actuaries.**The random variable X is exponentially distributed with mean 1/Y. The random variable Y follows a gamma distribution with parameters α = θ = 5. Find Pr(X

**Relevant properties of gamma distributions:** f(x) = (x/θ)α*(e-x/θ)/(x*Γ(α)).

**Solution S4C105-1.** The conditional cumulative distribution of X is F(x│Y = y) = e-xy. We are in particular interested in F(1/3│Y = y) = 1-e-y/3. F(1/3) = 0∞∫F(1/3│Y = y)*fY(y)*dy =

0∞∫e-y/3*fY(y)*dy. We find fY(y) = (y/5)5*(e-y/5)/(y*Γ(5)) = (y4/3125)*(e-y/5)/4 =

y4*(e-y/5)/12500.

Thus, our integral is 0∞∫(1-e-y/3)*y4*(e-y/5)/12500)dy =

0∞∫(y4*(e-y/5)/12500)dy – 0∞∫((e-8y/15*y4)/12500)dy.

The former integral is the integral of the gamma pdf over all of its possible values, and so has to equal 1.

To solve the latter integral, we use the Tabular Method of integration by parts:

**Sign………..u……………dv**

+………….. y4…………..(1/12500)e-8y/15

-……………4y3…………-(3/20000)e-8y/15

+…………..12y2…………(9/32000)e-8y/15

-……………24y………….-(27/51200)e-8y/15

+……………24…………..(81/81920)e-8y/15

-……………..0…………..-(243/131072)e-8y/15

Thus, the integral is

(-y4(3/20000)e-8y/15 – y3(36/32000)e-8y/15 – y2(324/51200)e-8y/15 – y(1944/81920)e-8y/15 – (5832/131072)e-8y/15)│0∞ = 1 – F(1/3) = 5832/131072 = 0.0444946289.

Thus, F(1/3) = 1 – 0.0444946289 = **F(1/3) = 0.9555053711**.

**Problem S4C105-2. Similar to Question 205 of the** **Exam C Sample Questions** **from the Society of Actuaries.**

There are 340 elephants in group A. The probability that an elephant in group A is hungry and will want to eat some peanuts is 0.4. If an elephant in group A eats some peanuts, the number of peanuts eaten follows a geometric distribution with mean 5.

There are 500 elephants in group B. The probability that an elephant in group B is hungry and will want to eat some peanuts is 0.1. If an elephant in group B eats some peanuts, the number of peanuts eaten follows a geometric distribution with mean 10.

Use a normal approximation to find the probability that the number of peanuts eaten by all elephants exceeds 1200. Refer to the Exam 4 / C Tables, as necessary. You do not need to use a continuity correction.

**Solution S4C105-2.** Let N be the number of occurrences of elephants eating peanuts, and let X be the amount of peanuts eaten per such occurrence. Let S be aggregate peanuts eaten.

For group A,

E(N) = 340*0.4 = 136.

We note that N is binomially distributed with m = 340 and q = 0.4, as elephants can either eat peanuts or not eat them.

Thus, for group A, Var(N) = 340*0.4*(1-0.4) = 81.6.

For group A, E(X) = β for the geometric distribution = 5; Var(X) = β(1+β) = 5*6 = 30.

Thus, for group A, E(S) = E(N)*E(X) = 136*5 = 680, and

Var(S) = E(N)*Var(X) + Var(N)*E(X)2 = 136*30 + 81.6*52 = Var(S) = 6120.

For group B,

E(N) = 500*0.1 = 50.

We note that N is binomially distributed with m = 500 and q = 0.1, as elephants can either eat peanuts or not eat them.

Thus, for group A, Var(N) = 500*0.1*(1-0.1) = 45.

For group A, E(X) = β for the geometric distribution = 10; Var(X) = β(1+β) = 10*11 = 110.

Thus, for group A, E(S) = E(N)*E(X) = 50*10 = 500, and

Var(S) = E(N)*Var(X) + Var(N)*E(X)2 = 50*110 + 45*102 = Var(S) = 10000.

For the two groups, we add the values of E(S) from each group to get E(S) = 680 + 500 = 1180.

We perform an analogous procedure to get Var(S) = 6120 + 10000 = 16120, implying that SD(S) = 126.964562. We want to find Pr(S > 1200) ≈ 1 – Φ((1200-1180)/126.964562) = 1 – Φ(0.1575242705), which we find via the Excel input “=1-NORMSDIST(0.1575242705)”, getting as our answer **0.437415841**.

**Problem S4C105-3. Similar to Question 206 of the** **Exam C Sample Questions** **from the Society of Actuaries.** Frequency of losses (N) follows a geometric distribution with mean 6. Severity of each loss (X) follows a distribution with the following probabilities:

Pr(X = 4) = 0.6;

Pr(X = 8) = 0.3;

Pr(X = 12) = 0.1.

Let S denote aggregate losses.

Find E((S-12)+), the expected value of the left-censored and shifted random variable at 12.

**Relevant properties of geometric distributions:** E(N) = β; Pr(N = k) = βk/(1+β)k+1.

**Solution S4C105-3.** We note that S can only occur in intervals of 4. We compare possible values of three random variables: **S:**………..0, 4, 8, 12, 16, 20, 24, ….

**(S-12)+:** 0, 0, 0, 0, 4, 8, 12,….

**S-12:** -12, -8, -4, 0, 4, 8, 12, …

We note that to turn E(S-12) into E((S-12)+), we need to add 12*fS(0) + 8*fS(4) + 4*fS(8).

Thus, E((S-12)+) = E(S) – 12 + 12*fS(0) + 8*fS(4) + 4*fS(8).

First, we find E(X) = 4*0.6 + 8*0.3 + 12*0.1 = 6.

E(S) = E(N)*E(X) = 6*6 = 36.

Now we find fS(0) = Pr(N = 0) = 1/(1+6) = 1/7.

We find fS(4) = Pr(N = 1)*Pr(X = 4) = (6/72)(0.6) = 18/245.

We find fS(8) = Pr(N = 1)*Pr(X = 8) + Pr(N = 2)*Pr(Both X are 4) =

(6/72)(0.3) + (62/73)(0.62) = fS(8) = 639/8575.

Thus, E((S-12)+) = 36 – 12 + 12*(1/7) + 8*(18/245) + 4*(639/8575) = **E(****(S-12)+) = 26.60011662**.

**Problem S4C105-4. Similar to Question 207 of the** **Exam C Sample Questions** **from the Society of Actuaries.** The size of each loss (X) follows a distribution with probability density function f(x) = 0.05x, for 0 P be the per-payment loss payment random variable. Find E(YP).

**Solution S4C105-4.** We first find the pdf of X, given that X exceeds the deductible of 2. This is f(x)/(1-F(2)), where F(2) = 02∫0.05x*dx = 0.025*22 = 0.1. Thus, our pdf of X, given that X exceeds the deductible of 2, is 0.05x/(1-0.1) = x/18. We find the expected value of X, given that X exceeds 2: 2√(40)∫(x)(x/18)*dx = (x3/54)│2√(40) = (403/2/54) – 8/54 = 4.536707645. To find E(YP), we subtract the deductible of 2 from this value and get **E(YP) = 2.536707645**.

**Problem S4C105-5. Similar to Question 212 of the** **Exam C Sample Questions** **from the Society of Actuaries.** Frequency of losses (N) follows a Poisson distribution with mean 30. Severity of each loss (X) follows a uniform distribution on the interval from 0 to 50. A deductible of 10 is applied to each loss. Find the variance of aggregate payments after the deductible has been applied.

**Solution S4C105-5.** The distribution of payment severity is different from the distribution of loss severity; it is still uniform, but it is uniform from 0 to (50-10) = 40, corresponding to the distribution of losses from 10 to 50. The pdf of this uniform distribution is 1/40. Since (1/5) of losses now do not result in payment, payment frequency is (4/5) of loss frequency, and so expected number of payments is 30*4/5 = 24.

Let R denote number of payments, and let Q denote payment severity. Let S denote aggregate payments. Since R is Poisson, Var(S) = E(R)*E(Q2). We find E(Q2) = 040∫(q2/40)dq =

(q3/120)│040 = 533.3333333.

Our answer is thus 24*533.33333333 = **Var(S) = 12800**.

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