Assorted Exam-Style Questions for Exam 4/C - Part 4

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 S4C86-1. Similar to Question 58 of the Exam C Sample Questions from the Society of Actuaries. The number of pet grooming liability claims per insured pet groomer follows a Poisson distribution with mean Λ. The prior distribution of Λ has the following probability density function: f_Λ(λ) = (30λ)¹²*e^-30λ/(λ*Γ(12)).

In year 1, there were 23 insured pet groomers and 4 pet grooming liability claims.

In year 2, there were 67 insured pet groomers and 20 pet grooming liability claims.

It is expected that there will be 100 insured pet groomers in year 3. Find the expected number of pet grooming liability claims for this group of 100 insured pet groomers. Refer to the Exam 4 / C Tables as necessary.

Solution S4C86-1. From the exam tables, we note that f_Λ(λ) has the format of a pdf of a gamma distribution with α = 12 and θ = (1/30). Since the prior distribution is gamma and the model distribution is Poisson, we are working with a common case of conjugate prior distributions (Case 1 from Section 65):
Prior distribution of λ is gamma with parameters α and θ.

Model distribution is Poisson with parameter λ.

Posterior distribution is gamma with parameters α + Σx_i and θ/(nθ+1).

We find n = number of exposures in the two years for which we have observations = 67 + 23 = 90.

We also find Σx_i = number of claims in the two years for which we have observations = 4 + 20 = 24.

Thus, the posterior distribution is gamma and has parameters α* = α + Σx_i = 12 + 24 = 36 and θ* = θ/(nθ+1) = (1/30)/(90/30 + 1) = 1/120.

The mean of the posterior distribution per insured is (α*)(θ*) = 36/120 = 0.3, and so for 100 insureds, the expected number of claims is 100*0.3 = 30 claims.

Problem S4C86-2. Similar to Question 59 of the Exam C Sample Questions from the Society of Actuaries. Construct a p-p plot by adhering to the following instructions. Remember that you are plotting two cumulative distribution functions (cdfs) against one another:

- Draw the horizontal axis and label it "Sample distribution." The values on the axis should range from 0 to 1.

- Draw the vertical axis and label it "Fitted distribution." The values on the axis should range from 0 to 1.

- Draw a 45-degree (slope = 1) line from the origin to the point (1, 1).

- Draw two points on the 45-degree line at (0.2, 0.2) and (0.8, 0.8).

- Now you will draw the actual p-p plot. Draw an always increasing curve between the origin and (0.2, 0.2) such that the entire curve is below the 45-degree line.

- Draw an always increasing curve between (0.2, 0.2) and (0.8, 0.8) such that the entire curve is above the 45-degree line.

- Draw an always increasing curve between (0.8, 0.8) and (1, 1) such that the entire curve is below the 45-degree line.

Answer the following questions about the fitted distribution:

(a) Does the fitted distribution have a lighter or heavier left tail than the sample distribution in the region from the origin to (0.2, 0.2)?
(b) Does the fitted distribution place greater or smaller probabilities on the "middle range" between (0.2, 0.2) and (0.8, 0.8) than the sample distribution?

(c) Does the fitted distribution have a lighter or heavier right tail than the sample distribution in the region from (0.8, 0.8) to (1, 1)?

Solution S4C86-2. We answer the questions in sequence:
(a) In the region from the origin to (0.2, 0.2), the curve is below the 45-degree line, meaning that the fitted probabilities are smaller than the sample probabilities, which means the fitted distribution emphasizes the left tail less and has a lighter left tail.

(b) In the region between (0.2, 0.2) and (0.8, 0.8), the curve is above the 45-degree line, meaning that the fitted probabilities are larger than the sample probabilities, which means that the fitted distribution emphasizes the "middle range" more and places greater probabilities on the "middle range" than the sample distribution.

(c) In the region between (0.8, 0.8) and (1, 1), the fitted distribution assigns smaller probabilities than the sample distribution for being below a particular value (as the p-p plot is a plot of cumulative distribution functions). This means that the fitted distribution assigns larger probabilities for being above the value in question, meaning that the fitted distribution emphasizes the right tail more than the sample distribution and so has a heavier right tail.

Problem S4C86-3. Similar to Question 60 of the Exam C Sample Questions from the Society of Actuaries. You know the following about three dice:
Die A has 0.2 probability of value 0, 0.34 probability of value 1, 0.46 probability of value 2.

Die B has 0.5 probability of value 0, 0.4 probability of value 1, 0.1 probability of value 2.

Die C has 0.43 probability of value 0, 0.55 probability of value 1, 0.02 probability of value 2.
You pick one die at random and observe the following sequence of rolls S: S = {0, 2, 1, 1}.

Find the posterior expected value of the next roll of the die.

Solution S4C86-3. We want to find E(Next roll │S). To do this, we need to find the posterior probability of each possible outcome.

We note that Pr(S and Die A) = Pr(Die A)*Pr(S │ Die A) = (1/3)(0.2*0.46*0.34²) = 0.0035450667.

Pr(S and Die B) = Pr(Die B)*Pr(S │ Die B) = (1/3)(0.5*0.1*0.4²) = 0.00266666667.

Pr(S and Die C) = Pr(Die C)*Pr(S │ Die C) = (1/3)(0.43*0.02*0.55²) = 0.00086716666667.

Posterior probability of A is Pr(S and Die A)/Pr(S) =

0.0035450667/(0.0035450667 + 0.00266666667 + 0.00086716666667) = 0.5007934378.

Posterior probability of B is Pr(S and Die B)/Pr(S) =

0.00266666667/(0.0035450667 + 0.00266666667 + 0.00086716666667) = 0.3767063621.

Posterior probability of C is Pr(S and Die C)/Pr(S) =

0.00086716666667/(0.0035450667 + 0.00266666667 + 0.00086716666667) = 0.1225002001.

Moreover, E(A) = 0.2*0 + 0.34*1 + 0.46*2 = 1.26.

E(B) = 0.5*0 + 0.4*1 + 0.1*2 = 0.6.

E(C) = 0.43*0 + 0.55*1 + 0.02*2 = 0.59.

Thus, our posterior expected value is

0.5007934378*1.26 + 0.3767063621*0.6 + 0.1225002001*0.59 = 0.9292986669.

Problem S4C86-4. Similar to Question 61 of the Exam C Sample Questions from the Society of Actuaries. You observe the following twelve raw claim amounts:
20, 20, 25, 26, 27, 35, 37, 45, 67, 89, 100, 103

Each claim is subject to a deductible of 10. You also know that raw claim amounts follow an exponential distribution. Use maximum likelihood estimation to find the average amount the insurer will have to pay out per claim.

Solution S4C86-4. The exponential distribution has the memoryless property, which means that the excess over a certain value (say, our deductible of 10) follows an exponential distribution if the original values follow an exponential distribution. This means that we can present our sample data as data for claims minus deductible:
10, 10, 15, 16, 17, 25, 27, 35, 57, 79, 90, 93.

This sample is also presumed to follow an exponential distribution. The maximum likelihood estimate for the mean θ of an exponential distribution is equal to the sample mean. In this case, our desired estimate is (10 + 10 + 15 + 16 + 17 + 25 + 27 + 35 + 57 + 79 + 90 + 93)/12 = 39.5.

Problem S4C86-5. Similar to Question 68 of the Exam C Sample Questions from the Society of Actuaries.

You have the following data for two groups of hippopotami, A and B:
For Group A:

In year 1, 20 hippopotami fly to the moon, out of 90 that could have.

In year 2, 40 hippopotami fly to the moon, out of 70 that could have.

In year 3, 10 hippopotami fly to the moon, out of30 that could have.

For Group B:

In year 1, 230 hippopotami fly to the moon, out of 1000 that could have.

In year 2, 320 hippopotami fly to the moon, out of 770 that could have.

In year 3, 150 hippopotami fly to the moon, out of450 that could have.

Let Ŝ_T(3) be the Kaplan-Meier product-limit estimate of the survival function (i.e., the proportion of hippopotami from both groups that did not fly to the moon after all three years have elapsed.

Let Ŝ_B(3) be the Kaplan-Meier product-limit estimate of the survival function (i.e., the proportion of hippopotami from Group B that did not fly to the moon after all three years have elapsed.

Find Ŝ_T(3) - Ŝ_B(3).

Solution S4C86-5. In each year, our risk set is the number of hippopotami that could have flown to the moon; our s_i is the number of hippopotami that actually flew to the moon.

We use the formula for the Kaplan-Meier product-limit estimator:

S_n(t) = 1 if 0 ≤ t < y₁;
S_n(t) = _i=1^j-1Π((r_i - s_i)/r_i) if y_j-1 ≤ t < y_j for j = 2, ..., k;
S_n(t) = _i=1^kΠ((r_i - s_i)/r_i) or 0 for y_k ≤ t.

Thus, we find Ŝ_B(3) = ((1000-230)/1000)*((770-320)/770)*((450-150)/450) = 300/1000 = 0.3.

We find Ŝ_T(3) = (((1000+90)-(230+20))/(1000+90))*((770+70)-(40+320))/(770+70))* ((450+30)-(150+10))/(450+30)) = ((450+30)-(150+10))/(1000+90) = 0.2935779817.

Thus, Ŝ_T(3) - Ŝ_B(3) = 0.2935779817 - 0.3 = -0.0064220183.

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