Download Practice Test 2 Solutions - Actuarial Problem Solving - Spring 2008 | MATH 370 and more Exams Mathematics in PDF only on Docsity!
Math 370, Spring 2008
Prof. A.J. Hildebrand
Practice Test 2 Solutions
About this test. This is a practice test made up of a random collection of 15 problems from past Course 1/P actuarial exams. Most of the problems have appeared on the Actuarial Problem sets passed out in class, but I have also included some additional problems.
Topics covered. This test covers the topics of Chapters 1–5 in Hogg/Tanis and Actuarial Problem Sets 1–5.
Ordering of the problems. In order to mimick the conditions of the actual exam as closely as possible, the problems are in no particular order. Easy problems are mixed in with hard ones. In fact, I used a program to select the problems and to put them in random order, with no human intervention. If you find the problems hard, it’s the luck of the draw!
Suggestions on taking the test. Try to take this test as if it were the real thing. Take it as a closed books, notes, etc., time yourself, and stop after 2 hours. In the actuarial exam you have 3 hours for 30 problems, so 2 is an appropriate time limit for a 20 problem test.
Answers/solutions. Answers and solutions will be posted on the course webpage, www. math.uiuc.edu/∼hildebr/370.
1. [4-115]
Let X and Y denote the values of two stocks at the end of a five-year period. X is uniformly distributed on the interval (0, 12). Given X = x, Y is uniformly distributed on the interval (0, x). Determine Cov(X, Y ) according to this model.
(A) 0 (B) 4 (C) 6 (D) 12 (E) 24
Answer: C: 6 Solution: To find Cov(X, Y ), we need to compute the three expectations E(XY ), E(X) and E(Y ). Now E(X) = 6, since X is uniform on the interval (0, 12). To find the remaining two expectations we need to first compute the joint density f (x, y). To this end we use the formula f (x, y) = h(y|x)fX (x). Since fX (x) is uniform on the interval (0, 12), we have fX (x) = 1/12 for 0 < x < 12. Since the conditional density of Y given X = x is uniform on [0, x], we have h(y|x) = 1/x for 0 ≤ y ≤ x ≤ 12. Hence
f (x, y) = h(y|x)fX (x) =
12 x
, 0 ≤ y ≤ x ≤ 12.
The rest is now routine:
E(XY ) =
x=
∫ (^) x
y=
xy
12 x
dydx =
x=
x^2 dx = 24,
E(Y ) =
x=
∫ (^) x
y=
y
12 x
dydx =
x=
xdx = 3,
Cov(X, Y ) = E(XY ) − E(X)E(Y ) = 24 − 6 · 3 = 6.
- [4-17]
A device contains two circuits. The second circuit is a backup for the first, so the second is used only when the first has failed. The device fails when and only when the second circuit fails. Let X and Y be the times at which the first and second circuits fail, respectively. X and Y have joint probability density function
f (x, y) =
6 e−xe−^2 y^ for 0 < x < y < ∞, 0 otherwise.
What is the expected time at which the device fails?
(A) 0. 33 (B) 0. 50 (C) 0. 67 (D) 0. 83 (E) 1. 50
Answer: D: 0. Solution: The time point at which the device fails is given by Y , the failure time of the second (back-up) device, so we need to compute E(Y ). Now, E(Y ) =
R yf^ (x, y)dxdy, where f (x, y) is the given density and R the region given by 0 ≤ y ≤ ∞, 0 ≤ x ≤ y. Therefore,
E(Y ) =
y=
∫ (^) y
x=
y 6 e−xe−^2 y^ dxdy
y=
6 ye−^2 y^ (1 − e−y^ )dy = 6
y=
y
e−^2 y^ − e−^3 y^
dy
[
y
e−^2 y − 2
e−^3 y − 3
)]∞
y=
y=
e−^2 y 2
e−^3 y 3
dy
(D) 2 + 3e−^2 /^3 (E) 5
Answer: D Solution: Since T is exponentially distributed with mean 3, the density of T is f (t) = (1/3)e−t/^3 for t > 0. Since X = max(T, 2), we have X = 2 if 0 ≤ T ≤ 2 and X = T if 2 < T < ∞. Thus,
E(X) =
0
e−t/^3 dt +
2
t ·
e−t/^3 dt
= 2(1 − e−^2 /^3 ) − te−t/^3
∞ 2
2
e−t/^3 dt
= 2(1 − e−^2 /^3 ) + 2e−t/^3 + 3e−^2 /^3 = 2 + 3e−^2 /^3
6. [4-2]
A car dealership sells 0, 1, or 2 luxury cars on any day. When selling a car, the dealer also tries to persuade the customer to buy an extended warranty for the car. Let X denote the number of luxury cars sold in a given day, and let Y denote the number of extended warranties sold, and suppose that
P (X = x, Y = y) =
1 / 6 for (x, y) = (0, 0), 1 / 12 for (x, y) = (1, 0), 1 / 6 for (x, y) = (1, 1), 1 / 12 for (x, y) = (2, 0), 1 / 3 for (x, y) = (2, 1), 1 / 6 for (x, y) = (2, 2).
What is the variance of X?
(A) 0. 47 (B) 0. 58 (C) 0. 83 (D) 1. 42 (E) 2. 58
Answer: B: 0. Solution: We first compute the marginal distribution of X by adding appropriate prob- abilities: P (X = 0) = 1/6, P (X = 1) = 1/12 + 1/6 = 1/4, P (X = 2) = 1/12 + 1/3 + 1/6 = 7 /12. Therefore
E(X) = 0 ·
E(X^2 ) = 0^2 ·
+ 1^2 ·
+ 2^2 ·
Var(X) = E(X^2 ) − E(X)^2 = 0. 58.
7. [2-53]
A company prices its hurricane insurance using the following assumptions:
(i) In any calendar year, there can be at most one hurricane. (ii) In any calendar year, the probability of a hurricane is 0.05.
(iii) The number of hurricanes in any calendar year is independent of the number of hurricanes in any other calendar year.
Using the company’s assumptions, calculate the probability that there are fewer than 3 hurricanes in a 20-year period.
(A) 0. 06 (B) 0. 19 (C) 0. 38 (D) 0. 62 (E) 0. 92
Answer: E: 0. Solution: By the given assumptions the occurrence of hurricanes can be modeled as success/failure trials, with success meaning that a hurricane occurs in a given year and occurring with probability p = 0.05. Thus, the probability that there are fewer than 3 hurricanes in a 20-year period is equal to the probability of having less than 3 successes in 20 success/failure trials with p = 0.05. By the binomial distribution, this probability is
∑^2
x=
x
- 05 x 0. 9520 −x
8. [3-51]
The loss due to a fire in a commercial building is modeled by a random variable X with density function
f (x) =
0 .005(20 − x) for 0 < x < 20, 0 otherwise. Given that a fire loss exceeds 8, what is the probability that it exceeds 16?
(A) 1/ 25 (B) 1/ 9 (C) 1/ 8 (D) 1/ 3 (E) 3/ 7
Answer: B: 1/ Solution: This is an easy integration exercise. We need to compute P (X ≥ 16 |X ≥ 8), which is the same as P (X ≥ 16)/P (X ≥ 8). Integrating the given density function, we get
P (X ≥ x) =
x
0 .005(20 − t)dt = 0.0025(20 − x)^2 , 0 < x < 20.
Taking the ratio of these expressions with x = 16 and x = 8 gives the answer: (20 − 16)^2 /(20 − 8)^2 = 1/9.
- [1-101]
A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This month, the insurance company writes 4 new policies for adults earning their first driver’s license. What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers?
(A) 0. 006 (B) 0. 012 (C) 0. 018 (D) 0. 049 (E) 0. 073
Answer: D: 0.
(D)
0
0
f (s, t)dsdt +
0
0
f (s, t)dsdt
(E)
0
- 5
f (s, t)dsdt +
0
0
f (s, t)dsdt
Answer: E Solution: The key to this problem is the correct interpretation of the event (∗) “the device fails within the first half hour” in terms of the variables s and t. Since failure occurs if either (i.e., the first, the second, or both) of the components fails, (∗) translates into (∫∫∗∗) “s ≤ 1 / 2 or t ≤ 1 /2”. The probability for failure is therefore given by double integral R f^ (s, t)dsdt, where^ R^ is the region inside the unit square 0^ ≤^ s^ ≤^1 ,^0 ≤^ t^ ≤^ 1 described by (∗∗). A sketch shows that R splits into two disjoint parts, the rectangle 0 ≤ s ≤ 0 .5, 0 ≤ t ≤ 1, and the square 0. 5 ≤ s ≤ 1, 0 ≤ t ≤ 0 .5. Thus, ∫ ∫
R
f (s, t)dsdt =
0
- 5
f (s, t)dsdt +
0
0
f (s, t)dsdt,
which is answer (E).
12. [3-7]
The time to failure of a component in an electronic device has an exponential distribution with a median of four hours. Calculate the probability that the component will work without failing for at least five hours.
(A) 0. 07 (B) 0. 29 (C) 0. 38 (D) 0. 42 (E) 0. 57
Answer: D: 0. Solution: The general form of the c.d.f. of an exponential distribution is F (x) = 1−e−x/θ for x > 0. Since the median is 4 hours, we have F (4) = 1/2. This allows us to determine the parameter θ: 1 − e−^4 /θ^ = 1/2, so θ = 4/ ln 2. The probability that the component will work for at least 5 hours is given by 1 − F (5) = e−^5 /θ^ = e−(5/4) ln 2^ = 0.42.
13. [1-13]
A health study tracked a group of persons for five years. At the beginning of the study, 20% were classified as heavy smokers, 30% as light smokers, and 50% as nonsmokers. Results of the study showed that light smokers were twice as likely as nonsmokers to die during the five-year study, but only half as likely as heavy smokers. A randomly selected participant from the study died over the five-year period. Calculate the probability that the participant was a heavy smoker.
(A) 0. 20 (B) 0. 25 (C) 0. 35 (D) 0. 42 (E) 0. 57
Answer: D: 0. Solution: A standard exercise in Bayes’ Rule; the only non-routine part here is to properly interpret the phrase “light smokers were twice as likely as nonsmokers, and half as likely as heavy smokers to die ...”. This boils down to relations between the conditional probabilities of dying given a light smoker, a nonsmoker, and a heavy smoker.
14. [4-108]
The stock prices of two companies at the end of any given year are modeled with random variables X and Y that follow a distribution with joint density function
f (x, y) =
2 x for 0 < x < 1, x < y < x + 1, 0 otherwise. What is the conditional variance of Y given that X = x?
(A) 1/ 12 (B) 7/ 6 (C) x + 1/ 2 (D) x^2 − 1 / 6 (E) x^2 + x + 1/ 3
Answer: A: 1/ Solution: The conditional density of Y given X = x is h(y|x) = f (x, y)/fX (x). Here f (x, y) is given, and the marginal density fX (x) can be computed by integrating f (x, y) with respect to y: fX (x) =
∫ (^) y=x+
y=x
2 xdy = 2x, 0 < x < 1.
It follows that h(y|x) = (2x)/(2x) = 1 for 0 < x < 1, x ≤ y ≤ x + 1. This means that the conditional density of Y given X = x is uniform on the interval [x, x + 1]. By the formula for the variance of a uniform distribution, it follows that Var(Y |x) = 1/12, so the answer is 1/12. Alternatively, one can compute Var(Y |x) directly, without resorting to variance formulas for the uniform distribution, by computing the appropriate integrals over the conditional density h(y|x):
E(Y |x) =
∫ (^) x+
y=x
yh(y|x)dy =
(x + 1)^2 − x^2
E(Y 2 |x) =
∫ (^) x+
y=x
y^2 h(y|x)dy. =
(x + 1)^3 − x^3
Var(X|x) = E(Y 2 |x) − E(Y |x)^2
=
3 x^2 + 3x + 1
(2x + 1)^2 =
15. [3-17]
An insurance company sells an auto insurance policy that covers losses incurred by a poli- cyholder, subject to a deductible of 100. Losses incurred follow an exponential distribution with mean 300. What is the 95th percentile of actual losses that exceed the deductible?
(A) 600 (B) 700 (C) 800 (D) 900 (E) 1000
Answer: E: 1000 Solution: The main difficulty here is the correct interpretation of the “95th percentile of actual losses that exceed the deductible”. The proper interpretation involves a conditional probability: we seek the value x such that the conditional probability that the loss is at most x, given that it exceeds the deductible, is 0.95, i.e., that P (X ≤ x|X ≥ 100) = 0.95, where X denotes the loss. By the complement formula for conditional probabilities, this is equivalent to P (X ≥ x|X ≥ 100) = 0.05. Since X is exponentially distributed with mean 300, we have P (X ≥ x) = e−x/^300 , so for x > 100,
P (X ≥ x|X ≥ 100) = P (X ≥ x) P (X ≥ 100
e−x/^300 e−^100 /^300
= e−(x−100)/^300.
Setting this equal to 0.05 and solving for x, we get (x − 100)/300 = − ln(0.05), so x = −300 ln(0.05) + 100 = 1000.
To determine the constant K, we use the fact that the (overall) probability of a loss is 0 .05. Thus, the sum of the probabilities for the possible loss amounts has to equal 0.05:
∑^5
N =
K
N
K.
Hence K = 3/137, and the above expectation betcomes (43/30)(3/137) = 0.03138.
- [5-10]
An insurance policy pays a total medical benefit consisting of a part paid to the surgeon, X, and a part paid to the hospital, Y , so that the total benefit is X + Y. It is known that Var(X) = 5, 000, Var(Y ) = 10, 000, and Var(X + Y ) = 17, 000. Due to increasing medical costs, the company that issues the policy decides to increase X by a flat amount of 100 per claim and to increase Y by 10% per claim. Calculate the variance of the total benefit after these revisions have been made.
(A) 18, 200 (B) 18, 800 (C) 19, 300 (D) 19, 520 (E) 20, 670
Answer: C: 19, Solution: We need to compute Var(X + 100 + 1. 1 Y ). Since adding constants does not change the variance, this is the same as Var(X + 1. 1 Y ), which expands as follows:
Var(X + 1. 1 Y ) = Var(X) + Var(1. 1 Y ) + 2 Cov(X, 1. 1 Y ) = Var(X) + 1. 12 Var(Y ) + 2 · 1 .1 Cov(X, Y ).
We are given that Var(X) = 5, 000, Var(Y ) = 10, 000, so the only remaining unknown quantity is Cov(X, Y ), which can be computed via the general formula for Var(X + Y ):
Cov(X, Y ) =
(Var(X + Y ) − Var(X) − Var(Y ))
=
Substituting this into the above formula, we get the answer:
Var(X + 1. 1 Y ) = 5, 000 + 1. 12 · 10 , 000 + 2 · 1. 1 · 1 , 000 = 19, 520
19. [5-54]
Let T 1 be the time between a car accident and reporting a claim to the insurance company. Let T 2 be the time between the report of the claim and payment of the claim. The joint density function of T 1 and T 2 , f (t 1 , t 2 ), is constant over the region 0 < t 1 < 6 , 0 < t 2 < 6 , 0 < t 1 + t 2 < 10, and zero otherwise. Determine E(T 1 + T 2 ), the expected time between a car accident and payment of the claim.
(A) 4. 9 (B) 5. 0 (C) 5. 7 (D) 6. 0 (E) 6. 7
Answer: C: 5. Solution: We work in the t 1 t 2 -plane. A sketch shows that the given region is the square [0, 6] × [0, 6] with a right triangle of side length 2 deleted in the upper right hand corner of the square. Thus, the total area of this region is 6^2 − (1/2)2^2 = 34, and because of the uniform distribution the joint density function of T 1 and T 2 is constant and equal to f (t 1 , t 2 ) = 1/34 in this region. Now, E(T 1 + T 2 ) = E(T 1 ) + E(T 2 ). By symmetry, E(T 2 )
is the same as E(T 1 ), so it suffices to compute the latter expectation and then double the result. We use the double integral formula E(T 1 ) =
R t^1 f^ (t^1 , t^2 )dt^2 dt^1 , where^ R^ is the above region and f (t 1 , t 2 ) = 1/34 in this region. A sketch shows that R can be broken into two disjoint pieces, described by the inequalities 0 ≤ t 1 ≤ 4, 0 ≤ t 2 ≤ 6, and 4 ≤ t 1 ≤ 6, 0 ≤ t 2 ≤ 10 − t 1 , respectively. It follows that
E(T 1 ) =
t 1 =
t 2 =
t 1 f (t 1 , t 2 )dt 2 dt 1 +
t 1 =
∫ (^10) −t 1
t 2 =
t 1 f (t 1 , t 2 )dt 2 dt 1
t 1 =
6 t 1 dt 1 +
t 1 =
t 1 (10 − t 1 )dt 1 = 2. 86
Hence E(T 1 + T 2 ) = 2 · 2 .86 = 5.72.
20. [5-61]
A company agrees to accept the highest of four sealed bids on a property. The four bids are regarded as four independent random variables with common cumulative distribution function F (x) =
(1 + sin πx) for 3/ 2 ≤ x ≤ 5 /2.
Which of the following represents the expected value of the accepted bid?
(A) π
3 / 2
x cos πxdx
(B)
3 / 2
(1 + sin πx)^4 dx
(C)
3 / 2
x(1 + sin πx)^4 dx
(D)
π 4
3 / 2
(cos πx)(1 + sin πx)^3 dx
(E)
π 4
3 / 2
x(cos πx)(1 + sin πx)^3 dx
Answer: E Solution: Let X 1 ,... , X 4 denote the four bids, and let X∗^ = max(X 1 ,... , X 4 ) denote the largest of these bids, i.e., the bid that is accepted. The c.d.f. of X∗^ then is given by (for 3/ 2 ≤ x ≤ 5 /2) (note the “maximum trick” here!)
F ∗(x) = P (X∗^ ≤ x) = P (max(X 1 ,... , X 4 ) ≤ x) = P (X 1 ≤ x,... , X 4 ≤ x) = P (X 1 ≤ x)... P (X 4 ≤ x) = F (x)^4.
Differentiating, we get the density of X∗:
f ∗(x) =
d dx
F (x)^4 = 4F (x)^3 F ′(x)
= 4
(1 + sin πx)^3
(cos πx)π
=
π 4 (1 + sin πx)^3 (cos πx) (3/ 2 ≤ x ≤ 5 /2).
