北美精算考试P练习题
北美精算师 exam P 2003 真题
Which of the following represents g?
(A)
15 y for 0 < y < 1 g ( y) = otherwise 0 15 y 2 for x 2 < y < x g ( y) = 2 0 otherwise 15 y 2 for 0 < y < 1 g ( y) = 2 0 otherwise
(A) (B) (C) (D) (E)
0.07 0.29 0.38 0.42 0.57
May 2003
9
Course 1
5.
An insurance company examines its pool of auto insurance customers and gathers the following information:
Course 1
10
Form 03A
6.
Let X and Y be continuous random variables with joint density function
8 xy f ( x, y ) = 3 0
for 0 ≤ x ≤ 1, x ≤ y ≤ 2 x otherwise.
Let X be a continuous random variable with density function
x for − 2 ≤ x ≤ 4 f ( x ) = 10 0 otherwise.
Calculate the expected value of X.
(A)
1 5 3 5
Calculate g′(3).
(A)
−34e −3 −29e −3 −5e −3 −4e −3 63e −3
美国数学竞赛试题-部分
2000AMC12ProblemsProblem1In the year,the United States will host the International Mathematical Olympiad.Let and be distinct positive integers such that the product.What is the largest possible value of the sum?Problem2Problem3Each day,Jenny ate of the jellybeans that were in her jar at the beginning of that day. At the end of the second day,remained.How many jellybeans were in the jar originally?Problem4The Fibonacci sequence starts with two1s,and each term afterwards is the sum of its two predecessors.Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?Problem5If where thenProblem6Two different prime numbers between and are chosen.When their sum is subtracted from their product,which of the following numbers could be obtained?Problem7How many positive integers have the property that is a positive integer?Problem8Figures,,,and consist of,,,and non-overlapping squares.If the pattern continued,how many non-overlapping squares would there be in figure?Problem9Mrs.Walter gave an exam in a mathematics class of five students.She entered the scores in random order into a spreadsheet,which recalculated the class average after each score was entered.Mrs. Walter noticed that after each score was entered,the average was always an integer.The scores (listed in ascending order)were71,76,80,82,and91.What was the last score Mrs.Walters entered?Problem10The point is reflected in the-plane,then its image is rotatedby about the-axis to produce,and finally,is translated by5units in the positive-direction to produce.What are the coordinates of?Problem11Two non-zero real numbers,and satisfy.Which of the following is a possible value of?Problem12Let A,M,and C be nonnegative integers such that.What is the maximum value of+++?Problem13One morning each member of Angela’s family drank an8-ounce mixture of coffee with milk.The amounts of coffee and milk varied from cup to cup,but were never zero.Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee.How many people are in the family?Problem14When the mean,median,and modeof the listare arranged in increasing order,they form a non-constant arithmetic progression.What is the sum of all possible real values of?Problem15Let be a function for which.Find the sum of all values of for which.Problem16A checkerboard of rows and columns has a number written in each square,beginning in the upper left corner,so that the first row is numbered,the second row, and so on down the board.If the board is renumbered so that the left column,top to bottom, is,the second column and so on across the board,some squares have the same numbers in both numbering systems.Find the sum of the numbers in these squares (under either system).Problem17A centered at has radius and contains the point.The segment is tangent to the circle at and.If point lies on and bisects,thenProblem18In year,the day of the year is a Tuesday.In year,the day is also a Tuesday.On what day of the week did th day of year occur?Problem19triangle,,,.Let denote the midpointof and let denote the intersection of with the bisector of angle.Which of the following is closest to the area of the triangle?Problem20If and are positive numbers satisfyingThen what is the value of latex?Problem21Through a point on the hypotenuse of right triangle,lines are drawn parallel to the legs of the triangle so that the triangle is divided into asquare and two smaller right triangles.The area of one of the two small right triangles times the area of the square.The ratio of the area of the other small right triangle to the area of the square isProblem22The graph below shows a portion of the curve defined by the quarticpolynomial.Which of the following is the smallest?Problem23Professor Gamble buys a lottery ticket,which requires that he pick six different integers from through,inclusive.He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer.It so happens that the integers on the winning ticket have the same property—the sum of the base-ten logarithms is an integer.What is the probability that Professor Gamble holds the winning ticket?Problem24If circular arcs and centers at and,respectively,then there exists a circletangent to both and,and to.If the length of is,then the circumference of the circle isProblem25Eight congruent Equilateral triangle each of a different color,are used to construct a regular octahedron.How many distinguishable ways are there to construct the octahedron?(Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)Problem6For how many positive integers does there exist at least one positive integer n such that?infinitely manyProblem7A arc of circle A is equal in length to a arc of circle B.What is the ratio of circle A's area and circle B's area?Problem8Betsy designed a flag using blue triangles,small white squares,and a red center square,as shown. Let be the total area of the blue triangles,the total area of the white squares,and the area of the red square.Which of the following is correct?Problem9Jamal wants to save30files onto disks,each with1.44MB space.3of the files take up0.8MB, 12of the files take up0.7MB,and the rest take up0.4MB.It is not possible to split a file onto2different disks.What is the smallest number of disks needed to store all30files?Problem10Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size.She then pours half the coffee from the first cup to the second and,after stirring thoroughly,pours half the liquid in the second cup back to the first.What fraction of the liquid in the first cup is now cream?Problem11Mr.Earl E.Bird gets up every day at8:00AM to go to work.If he drives at an average speed of40miles per hour,he will be late by3minutes.If he drives at an average speed of60milesper hour,he will be early by3minutes.How many miles per hour does Mr.Bird need to drive to get to work exactly on time?Problem12Both roots of the quadratic equation are prime numbers.The number of possible values of isProblem13Two different positive numbers and each differ from their reciprocals by.What is?Problem14For all positive integers,let.Let. Which of the following relations is true?Problem15The mean,median,unique mode,and range of a collection of eight integers are all equal to8. The largest integer that can be an element of this collection isProblem16Tina randomly selects two distinct numbers from the set{1,2,3,4,5},and Sergio randomly selects a number from the set{1,2,...,10}.What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?Problem17Several sets of prime numbers,such as use each of the nine nonzero digits exactly once.What is the smallest possible sum such a set of primes could have?Problem18Let and be circles definedby and respectively.What is the length of the shortest line segment that is tangent to at and to at?Problem19The graph of the function is shown below.How many solutions does theequation have?Problem20Suppose that and are digits,not both nine and not both zero,and the repeatingdecimal is expressed as a fraction in lowest terms.How many different denominators are possible?Problem21Consider the sequence of numbers:For,the-th term of the sequence is the units digit of the sum of the two previous terms.Let denote the sum of the first terms of this sequence.The smallest value of for which is:Problem22Triangle is a right triangle with as its right angle,, and.Let be randomly chosen inside,and extend to meet at. What is the probability that?Problem23In triangle,side and the perpendicular bisector of meet in point,and bisects.If and,what is the area of triangle?SAT II数学词汇表代数部分1.基础add，plus加subtract减difference差multiply times乘product积divide除divisible可被整除的divided evenly被整除dividend被除数divisor因子，除数quotient商remainder余数factorial阶乘power乘方radical sign,root sign根号round to四舍五入to the nearest四舍五入2.有关集合union并集proper subset真子集solution set解集3.有关代数式、方程和不等式algebraic term代数项like terms,similar terms同类项numerical coefficient数字系数literal coefficient字母系数inequality不等式triangle inequality三角不等式range值域original equation原方程equivalent equation同解方程等价方程linear equation线性方程(e.g.5x+6=22)4.有关分数和小数proper fraction真分数improper fraction假分数mixed number带分数vulgar fraction，common fraction普通分数simple fraction简分数complex fraction繁分数numerator分子denominator分母(least)common denominator(最小)公分母quarter四分之一decimal fraction纯小数infinite decimal无穷小数recurring decimal循环小数tenths unit十分位5.基本数学概念arithmetic mean算术平均值weighted average加权平均值geometric mean几何平均数exponent指数，幂base乘幂的底数,底边cube立方数，立方体square root平方根cube root立方根common logarithm常用对数digit数字constant常数variable变量inverse function反函数complementary function余函数linear一次的，线性的factorization因式分解absolute value绝对值，e.g.｜-32｜=32round off四舍五入6.有关数论natural number自然数positive number正数negative number负数odd integer奇整数,odd number奇数even integer,even number偶数integer,whole number整数positive whole number正整数negative whole number负整数consecutive number连续整数real number,rational number实数,有理数irrational(number)无理数inverse倒数composite number合数e.g.4,6,8,9,10,12,14,15……reciprocal 倒数common divisor公约数multiple倍数(least)common multiple(最小)公倍数(prime)factor(质)因子common factor公因子prime number质数e.g.2,3,5,7,11,13,15……ordinary scale,decimal scale十进制nonnegative非负的tens十位units个位mode众数median中数common ratio公比7.数列arithmetic progression(sequence)等差数列geometric progression(sequence)等比数列8.其它approximate近似(anti)clockwise(逆)顺时针方向cardinal基数ordinal序数direct proportion正比distinct不同的estimation估计，近似parentheses括号proportion比例permutation排列combination组合table表格trigonometric function三角函数unit单位,位几何部分1.所有的角alternate angle内错角corresponding angle同位角vertical angle对顶角central angle圆心角interior angle内角exterior angle外角supplementary angles补角complementary angle余角adjacent angle邻角acute angle锐角obtuse angle钝角right angle直角round angle周角straight angle平角included angle夹角2.所有的三角形equilateral triangle等边三角形scalene triangle不等边三角形isosceles triangle等腰三角形right triangle直角三角形oblique斜三角形inscribed triangle内接三角形3.有关收敛的平面图形，除三角形外semicircle半圆concentric circles同心圆quadrilateral四边形pentagon五边形hexagon六边形heptagon七边形octagon八边形nonagon九边形decagon十边形polygon多边形parallelogram平行四边形equilateral等边形plane平面square正方形，平方rectangle长方形regular polygon正多边形rhombus菱形trapezoid梯形4.其它平面图形arc弧line,straight line直线line segment线段parallel lines平行线segment of a circle弧形5.有关立体图形cube立方体，立方数rectangular solid长方体regular solid/regular polyhedron正多面体circular cylinder圆柱体cone圆锥sphere球体solid立体的6.有关图形上的附属物altitude高depth深度side边长circumference,perimeter周长radian弧度surface area表面积volume体积arm直角三角形的股cross section横截面center of acircle圆心chord弦radius半径angle bisector角平分线diagonal对角线diameter直径edge棱face of a solid立体的面hypotenuse斜边included side夹边leg三角形的直角边median of a triangle三角形的中线base底边，底数(e.g.2的5次方，2就是底数) opposite直角三角形中的对边midpoint中点endpoint端点vertex(复数形式vertices)顶点tangent切线的transversal截线intercept截距7.有关坐标coordinate system坐标系rectangular coordinate直角坐标系origin原点abscissa横坐标ordinate纵坐标Number line数轴quadrant象限slope斜率complex plane复平面8.其它plane geometry平面几何trigonometry三角学bisect平分circumscribe外切inscribe内切intersect相交perpendicular垂直Pythagorean theorem勾股定理congruent全等的multilateral多边的其它相关词汇cent美分penny一美分硬币nickel5美分硬币dime一角硬币dozen打(12个)score廿(20个)Centigrade摄氏Fahrenheit华氏quart夸脱gallon加仑(1gallon=4quart)yard码meter米micron微米inch英寸foot英尺minute分(角度的度量单位，60分=1度) square measure平方单位制cubic meter立方米pint品脱(干量或液量的单位)。
09年SOA北美精算师考试第二门FM官方样题第一部分（主要是金融数学）答案
09年SOA北美精算师考试第二门FM官方样题第一部分（主要是金融数学）答案SAMPLE SOLUTIONS FOR DERIVATIVES MARKETSQuestion #1Answer is DIf the call is at-the-money, the put option with the same cost will have a higher strike price.A purchased collar requires that the put have a lower strike price. (Page 76)Question #2Answer is C66.59 – 18.64 = 500 – K exp(–0.06) for K = 480 (Page 69)Question #3Answer is DThe accumulated cost of the hedge is (84.30-74.80)exp(.06) = 10.09.Let x be the market price.If x < 0.12 the put is in the money and the payoff is 10,000(0.12 – x) = 1,200 – 10,000x. The sale of the jalapenos has a payoff of 10,000x –1,000 for a profit of 1,200 –10,000x + 10,000x – 1,000 – 10.09 = 190.From 0.12 to 0.14 neither option has a payoff and the profit is 10,000x – 1,000 – 10.09 = 10,000x – 1,010.If x >0.14 the call is in the money and the payoff is –10,000(x – 0.14) = 1,400 – 10,000x. The profit is 1,400 – 10,000x + 10,000x – 1,000 – 10.09 = 390.The range is 190 to 390. (Pages 33-41)Question #4Answer is BThe present value of the forward prices is 10,000(3.89)/1.06 + 15,000(4.11)/1.0652 +20,000(4.16)/1.073 = 158,968. Any sequence of payments with that present value is acceptable. All but B have that value. (Page 248)Question #5Answer is EIf the index exceeds 1,025, you will receive x – 1,025. After buying the index for x you will have spent 1,025. If the index is below 1,025, you will pay 1,025 – x and after buying the index for x you will have spent 1,025. One way to get the cost is to note that the forward price is 1,000(1.05) = 1,050. You want to pay 25 less and so must spend 25/1.05 = 23.81 today. (Page 112) Question #6Answer is EIn general, an investor should be compensated for time and risk. A forward contract has no investment, so the extra 5 represents the risk premium. Those who buy the stock expect to earn both the risk premium and the time value of their purchase and thus the expected stock value is greater than 100 + 5 = 105. (Page 140)Question #7Answer is CAll four of answers A-D are methods of acquiring the stock. The prepaid forward has the payment at time 0 and the delivery at time T. (Pages 128-129)Question #8Answer is BOnly straddles use at-the-money options and buying is correct for this speculation. (Page 78)Question #9Answer is DThis is based on Exercise 3.18 on Page 89. To see that D does not produce the desired outcome, begin with the case where the stock price is S and is below 90. The payoff is S + 0 + (110 – S) –2(100 – S) = 2S – 90 which is not constant and so cannot produce the given diagram. On the other hand, for example, answer E hasa payoff of S + (90 – S) + 0 – 2(0) = 90. The cost is 100 + 0.24 +2.17 – 2(6.80) = 88.81. With interest it is 93.36. The profit is 90 –93.36 = –3.36 which matches the diagram.Question #10Answer is D[rationale-a] True, since forward contracts have no initial premium[rationale-b] True, both payoffs and profits of long forwards are opposite to short forwards.[rationale-c] True, to invest in the stock, one must borrow 100 at t=0, and then pay back 110 = 100*(1+.1) at t=1, which is like buying a forward at t=1 for 110. [rationale-d] False, repeating the calculation shown above in part c), but with 10% as a continuously compounded rate, the stock investor must now pay back100*e.1 = 110.52 at t=1; this is more expensive than buying a forward at t=1for 110.00.[rationale-e] True, the calculation would be the same as shown above in part c), but now the stock investor gets an additional dividend of 3.00 at t=.5, which theforward investor does not receive (due to not owning the stock until t=1). [This is based on Exercise 2-7 on p.54-55 ofMcDonald][McDonald, Chapter 2, p.21-28]Question #11Answer is CSolution: The 35-strike call has future cost (at t=1) of 9.12*(1+.08) = 9.85The 40-strike call has future cost (at t=1) of 6.22*(1+.08) = 6.72The 45-strike call has future cost (at t=1) of 4.08*(1+.08) = 4.41If S1<35, the profits of the 3 calls, respectively, are -9.85, -6.72, and -4.41.If 35<s1<="" -6.72,="" 3="" and="" are="" bdsfid="114" calls,="" of="" p="" profits="" respectively,="" s1-44.85,="" the=""></s1If 40<s1<="" 3="" and="" are="" bdsfid="116" calls,="" of="" p="" profits="" respectively,="" s1-44.85,="" s1-46.72,="" the=""></s1If S1>45, the profits of the 3 calls, respectively, are S1-44.85, S1-46.72, and S1-49.41.The cutoff points for when the relative profit ranking of the 3 calls change are:S1-44.85=-6.72, S1-44.85=-4.41, and S1-46.72=-4.41, yielding cutoffs of 38.13, 40.44, and 42.31.If S1<38.13, the 45-strike call has the highest profit, and the 35-strike call the lowest.If 38.13<s1<="" p="" profit,="" the=""></s1If 40.44<s1<="" p="" profit,="" the=""></s1If S1<42.31, the 35-strike call has the highest profit, and the 45-strike call the lowest.We are looking for the case where the 35-strike call has the highest profit, and the 40-strike call has the lowest profit, which occurs when S1 is between 40.44 and 42.31.[This is based on Exercise 2-13 on p.55-56 of McDonald][McDonald, Chapter 2, p.33-37]Question #12Answer is BSolution: The put premium has future value (at t=.5) of 74.20 * (1+(.04/2)) = 75.68 Then, the 6-month profit on a long put position is: max(1,000-S.5,0)-75.68. Correspondingly, the 6-month profit on a short put position is 75.68-max(1,000-S.5,0). These two profits are opposites (naturally, since long and short positions have opposite payoff and profit). Thus, they can only be equal if producing 0 profit. 0 profit is only obtained if 75.68 = max(1,000-S.5,0), or 1,000-S.5 = 75.68, or S.5 = 924.32. [McDonald, Chapter 2, p.39-42]Question #13Answer is DSolution: Buying a call, in conjunction with a short position in a stock index, is a form of insurance called a cap. Answers (A) and (B) are incorrect because they relate to a floor, which is the purchase of a put to insure against a long position in a stock index. Answer (E) is incorrect because it relates to writing a covered call, which is the sale of a call along with a long position in the stock index, so that the investor is selling rather than buying insurance. Note that a cap can also be thought of as ‘buying’ a covered call. Now, let’s calculate the profit: 2-year profit = payoff at time 2 – the future value of the initial cost to establish the position = (-75 + max(75-60,0)) –(-50 + 10)*(1+.03)2 = -75+15+40*(1.03)2 = 42.44-60 = -17.56. Thus,we’ve lost more from holding the short position in the index (since the index went up) than we’ve gained from owning the long call option.[McDonald, Chapter 3, p.59-65]Question #14Answer is ASolution: This consists of standard applications of the put-call parity equation on p.69. Let C be the price for the 40-strike call option. Then, C + 3.35 is the price for the 35-strike call option. Similarly, let P be the price for the 40-strike put option. Then, P –x is the price for the 35-strike put option, where x is what we’re trying to find. Using put-call parity, we have:(C + 3.35) + 35*e-.02 - 40 = P – x (this is for the 35-strike options)C + 40*e-.02 – 40 = P (this is for the 40-strike options)Subtracting the first equation from the second, 5*e-.02 – 3.35 = x = 1.55.[McDonald, Chapter 3, p.68-69]Answer is CSolution: The initial cost to establish this position is 5*2.78 –3*6.13 = -4.49. Thus, you are receiving 4.49 up front. This grows to 4.49*e .08*.25 = 4.58 after 3 months. Then, the following payoff/profit table can be constructed at T=.25 years: S T : 5*max(S T – 40, 0) – 3*max(S T – 35, 0) + 4.58 = Profit S T <35 0 - 0 + 4.58 = 4.58 35 <= S T <= 40 0 - 3*(S T – 35) + 4.58 = 109.58-3S T S T > 40 5*(S T -40) - 3*(S T – 35) + 4.58 = 2S T -90.42Thus, the maximum profit is unlimited (as S T increases beyond 40, so does the profit) Also, the maximum loss is 10.42(occurs at S T = 40, where profit = 109.58-120 = -10.42) [Notes] The above problem is an example of a ratio spread.[McDonald, Chapter 3, p.73]Question #16Answer is DSolution: The ‘straddle’ consists of buying a 40-strike call and buying a 40-strike put. This costs 2.78 + 1.99 = 4.77 at t=0, and grows to 4.77*e .02 = 4.87 at t=.25. The ‘strangle’ consists of buying a 35-strike put and a 45-strike call. This costs 0.44 + 0.97 = 1.41 at t=0, and grows to 1.41*e .02 = 1.44 at t=.25. For S T <40, the ‘straddle’ has a profit of 40-S T -4.87 = 35.13, and for S T >=40, the ‘straddle’ has a profit of S T -40-4.87 = 44.87. For S T <35, the ‘strangle’ has a profit of 35-S T -1.44 = 33.56, and for S T >45, the ‘strangle’ has a profit of S T -45-1.44 = 46.44. However, for 35<=S T <=45, the ‘strangle’ has a profit of -1.44 (since both options would not be exercised). Comparing the payoff structures between the ‘straddle’ and ‘strangle,’ we see that if S T <35 or if S T >45, the ‘straddle’ would outperform the ‘strangle’ (since 35.13 > 33.56,and since -44.87 > -46.44). However, if 35<=S T <=45, we can solve for the two cutoff points for S T , where the ‘strangle’ would outperform the ‘straddle’ as follows:-1.44 > 35.13 – S T, and -1.44 > S T - 44.87. The first inequality gives S T > 36.57, and the second inequality gives S T < 43.43. Thus, 36.57 < S T < 43.43.[McDonald, Chapter 3, p.78-80]Answer is B[rationale I] Yes, since Strategy I is a bear spread using calls, and bear spreads perform better when the prices of the underlying asset goes down.[rationale II] Yes, since Strategy II is also a bear spread – it just uses puts instead! [rationale III] No, since Strategy III is a box spread, which has no price risk; thus, the payoff is the same (1,000-950 = 50), no matter what the price of theunderlying asset.[Note]: An alternative, but much longer, solution is to develop payoff tables for all 3 strategies.[McDonald, Chapter 3, p.70-73]Question #18Answer is BSolution: First, let’s calculate the expected one-year profit without using the forward. This would be .2*(700+150-750) + .5(700+170-850) + .3*(700+190-950) = 20 + 10 - 18 = 12. Next, let’s calculate the expected one-year profit when buying the 1-year forward for 850. This would be 1*(700+170-850) = 20. Thus, the expected profit increases by 20 - 12 = 8 as a result of using the forward.[This is based on Exercise 4-7 on p.122 of McDonald][McDonald, Chapter 4, p.98-100]Question #19Answer is DSolution: There are 3 cases, one for each row in the above probability table.For all 3 cases, the future value of the put premium (at t=1) = 100*e.06 = 106.18.In Case 1, the 1-year profit would be: 750 - 800 - 106.18 + max(900-750,0) = -6.18In Case 2, the 1-year profit would be: 850 - 800 - 106.18 + max(900-850,0) = -6.18In Case 3, the 1-year profit would be: 950 - 800 - 106.18 +max(900-950,0) = 43.82 Thus, the expected 1-year profit = .7 * -6.18 + .3 * 43.82 = -4.326 + 13.146 = 8.82.[This is based on Exercise 4-3 on p.121 of McDonald][McDonald, Chapter 4, p.92-96]Answer is BSolution: This is an example of pricing a forward contract using discrete dividends. Thus, we need the future value of the current stock price minus the future value of each of the 12 dividends, where the valuation date is T=3. Thus, the valuation equation is: Forward price = 200*e.04(3) –[1.50*e.04(2.75) + 1.50*1.01*e.04(2.5) + 1.50*(1.01)2*e.04(2.25) + …1.50*(1.01)12] = 200*e.12 - 1.50*e.11{[1-(1.01*e-.01)12]/[1-(1.01*e.01)]}, using the geometric series formula from interest theory. This simplifies numerically to 225.50 -1.67442*11.99666 = 205.41.[This problem combines the material from interest theory and derivatives, although one could also simplify the above expression by brute force (instead of geometric series), since there are only 12 dividends to accumulate forward to T=3.] [McDonald, Chapter 5, p.133-134]Question #21Answer is ESolution: Here, the fair value of the forward contract is given by S0 * e(r-d)T =110 * e(.05-.02).5 = 110 * e.015 = 111.66. This is 0.34 less than the observed price. Thus, one could exploit this arbitrage opportunity by selling the observed forward at 112 and buying a synthetic forward at 111.66, making 112-111.66 = 0.34 profit.[This is based on Exercise 5-8 on p.163-164 of McDonald][McDonald, Chapter 5, p.135-138]Answer is BSolution: First, we must determine the present value of the forward contracts. On a per ton basis, this is: 1,600/1.05 + 1,700/(1.055)2 + 1,800/(1.06)3 = 4,562.49.Then, we must solve for the level swap price, which is labeled x below, as follows:4,562.49 = x/1.05 + x/(1.055)2 + x/(1.06)3 = x*[1/1.05 + 1/(1.055)2 + 1/(1.06)3] =2.69045*x.Thus, x = 4,562.49 / 2.69045 = 1,695.81.Thus, the amount he would receive each year is 50*1,695.81 = 84,790.38. [McDonald, Chapter 8, p.247-248]Question #23Answer is ESolution: First, note that the notional amount and the future 1-year LIBOR rates (not given) do not factor into the calculation of the swap’s fixed rate. All we need at the various zero-coupon bond prices P(0, t) for t=1,2,3,4,5, along with the 1-year implied forward rates, which are given by r0(t-1,t), for t=1,2,3,4,5. These calculations are shown in the following table:t 1 2 3 4 5P(0,t) (1.04)-1(1.045)-2 (1.0525)-3 (1.0625)-4 (1.075)-5=.96154 =.91573 =.85770 =.78466 =.69656 r0(t-1,t) s1[1.0452/1.04]-1 [1.05253/1.0452]-1 [1.06254/1.05253]-1 [1.0755/1.06254]-1 =.04000 =.05002 =.06766 =.09307 =.12649 Thus, the fixed swap rate = R = [(.96154)*(.04)+…+(.69656)*(.12649)] / [.96154 +…+.69656]= [.03846 + .04580 + .05803 + .07303 + .08811]/[.96154 + .91573 + .85770 + .78466 +.69656]= .30344 / 4.21619 = .07197 = 7.20% (approximately).[Note: This is much less calculation-intensive if you realize that the numerator (.30344) for R can also be obtained by taking 1- P(0,n) = 1 – P(0,5) = 1 - .69656 = .30344. Then, you would not need to calculate any of the implied forward rates!][McDonald, Chapter 8, p.255-258]Answer is D[rationale-a] True, hedging reduces the risk of loss, which is a primary function of derivatives.[rationale-b] True, derivatives can be used the hedge some risks that could result in bankruptcy.[rationale-c] True, derivatives can provide a lower-cost way to effect a financialtransaction.[rationale-d] False, derivatives are often used to avoid these types of restrictions. [rationale-e] True, an insurance contract can be thought of as a hedge against the risk of loss.[McDonald, Chapter 1, p.2-3]Question #25Answer is C[rationale-a] True, both types of individuals are involved in the risk-sharing process. [rationale-b] True, this is the primary reason reinsurance companies exist.[rationale-c] False, reinsurance companies share risk by issuing rather than investing in catastrophic bonds. In effect, they are ceding this excess risk to thebondholder.[rationale-d] True, it is diversifiable risk which is reduced or eliminated when risks are shared.[rationale-e] True, this is a fundamental idea underlying risk management andderivatives.[McDonald, Chapter 1, p.5-6]Question #26Answer is B[rationale-I] True, the forward seller has unlimited exposure if the underlying asset’s price increases.[rationale-II] True, the call issuer has unlimited exposure if the underlying asset’s price rises.[rationale-III] False, the maximum loss on selling a put is FV(put premium) – strike price. [McDonald, Chapter 2, p.43 (Table 2.4)]Answer is A[rationale-I] True, as prices go down, the value of holding a put option increases.canbe thought of as a put option.insuranceHomeowner’s[rationale-II] False, returns from equity-linked CDs are zero if prices decline, but positive if prices rise. Thus, owners of these CDs benefit from rising prices. [rationale-III] False, a synthetic forward consists of a long call and a short put, both of which benefit from rising prices (so the net position also benefits as such). [McDonald, Chapter 2, p.44-48]Question #28Answer is E[rationale-a] True, derivatives are used to shift income, thereby potentially lowering taxes.[rationale-b] True, as with taxes, the transfer of income lowers the probability ofbankruptcy.[rationale-c] True, hedging can safeguard reserves, and reduce the need for external financing, which has both explicit (e.g. – fees) and implicit (e.g. –reputational) costs.[rationale-d] True, when operating internationally, hedging can reduce exchange rate risk. [rationale-e] False, a firm that credibly hedges will reduce the riskiness of its cash flows, and will be able to increase debt capacity, which will lead to tax savings, since interest is deductible.[McDonald, Chapter 4, p.103-106]Question #29Answer is ASolution: If S0 is the price of the stock at time-0, then the following payments are required: Outright purchase – payment at time 0 – amount of payment = S0.Fully leveraged purchase – payment at time T – amount of payment = S0*e rT.Prepaid forward contract – payment at time 0 – amount of payment = S0*e-dT.Forward contract – payment at time T – amount of payment = S0*e(r-d)T.Since r>d>0, it must be true that S0*e-dT < S0 < S0*e(r-d)T < S0*e rT.Thus, the correct ranking is given by choice (A).[McDonald, Chapter 5, p.127-134]Answer is C[rationale-a] True, marking to market is done for futures, andcan lead to pricedifferences relative to forward contracts.[rationale-b] True, futures are more liquid; in fact, if you use the same broker to buy and sell, your position is effectively cancelled.[rationale-c] False, forwards are more customized, and futures are more standardized. [rationale-d] True, because of the daily settlement, credit risk is less with futures (v.forwards).[rationale-e] True, futures markets, like stock exchanges, do have daily price limits. [McDonald, Chapter 5, p.142]。
北美精算师考试官方样题2014-10-exam-fm-ques
EXAM FM SAMPLE QUESTIONS Financial Economics
June 2014 changes Questions 1-30 are from the prior version of this document. They have been edited to conform more closely to current question writing style, but are unchanged in content. Question 31 is the former Question 58 from the interest theory question set. Questions 32-34 are new. January 2015 changes Questions 35-46 are new. May 2015 changes Question 32 was modified (and re-modified in June) Questions 47-62 are new. Some of the questions in this study note are taken from past examinations. These questions are representative of the types of questions that might be asked of candidates sitting for the Financial Mathematics (FM) Exam. These questions are intended to represent the depth of understanding required of candidates. The distribution of questions by topic is not intended to represent the distribution of questions on future exams.
北美精算师(SOA)考试P 2000 November年真题
(A) (B) (C) (D) (E)
208 260 270 312 374
November 2000
1
Course 1
2.
An investor purchases two assets, each having an initial value of 1000 . The value Vn of the first asset after n years can be modeled by the relationship:
0.06 0.16 0.19 0.22 0.25
Course 1
12
Form 00B
13.
An actuary believes that the demand for life insurance, L, and the demand for health insurance, H, can be modeled as functions of time, t:
What is the approximate probability that there is a total of between 2450 and 2600 claims during a one-year period?
(A) (B) (C) (D) (E)
0.68 0.82 0.87 0.95 1.00
Course 1
2
Form 00B
3.
An auto insurance company has 10,000 policyholders. Each policyholder is classified as
(i) (ii) (iii)
young or old; male or female; and married or single.
2003年5月北美精算第四门考试试题
Course 4Fall 2003 Society of Actuaries**BEGINNING OF EXAMINATION** 1. You are given the following information about a stationary AR(2) model:=.(i) ρ105(ii) ρ201=.Determine φ2.(A) –0.2(B) 0.1(C) 0.4(D) 0.7(E) 1.0(i) Losses follow a loglogistic distribution with cumulative distribution function:F x x x b g b g b g =+//θθγγ1(ii)The sample of losses is:10 35 80 86 90 120 158 180 200 210 1500Calculate the estimate of θ by percentile matching, using the 40th and 80th empirically smoothed percentile estimates.(A) Less than 77(B) At least 77, but less than 87(C) At least 87, but less than 97(D) At least 97, but less than 107(E) At least 107(i) The number of claims has a Poisson distribution.(ii) Claim sizes have a Pareto distribution with parameters θ=0.5 and α=6.(iii) The number of claims and claim sizes are independent.(iv) The observed pure premium should be within 2% of the expected pure premium 90% of the time.Determine the expected number of claims needed for full credibility.(A) Less than 7,000(B) At least 7,000, but less than 10,000(C) At least 10,000, but less than 13,000(D) At least 13,000, but less than 16,000(E) At least 16,0004. You study five lives to estimate the time from the onset of a disease to death. The times todeath are:2 3 3 3 7Using a triangular kernel with bandwidth 2, estimate the density function at 2.5.(A) 8/40(B) 12/40(C) 14/40(D) 16/40(E) 17/405. For the model i i i Y X αβε=++, where 1,2,...,10i =, you are given:(i) X i i =R S T1, if the th individual belongs to a specified group 0, otherwise(ii) 40 percent of the individuals belong to the specified group.(iii) The least squares estimate of β is β=4.(iv) ()2ˆˆ92i i Y X αβ−−=∑Calculate the t statistic for testing H 00:β=.(A) 0.9(B) 1.2(C) 1.5(D) 1.8(E) 2.1(i) Losses follow a Single-parameter Pareto distribution with density function:()()1,1f x x xαα+=>, 0 < α < ∞ (ii) A random sample of size five produced three losses with values 3, 6 and 14, and twolosses exceeding 25.Determine the maximum likelihood estimate of α.(A) 0.25(B) 0.30(C) 0.34(D) 0.38(E) 0.42(i) The annual number of claims for a policyholder has a binomial distribution withprobability function:()()221x x p x q q q x −⎛⎞=−⎜⎟⎝⎠, x = 0, 1, 2(ii) The prior distribution is:()34,01q q q π=<<This policyholder had one claim in each of Years 1 and 2.Determine the Bayesian estimate of the number of claims in Year 3.(A) Less than 1.1(B) At least 1.1, but less than 1.3(C) At least 1.3, but less than 1.5(D) At least 1.5, but less than 1.7(E) At least 1.78. For a sample of dental claims 1210,,...,x x x , you are given:(i) 23860 and 4,574,802i i x x ==∑∑(ii) Claims are assumed to follow a lognormal distribution with parameters µ and σ.(iii)µ and σ are estimated using the method of moments.Calculate ∧ for the fitted distribution.(A) Less than 125(B) At least 125, but less than 175(C) At least 175, but less than 225(D) At least 225, but less than 275(E) At least 2759. You are given:(i)Y tij is the loss for the j th insured in the i th group in Year t . (ii)ti Y is the mean loss in the i th group in Year t . (iii)X j i j i ij =R S T0, if the th insured is in the first group (=1)1, if the th insured is in the second group (=2) (iv)21ij ij ij ij Y Y X δφθε=+++, where 1,2i = and 1,2,...,j n = (v)Y Y Y Y 2122111230374041====,,, (vi) ˆ0.75φ=Determine the least-squares estimate of θ.(A) 5.25(B) 5.50(C) 5.75(D) 6.00(E) 6.2510. Two independent samples are combined yielding the following ranks:Sample I: 1, 2, 3, 4, 7, 9, 13, 19, 20Sample II: 5, 6, 8, 10, 11, 12, 14, 15, 16, 17, 18You test the null hypothesis that the two samples are from the same continuous distribution.The variance of the rank sum statistic is:()112n m n m ++Using the classical approximation for the two-tailed rank sum test, determine the p -value.(A) 0.015(B) 0.021(C) 0.105(D) 0.210(E) 0.420(i) Claim counts follow a Poisson distribution with mean θ. (ii) Claim sizes follow an exponential distribution with mean 10θ. (iii) Claim counts and claim sizes are independent, given θ. (iv) The prior distribution has probability density function:b g=5, θ>1πθθCalculate Bühlmann’s k for aggregate losses.(A) Less than 1(B) At least 1, but less than 2(C) At least 2, but less than 3(D) At least 3, but less than 4(E) At least 4(i) A survival study uses a Cox proportional hazards model with covariates Z 1 and Z 2,each taking the value 0 or 1.(ii) The maximum partial likelihood estimate of the coefficient vector is:, .,.ββ12071020e j b g=(iii) The baseline survival function at time t 0 is estimated as .S t 0065b g =.Estimate S t 0b gfor a subject with covariate values 121Z Z ==.(A) 0.34(B) 0.49(C) 0.65(D) 0.74(E) 0.84(i) Z 1 and Z 2 are independent N(0,1) random variables.(ii) a , b , c , d , e , f are constants.(iii) Y a bZ cZ X d eZ f Z =++=++1212 andDetermine ()E Y X .(A) a(B) ()()a b c X d ++−(C) a be cf X d ++−b gb g(D) a be cf e f +++g d /22(E) a be cf e f X d +++−g d g/22(i) Losses on a company’s insurance policies follow a Pareto distribution with probabilitydensity function:()(),0f x x x θθθ=<<∞+(ii) For half of the company’s policies θ=1, while for the other half θ=3.For a randomly selected policy, losses in Year 1 were 5.Determine the posterior probability that losses for this policy in Year 2 will exceed 8.(A) 0.11(B) 0.15(C) 0.19(D) 0.21(E) 0.2715. You are given total claims for two policyholders:Year1 2 3 4PolicyholderX 730 800 650 700Y 655 650 625 750Using the nonparametric empirical Bayes method, determine the Bühlmann credibilitypremium for Policyholder Y.(A) 655(B) 670(C) 687(D) 703(E) 71916. A particular line of business has three types of claims. The historical probability and thenumber of claims for each type in the current year are:Type HistoricalProbabilityNumber of Claimsin Current YearA 0.2744 112B 0.3512 180C 0.3744 138You test the null hypothesis that the probability of each type of claim in the current year is the same as the historical probability.Calculate the chi-square goodness-of-fit test statistic.(A) Less than 9(B) At least 9, but less than 10(C) At least 10, but less than 11(D) At least 11, but less than 12(E) At least 1217. Which of the following is false?(A) If the characteristics of a stochastic process change over time, then the process isnonstationary.(B) Representing a nonstationary time series by a simple algebraic model is often difficult.(C) Differences of a homogeneous nonstationary time series will always be nonstationary.(D) If a time series is stationary, then its mean, variance and, for any lag k, covariancemust also be stationary.(E) If the autocorrelation function for a time series is zero (or close to zero) for all lagsk>0, then no model can provide useful minimum mean-square-error forecasts offuture values other than the mean.18. The information associated with the maximum likelihood estimator of a parameter θ is 4n,where n is the number of observations.Calculate the asymptotic variance of the maximum likelihood estimator of 2θ.(A) 12n(B) 1n(C) 4n(D) 8n(E) 16n19. You are given:(i) The probability that an insured will have at least one loss during any year is p.(ii) The prior distribution for p is uniform on []0,0.5.(iii) An insured is observed for 8 years and has at least one loss every year.Determine the posterior probability that the insured will have at least one loss during Year 9.(A) 0.450(B) 0.475(C) 0.500(D) 0.550(E) 0.62520. At the beginning of each of the past 5 years, an actuary has forecast the annual claims for agroup of insureds. The table below shows the forecasts (X) and the actual claims (Y). Atwo-variable linear regression model is used to analyze the data.t X t Y t1 475 2542 254 4633 463 5154 515 5675 567 605You are given:(i) The null hypothesis is0:0,1Hαβ==.(ii) The unrestricted model fit yields ESS = 69,843.Which of the following is true regarding the F test of the null hypothesis?(A) The null hypothesis is not rejected at the 0.05 significance level.(B) The null hypothesis is rejected at the 0.05 significance level, but not at the 0.01 level.(C) The numerator has 3 degrees of freedom.(D) The denominator has 2 degrees of freedom.(E) TheF statistic cannot be determined from the information given.21-22. Use the following information for questions 21 and 22.For a survival study with censored and truncated data, you are given:Time (t) Number at Riskat Time t Failures at Time t1 30 52 27 93 32 64 25 55 20 4 21. The probability of failing at or before Time 4, given survival past Time 1, is31q.Calculate Greenwood’s approximation of the variance of 31 q.(A) 0.0067(B) 0.0073(C) 0.0080(D) 0.0091(E) 0.010521-22. (Repeated for convenience) Use the following information for questions 21 and 22.For a survival study with censored and truncated data, you are given:Time (t) Number at Riskat Time t Failures at Time t1 30 52 27 93 32 64 25 55 20 4 22. Calculate the 95% log-transformed confidence interval for H3b g, based on the Nelson-Aalenestimate.(A) (0.30,0.89)(B) (0.31,1.54)(C) (0.39,0.99)(D) (0.44,1.07)(E) (0.56,0.79)(i) Two risks have the following severity distributions:Amount of Claim Probability of ClaimAmount for Risk 1Probability of ClaimAmount for Risk 2250 0.5 0.72,500 0.3 0.260,000 0.2 0.1(ii) Risk 1 is twice as likely to be observed as Risk 2.A claim of 250 is observed.Determine the Bühlmann credibility estimate of the second claim amount from the same risk.(A) Less than 10,200(B) At least 10,200, but less than 10,400(C) At least 10,400, but less than 10,600(D) At least 10,600, but less than 10,800(E) At least 10,800(i) A sample x x x 1210,,,… is drawn from a distribution with probability density function:1211exp()exp(), 0[]x x x θθσσ−+−<<∞(ii)θσ>(iii) x x i i ==∑∑15050002 andEstimate θ by matching the first two sample moments to the corresponding population quantities.(A) 9(B) 10(C) 15(D) 20(E) 2125. You are given the following time-series model:115.028.0−−−++=t t t t y y εεWhich of the following statements about this model is false?(A) 10.4ρ=(B) 1,2,3,4,....k k ρρ<=(C) The model is ARMA(1,1).(D) The model is stationary.(E) The mean, µ, is 2.26. You are given a sample of two values, 5 and 9.You estimate Var(X ) using the estimator g (X 1, X 2) = 21().2i X X −∑Determine the bootstrap approximation to the mean square error of g .(A) 1(B) 2(C) 4(D) 8(E) 1627. You are given:(i) The number of claims incurred in a month by any insured has a Poisson distributionwith mean λ.(ii) The claim frequencies of different insureds are independent.(iii) The prior distribution is gamma with probability density function:()()6100100120efλλλλ−=(iv) Month Number of Insureds NumberofClaims1 100 62 150 83 200 114 300 ?Determine the Bühlmann-Straub credibility estimate of the number of claims in Month 4.(A) 16.7(B) 16.9(C) 17.3(D) 17.6(E) 18.028. You fit a Pareto distribution to a sample of 200 claim amounts and use the likelihood ratio testto test the hypothesis that 1.5α= and 7.8θ=.You are given:(i) The maximum likelihood estimates are α= 1.4 and θ = 7.6.(ii) The natural logarithm of the likelihood function evaluated at the maximum likelihoodestimates is −817.92.(iii) ()ln 7.8607.64i x +=∑Determine the result of the test.(A) Reject at the 0.005 significance level.(B) Reject at the 0.010 significance level, but not at the 0.005 level.(C) Reject at the 0.025 significance level, but not at the 0.010 level.(D) Reject at the 0.050 significance level, but not at the 0.025 level.(E) Do not reject at the 0.050 significance level.29. You are given:(i) The model is Y X i i i =+βε, i = 1, 2, 3.(ii)i X i Var εi b g11 12 2 93 316 (iii)The ordinary least squares residuals are εβi i i Y X =−, i = 1, 2, 3.Determine E X X X ,,ε12123d i.(A) 1.0(B) 1.8(C) 2.7(D) 3.7(E) 7.630. For a sample of 15 losses, you are given:(i)Interval Observed Number ofLosses(0, 2] 5(2, 5] 5(5, ∞) 5 (ii) Losses follow the uniform distribution on 0,θb g.Estimate θ by minimizing the function()231j jjjE OO=−∑, where j E is the expected number oflosses in the j th interval andjO is the observed number of losses in the j th interval.(A) 6.0(B) 6.4(C) 6.8(D) 7.2(E) 7.631. You are given:(i) The probability that an insured will have exactly one claim is θ.(ii) The prior distribution of θ has probability density function:πθθθb g=<<3201,A randomly chosen insured is observed to have exactly one claim.Determine the posterior probability that θ is greater than 0.60.(A) 0.54(B) 0.58(C) 0.63(D) 0.67(E) 0.7232. The distribution of accidents for 84 randomly selected policies is as follows:Number of Accidents Number of Policies0 321 262 123 74 45 26 1Total 84 Which of the following models best represents these data?binomial(A) Negativeuniform(B) Discrete(C) Poisson(D) Binomial(E) Either Poisson or Binomial33. A time series yt follows an ARIMA(1,1,1) model with φ107=., θ103=−. and σε210=..Determine the variance of the forecast error two steps ahead.(A) 1(B)5(C) 8(D)10(E) 12(i) Low-hazard risks have an exponential claim size distribution with mean θ. (ii) Medium-hazard risks have an exponential claim size distribution with mean 2θ. (iii) High-hazard risks have an exponential claim size distribution with mean 3θ. (iv) No claims from low-hazard risks are observed.(v) Three claims from medium-hazard risks are observed, of sizes 1, 2 and 3. (vi) One claim from a high-hazard risk is observed, of size 15.Determine the maximum likelihood estimate of θ.(A) 1(B) 2(C) 3(D) 4(E) 5(i)partial X =pure premium calculated from partially credible data(ii)partial E X µ⎡⎤=⎣⎦ (iii) Fluctuations are limited to ±k µ of the mean with probability P(iv) Z = credibility factorWhich of the following is equal to P ?(A) partial Pr k X k µµµµ⎡⎤−≤≤+⎣⎦(B) partial Pr +Z k Z X Z k µµ⎡⎤−≤≤⎣⎦(C) partial Pr +Z Z X Z µµµµ⎡⎤−≤≤⎣⎦(D) ()partial Pr 111k Z X Z k µ⎡⎤−≤+−≤+⎣⎦(E) ()partial Pr 1k Z X Z k µµµµµ⎡⎤−≤+−≤+⎣⎦36. For the model 1223344i i i i i Y X X X ββββε=++++, you are given:(i) N = 15(ii)(iii) ESS =28282.Calculate the standard error of 32ˆˆββ−.(A) 6.4(B) 6.8(C) 7.1(D) 7.5(E) 7.837. You are given:Assume a uniform distribution of claim sizes within each interval.Estimate E X X 2150c h g −∧.(A)Less than 200(B)At least 200, but less than 300(C)At least 300, but less than 400(D)At least 400, but less than 500(E)At least 50038. Which of the following statements about moving average models is false?(A) Both unweighted and exponentially weighted moving average (EWMA) models canbe used to forecast future values of a time series.(B) Forecasts using unweighted moving average models are determined by applying equalweights to a specified number of past observations of the time series.(C) Forecasts using EWMA models may not be true averages because the weights appliedto the past observations do not necessarily sum to one.(D) Forecasts using both unweighted and EWMA models are adaptive because theyautomatically adjust themselves to the most recently available data.(E) Using an EWMA model, the two-period forecast is the same as the one-periodforecast.39. You are given:(i) Each risk has at most one claim each year.(ii)Type of Risk Prior Probability Annual Claim ProbabilityI 0.7 0.1II 0.2 0.2III 0.1 0.4 One randomly chosen risk has three claims during Years 1-6.Determine the posterior probability of a claim for this risk in Year 7.(A) 0.22(B) 0.28(C) 0.33(D) 0.40(E) 0.4640. You are given the following about 100 insurance policies in a study of time to policysurrender:(i) The study was designed in such a way that for every policy that was surrendered, ar, is always equal to 100.new policy was added, meaning that the risk set,j(ii) Policies are surrendered only at the end of a policy year.(iii) The number of policies surrendered at the end of each policy year was observed to be:1 at the end of the 1st policy year2 at the end of the 2nd policy year3 at the end of the 3rd policy yearn at the end of the n th policy year(iv) The Nelson-Aalen empirical estimate of the cumulative distribution function at time n, F, is 0.542.)(ˆnWhat is the value of n?(A) 8(B) 9(C) 10(D) 11(E) 12**END OF EXAMINATION**Course 4, Fall 2003PRELIMINARY ANSWER KEYQuestion # Answer Question # Answer 1 A21 A 2 E22 D 3 E23 D 4 B24 D 5 D25 E 6 A26 D 7 C27 B 8 D28 C 9 E29 B 10 D30 E 11 C31 E 12 A32 A 13 E33 B 14 D34 B 15 C35 E 16 B36 C 17 C37 C 18 B38 C 19 A39 B 20 A40 E。
精算师考试练习题
精算师考试练习题精算师考试练习题SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETYEXAM FM FINANCIAL MATHEMATICSEXAM FM SAMPLE QUESTIONSCopyright 2005 by the Society of Actuaries and the CasualtyActuarial Society Some of the questions in this study note are taken from past SOA/CAS examinations. FM-09-05 PRINTED IN U.S.A. 11/08/04 2 These questions are representative of the types of questions that might be asked of candidates sitting for the new examination on Financial Mathematics (2/FM). These questions are intended to represent the depth of understanding required of candidates. The distribution of questions by topic is not intended to represent the distribution of questions on future exams.11/08/04 31.Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate of interest of 4% convertible semiannually.At the same time, Peter deposits 100 into a separate account. Peter’s account is credited interest at a force of interest of . After 7.25 years, the value of each account is the same.Calculate .(A) 0.0388(B) 0.0392(C) 0.0396(D) 0.0404(E) 0.041411/08/04 42.Kathryn deposits 100 into an account at the beginning of each 4-year period for 40 years. The account credits interest at an annual effective interest rate of i. The accumulated amount in the account at the end of 40 years is X, which is 5 times the accumulated amount in the account at the end of 20 years.Calculate X.(A) 4695(B) 5070(C) 5445(D) 5820(E) 619511/08/04 53.Eric deposits 100 into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually.(B) 9.26%(C) 9.46%(D) 9.66%(E) 9.86%11/08/04 64.John borrows 10,000 for 10 years at an annual effective interestrate of 10%. He can repay this loan using the amortization method with payments of 1,627.45 at the end of each year. Instead, John repays the 10,000 using a sinking fund that pays an annual effective interest rate of 14%. The deposits to the sinking fund are equal to 1,627.45 minus the interest on the loan and are made at the end of each year for 10 years. Determine the balance in the sinking fund immediately after repayment of the loan.(A) 2,130(B) 2,180(C) 2,230(D) 2,300(E) 2,37011/08/04 75.An association had a fund balance of 75 on January 1 and 60 on December 31. At the end of every month during the year, the associationdeposited 10 from membership fees. There were withdrawals of 5 on February 28, 25 on June 30, 80 on October 15, and 35 on October 31.Calculate the dollar-weighted (money-weighted) rate of return for the year.(A) 9.0%(B) 9.5%(C) 10.0%(D) 10.5%(E) 11.0%11/08/04 86.A perpetuity costs 77.1 and makes annual payments at the end of the year. The perpetuvtw paws 1 at the end of wear 2, 2 at the end of wear 3, …., n at the end of year (n+1). After year (n+1), the payments remain constant at n. The annual effective interest rate is 10.5%.Calculate n.(A) 17(B) 187.1000 is deposited into Fund X, which earns an annual effective rate of 6%. At the end of each year, the interest earned plus an additional 100 is withdrawn from the fund. At the end of the tenth year, the fund is depleted.The annual withdrawals of interest and principal are deposited into Fund Y, which earns an annual effective rate of 9%. Determine the accumulated value of Fund Y at the end of year 10.(A) 1519(B) 1819(C) 2085(D) 2273(E) 243111/08/04 108.You are given the following table of interest rates:Calendar Year Portfolioof Original RatesInvestment Year Rates (in %) Investment (in %)i1y i2y i3y i4y i5y y iy+51992 8.25 8.25 8.4 8.5 8.5 8.351993 8.5 8.7 8.75 8.9 9.0 8.61994 9.0 9.0 9.1 9.1 9.2 8.851995 9.0 9.1 9.2 9.3 9.4 9.11996 9.25 9.35 9.5 9.55 9.6 9.359.7 1997 9.5 9.5 9.6 9.71998 10.0 10.0 9.9 9.81999 10.0 9.8 9.72000 9.5 9.5P: under the investment year methodQ: under the portfolio yield methodR: where the balance is withdrawn at the end of everyyear and is reinvested at the new money rate Determine the ranking of P, Q, and R.P Q R (A)(B) P R Q(C) Q P R(D) R P Q(E) R Q P11/08/04 119.A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount of interest due. Each of the last ten payments is X.The lender charges interest at an annual effective rate of 10%.Calculate X.(A) 32(B) 57(C) 70(D) 97(E) 11711/08/04 1210.A 10,000 par value 10-year bond with 8% annual coupons is bought at a premium to yield an annual effective rate of 6%. Calculate the interest portion of the 7th coupon.(A) 632(B) 642(C) 651(D) 660(E) 66711/08/04 1311.A perpetuity-immediate pays 100 per year. Immediately after thefifth payment, the perpetuity is exchanged for a 25-year annuity-immediate that will pay X at the end of the first year. Each subsequent annual payment will be 8% greater than the preceding payment. The annual effective rate of interest is 8%. Calculate X.(C) 74(D) 84(E) 9411/08/04 1412.Jeff deposits 10 into a fund today and 20 fifteen years later. Interest is credited at a nominal discount rate of d compounded quarterly for the first 10 years, and at a nominal interest rate of 6% compounded semiannually thereafter. The accumulated balance in the fund at the end of 30 years is 100.Calculate d.(A) 4.33%(B) 4.43%(C) 4.53%(D) 4.63%(E) 4.73%11/08/04 1513.Ernie makes deposits of 100 at time 0, and X at time 3. The fund grows at a force of interest2 t t , t > 0.100The amount of interest earned from time 3 to time 6 is also X. Calculate X.(A) 385(B) 485(C) 585(D) 685(E) 78511/08/04 1614.Mike buys a perpetuity-immediate with varying annual payments.During the first 5 years, the payment is constant and equal to 10. Beginning in year 6, the payments start to increase. For year 6 and all future years, the current year’s payment is K% larger than the previous year’s payment.At an annual effective interest rate of 9.2%, the perpetuity has a present value of 167.50.Calculate K, given K < 9.2.(A) 4.0(B) 4.2(C) 4.411/08/04 1715.A 10-year loan of 2000 is to be repaid with payments at the end of each year. It can be repaid under the following two options: (i) Equal annual payments at an annual effective rate of 8.07%.Installments of 200 each year plus interest on the unpaid balance at an annual effective (ii)rate of i.The sum of the payments under option (i) equals the sum of the payments under option (ii). Determine i.(A) 8.75%(B) 9.00%(C) 9.25%(D) 9.50%(E) 9.75%11/08/04 1816.A loan is amortized over five years with monthly payments at a nominal interest rate of 9% compounded monthly. The first payment is 1000 and is to be paid one month from the date of the loan. Eachsucceeding monthly payment will be 2% lower than the prior payment. Calculate the outstanding loan balance immediately after the 40th payment is made.(A) 6751(B) 6889(C) 6941(D) 7030(E) 734411/08/04 1917.To accumulate 8000 at the end of 3n years, deposits of 98 are made at the end of each of the first n years and 196 at the end of each of the next 2n years.The annual effective rate of interest is i. You are given (l + i)n =2.0. Determine i.(A) 11.25%(B) 11.75%(C) 12.25%(D) 12.75%(E) 13.25%11/08/04 2018.month thereafter the payment increases by 2.The nominal interest rate is 9% convertible quarterly.Calculate X.(A) 2680(B) 2730(C) 2780(D) 2830(E) 288011/08/04 2119.You are given the following information about the activity in two different investment accounts:Account KFund value ActivityDate Deposit Withdrawal before activityJanuary 1, 1999 100.0125.0 July 1, 1999 X2X October 1, 1999 110.0December 31, 1999 125.0Account LActivity Fund valueDate Deposit Withdrawal before activityJanuary 1, 1999 100.0X July 1, 1999 125.0December 31, 1999 105.8During 1999, the dollar-weighted (money-weighted) return for investment account K equals the time-weighted return for investment account L, which equals i. Calculate i.(A) 10%(D) 18%(E) 20%11/08/04 2220.David can receive one of the following two payment streams:(i) 100 at time 0, 200 at time n, and 300 at time 2n600 at time 10 (ii)At an annual effective interest rate of i, the present values of the two streams are equal. Given vn = 0.76, determine i.(A) 3.5%(B) 4.0%(C) 4.5%(D) 5.0%(E) 5.5%11/08/04 2321.Payments are made to an account at a continuous rate of (8k + tk), where0 t 10 .1 Interest is credited at a force of interest δt = . 8 tAfter 10 years, the account is worth 20,000.Calculate k.(A) 111(B) 116(C) 121(D) 126(E) 13111/08/04 2422.You have decided to invest in Bond X, an n-year bond with semi-annual coupons and the following characteristics: Par value is 1000.r The ratio of the semi-annual coupon rate to the desired semi-annual yield rate, , is 1.03125. iThe present value of the redemption value is 381.50.Given vn = 0.5889, what is the price of bond X?(A) 1019(B) 1029(C) 105011/08/04 2523.Project P requires an investment of 4000 at time 0. The investment pays 2000 at time 1 and 4000 at time 2.Project Q requires an investment of X at time 2. The investment pays 2000 at time 0 and 4000 at time 1.The net present values of the two projects are equal at an interest rate of 10%. Calculate X.(A) 5400(B) 5420(C) 5440(D) 5460(E) 548011/08/04 2624.A 20-year loan of 20,000 may be repaid under the following two methods:i) amortization method with equal annual payments at an annual effectiverate of 6.5%ii) sinking fund method in which the lender receives an annual effectiverate of 8% and the sinking fund earns an annual effective rate of j Both methods require a payment of X to be made at the end of each year for 20 years. Calculate j.(A) j , 6.5%(B) 6.5% < j , 8.0%(C) 8.0% < j , 10.0%(D) 10.0% < j , 12.0%(E) j > 12.0%11/08/04 2725.A perpetuity-immediate pays X per year. Brian receives the first n payments, Colleen receives the next n payments, and Jeff receives the remaining payments. Brian's share of the present value of the original perpetuity is 40%, and Jeff's share is K. Calculate K.(A) 24%(B) 28%(C) 32%(D) 36%(E) 40%11/08/04 2826.Janice pays interest at the end of every six-month period as it accrues and the principal at the end of five years.Lori repays her loan with 10 level payments at the end of every six-month period. Calculate the total amount of interest paid on all three loans.(A) 8718(B) 8728(C) 8738(D) 8748(E) 875811/08/04 2927.Bruce and Robbie each open up new bank accounts at time 0. Bruce deposits 100 into his bank account, and Robbie deposits 50 into his. Each account earns the same annual effective interest rate.The amount of interest earned in Bruce's account during the 11th year is equal to X. The amount of interest earned in Robbie's account during the 17th year is also equal to X. Calculate X.(A) 28.0(B) 31.3(C) 34.6(D) 36.7(E) 38.911/08/04 3028.Ron is repaying a loan with payments of 1 at the end of each yearfor n years. The amount of interest paid in period t plus the amount of principal repaid in period t + 1 equals X. Calculate X.n t v 1 + (A)in t (B) 1 + vd1 + vn ti (C)1 + vn td (D)1 + vn t (E)11/08/04 3129.At an annual effective interest rate of i, i > 0%, the present value of a perpetuity paying `10 at the end of each 3-year period, with thefirst payment at the end of year 3, is 32. At the same annual effective rate of i, the present value of a perpetuity paying 1 at the end of each 4-month period, with first payment at the end of 4 months, is X.Calculate X.(C) 33.6(D) 34.6(E) 35.611/08/04 3230.As of 12/31/03, an insurance company has a known obligation to pay $1,000,000 on 12/31/2007. To fund this liability, the company immediately purchases 4-year 5% annual coupon bonds totaling $822,703 of par value. The company anticipates reinvestment interest rates to remain constant at 5% through 12/31/07. The maturity value of the bond equals the par value. Under the following reinvestment interest rate movement scenarios effective 1/1/2004, what best describes the insurance company’s profit or (loss) as of 12/31/2007 after the liability is paid?。
北美精算师(SOA)考试P 2001 November年真题
is independent of the number of accidents that occur in all other months.
Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs.
(A) (B) (C) (D) (E)
0.0000 0.0004 0.0027 0.0064 0.3679
Course 1, November 2001
9
9.
Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist and a chiropractor, whereas 12% visit neither of these. The probability that a patient visits a chiropractor exceeds by 0.14 the probability that a patient visits a physical therapist.
(–1)n n
− an = 1 2n n
Course 1, November 2001
11
11.
A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is
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0
for −2 ≤ x ≤ 4 otherwise
(c) 1
3 5 1 5
5
6. The time of 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) 0.38 (c) 0.42 (d) 0.29 (e) 0.57
where c is a constant. Calculate the moment generating function of X .
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1 8 4 3 23 12 7 72 5 36
1
2. An insurance policy pays for a random loss X subject to a deductible of C , where 0 < C < 1. The loss amount is modeled as a continuous random variable with density function f (x) = 2x for 0 < x < 1 0 otherwise
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Student #:
Quiz 3 Duration: 50 min Total: 10 Marks 1. A random variable X has thction: for x < 1 0 x2 −2x+2 F (x) = for 1 ≤ x < 2 2 1 for x ≥ 2 Find the variance of X. (a) (b) (c) (d) (e)
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7. The lifetime of a printer costing $200 is exponentially distributed with mode 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one–half refund if it fails during the second year. If the manufacturer sells 100 printers, how much should it expected to pay in refunds? (a) 6,321 (b) 7,358 (c) 10,256 (d) 12, 642 (e) None of above.
2
3. The lifetime of a machiine part has a continuous distribution on the interval (0, 40) with probability density function f , where f (x) is proportional to (10 + x)−2 . Calculate the probability that the lifetime of the machine is less than 6. (a) 0.04 (b) 0.15 (c) 0.47 (d) 0.53 (e) 0.94
6−k 15
0
for k = 1, 2, 3, 4, 5 otherwise
Calculate the expected payment for hospitalization under this policy. (a) 255 (b) 163 (c) 213 (d) 85 (e) 168
4
5. Let X be a continuous random variable with density function f (x) Calculated the expectation of X . (a) (b) (d) (e)
Given a random loss X, the probability that the insurance payment is less than 0.5 is equal to 0.64 . Calculate C . (a) 0.1 (b) 0.3 (c) 0.5 (d) 0.7 (e) None of above
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4. An insurance policy pays an individual 100 per day for up to 3 days of hospitalization and 25 per day for each day of hospitalization thereafter. The number of days of hospitalization, X ¡ is a discrete random variable with probability mass function: Pr{X = k } =
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8. An insurance policy reimburses dental expense, X , up to a maximum benefit of 250. The probability density function for X is: f (x) = ce−0.004x 0 for x ≥ 0 otherwise