www.coachingactuaries.com

Raise Your Odds

The Problem What is Adapt? How does Adapt work? Adapt statistics What are people saying about Adapt? So how will these flashcards help you?

The Problem Your confidence on exam day and not knowing how to study efficiently. FRONT

Our Solution We provide students online exam preparation tools for the preliminary SOA/CAS actuarial exams. Our goal is to help you raise your odds of passing the exams by providing the right coaching at the right time. Our team is dedicated to giving actuarial students the game plan, tools and techniques necessary to pass actuarial exams. Back

What is Adapt? FRONT

Adapt stands for Adaptive Dynamic Actuarial Practice Tests. It is a dynamic, Adaptive online practice exam tool using a large question bank to dynamically create practice exams that continually Adapt to your level or preparation. Adapt measures each student s knowledge on a scale of 1 10. This measurement is called an Earned Level. 90% of students that reach a Level 7 pass their exam! Back

How does Adapt work? FRONT

Adapt works through your Earned Level. Your Earned Level changes as you take more tests. If you perform well on your test, your Earned Level increases. As a result, you will see more difficult questions. If you perform poorly on a test, your Earned Level decreases. When you first begin Adapt, you start at a Level 3. The goal is for students to reach exam level difficulty questions and be able to master those questions. Back

Adapt Statistics FRONT

2013 Adapt Survey Introduction 100% Pass % Cumulative Pass% is 91.3% for level 7-10 92% 93% 97% Pass 85% 87% 75% 78% 50% 63% *Results from 2,300 Survey Responses for CBT Exams 43% 38% 25% Back 0% 0-2 3 4 5 6 7 8 9 10 Level

What are people saying about Adapt? FRONT

The Adapt software is an outstanding complement to the Financial Math Course that I teach. Students rave about how effective the Adapt software is at preparing them for the actuarial exams. I know of no student who has attainted Level 7 and then failed the exam. James Trimble Director of the Actuarial Science Program at UConn See more at: http://www.coachingactuaries.com/testimonials.aspx Back

So how will these flashcards help you? FRONT

In this PDF, there are 10 P/1 practice problems, 10 FM/2 practice problems and 10 MFE/3F practice problems. The goal is for them to aid in your studying and give you an idea of how beneficial Adapt can be during your exam preparation. Back

CON 41 CON 46 CON 52 CON 92 PT0031 PT0046 PT0048 PT0052 SOA6 SOA7

CON 41 One vase was broken in the house this year. The number of vases broken by any possible culprit during a year is a Poisson variable independent of all other possible culprits. The possible culprits are, with their average number of vases broken per year: family member 0.043, guest 0.012, pet 0.05, and natural forces 0.040. There are 8 individuals in the line-up: 4 family members, 2 guests, 1 pet, and 1 natural force (the wind). What is the probability that the guilty individual is a human being? A) 0.16 B) 0.39 C) 0.52 D) 0.69 E) 0.75 FRONT

Back Solution For Poisson variables, we can add the lambdas to calculate the new lambdas. λ = 0.043 ( 4) + 0.012( 2) = 0.196 human λ = 0.05( 1) + 0.04( 1) = 0.09 non-human λ total = λ human + λ non-human = 0.196 + 0.09 = 0.286 Pr human broke the vase 1 broken vase = = Pr human broke 1 vase AND non-human broke 0 vase e 0.196 ( 0.196) 1 1! e e 0.286 ( 0.286) 1 1! 0.09 ( 0.09) 1 0! Pr 1 broken vase = 0.161114 ( 0.913931) 0.214861 = 0.69

Canclled Days CON 46 The school board wants to determine how many school days should be designated as snow days. The weather bureau predicts the following number of school days will be cancelled due to snow. Probability 0 14% 1 16% 2 27% 3 27% 4 10% 5 6% Calculate the range of snow days that will be within one standard deviation of the mean. A) 2 B) 2-4 C) 1-3 D) 1-4 E) 0-4 FRONT

4 Solution E X = x Pr X = x = 2.21 E X 2 Var = x=1 4 x2 Pr X = x = 6.67 x=1 X = E X 2 σ = 1.89 = 1.37 2 = 6.77 2.21= 1.89 E X Range = 2.21± 1.37 = ( 0.84,3.58) or between 1 and 3 days Back

CON 52 A casualty insurer is issuing flood insurance to three categories of households: low elevation, medium elevation, and high elevation. Low elevation claims are 6% of the claims and have an exponential distribution with mean = 10. Medium elevation claims account for 71% of the claims and have an exponential distribution with mean = 4. The high elevation's average claim is insignificant. What is the probability that a randomly selected claim will be under 7? A) 18% B) 34% C) 49% D) 62% E) 85% FRONT

Solution For exponential distribution, mean = 1 λ. So, for low elevation λ = 1 10, for medium elevation λ = 1 4 For high elevation, we are assuming 100% of the claims are approximately 0. Pr ( X < 7) = F ( 7) = 1 e 7λ ( + 0.71 1 e 7 4 ) + 0.23 1 0.06 1 e 7 10 = 0.85 Back

CON 92 Accidents occur at a Poisson rate of two per week at a busy intersection. If three accidents occur in one week the roads will be policed by local enforcement, essentially reducing the accident rate for the rest of that week to zero. The distribution of accidents follows a Poisson function with a cap of three accidents. Find the expected number of accidents in a given week at this intersection. A) 1.2 B) 1.4 C) 1.6 D) 1.8 E) 2.0 FRONT

Solution Since the distribution is Poisson for less than 3 accidents. Pr ( K = k) = e 2 2 k k! for k < 3 note that k is capped at 3, but the sum of the must still be one. probabilities k 0 1 2 3 Pr ( K = k) e 2 2e 2 2e 2 1 Pr ( 0) Pr ( 1) Pr ( 2) = 1 5e 2 Back 3 k=0 E K = k Pr K = k = 9e 2 + 3 = 1.8

PT0031 A random variable X has a binomial distribution with mean 13.5 and variance 9.855. Determine the mode of X. A) 10 B) 11 C) 12 D) 13 E) 14 FRONT

Solution Introduction The pdf of a Binomial distribution with n trials and probability p: = n x Pr X = x p x Solving for n and p: ( 1 p) n x, x = 0,1,2, np = 13.5 np( 1 p) = 9.855 ( 1 p) = 0.73 p = 0.27,n = 50 mode = highest probability x Pr ( X = x) 10 0.0721 11 0.0970 12 0.1166 13 0.1261 mode Back 14 0.1233

PT0046 A machine consisting of two components will fail if one of the components fails. Let X be the time-to-failure random variable of the first component and Y be the time-to-failure random variable of the second component. The time-to-failure is measured in hours. The time-tofailure of the two components have the following joint moment generating function: M X,Y ( s,t) = 1 1 3s 2t + 6st Calculate the probability that the machine will fail within the first hour. A) 0.112 B) 0.332 C) 0.565 D) 0.677 E) 0.710 FRONT

Solution M X,Y ( s,t) = 1 1 3s 2t + 6st = 1 1 3s 1 1 2t = M X ( s ) M ( t) Y X and Y are independent where X exponential θ = 3 Y exponential θ = 2 Pr ( X 1) ( Y 1) = Pr ( X 1 ) + Pr ( Y 1) Pr ( X 1) ( Y 1) = Pr ( X 1) + Pr ( Y 1) Pr ( X 1) Pr ( Y 1) = F ( 1) + F ( 1) F ( 1) F ( 1) X Y X Y ( + 1 e 1 2 ) 1 e 1 3 = 1 e 1 3 = 0.5654 1 e 1 2 Back

PT0048 A discrete random variable X has the following probability density function: Pr X = x = e 4 4 x x!, x = 0,1,2, Find the probability that X is within 0.5 standard deviation from its mean. A) 0.51 B) 0.53 C) 0.55 D) 0.57 E) 0.59 FRONT

Solution X is a Poisson distribution with mean 4. E X = Var X = 4 The probability that X is 0.5 standard deviation from its mean is: Pr 0.5σ X E X X 0.5σ X = Pr 0.5( 2) X 4 0.5( 2) = Pr 3 X 5 5 x=3 = Pr X = x = e 4 4 3 3! + 44 4! + 45 5! = 0.547 Back

PT0052 A random variable T has the following probability density function: f T ( t) = 3θ 3 t 4 A) 0.030 B) 0.056 C) 0.064 D) 0.210, t > θ Calculate Pr T > 5 T > 2. E) 0.430 FRONT

Solution F T t ( t) = f ( s)ds T = θ t θ 3θ 3 s 4 ds = 3θ 3 t s 4 ds θ = 3θ 3 s 3 3 t θ = θ 3 t 3 θ 3 = 1 θ t 3 Pr T > 5 T > 2 Pr T > 5 = ( T > 2) Pr T > 5 = Pr T > 2 = Pr T > 2 1 Pr T 5 1 Pr T 2 1 F = T 5 1 F 2 T ( θ 5 ) 3 = ( θ 2) 3 = 0.064 Back

SOA6 A public health researcher examines the medical records of a group of 937 men who died in 1,999 and discovers that 210 of the men died from causes related to heart disease. Moreover 312 of the 937 men had at least one parent who suffered from heart disease, and, of these 312 men, 102 died from causes related to heart disease. Determine the probability that a man randomly selected from this group died of causes related to heart disease, given that neither of his parents suffered from heart disease. A) 0.115 B) 0.173 C) 0.224 D) 0.327 E) 0.514 FRONT

Solution H = event that a death is due to heart disease F = event that at least one parent suffered from heart disease = Pr H F C = Pr F C = Pr H F C 210 102 937 937 312 937 Pr H FC = 625 937 Pr F C = 108 937 = 108 625 = 0.173 Back

SOA 7 An insurance company estimates that 40% of policyholders who have only an auto policy will renew next year and 60% of policyholders who have only a homeowners policy will renew next year. The company estimates that 80% of policyholders who have both an auto and a homeowners policy will renew at least one of those policies next year. Company records show that 65% of policyholders have an auto policy, 50% of policyholders have a homeowners policy, and 15% of policyholders have both an auto and a homeowners policy. Using company estimates, calculate the percentage of policyholders that will renew at least one policy next year. A) 20 B) 29 C) 41 D) 53 E) 70 FRONT

Solution Let A = event that a policyholder has an auto policy B = event that a policyholder has a homeowners policy Then, based on the given information, Pr ( A H) = 0.15 = Pr A = Pr H Pr A H C Pr A C H Pr ( A H) = 0.65 0.15 = 0.50 Pr ( A H) = 0.50 0.15 = 0.35 The portion of policyholders that will renew at least 1 policy: ( + 0.6Pr A C ) H + 0.8Pr A H 0.4Pr A H C = ( 0.4) ( 0.5) + 0.6 = 0.53 ( 0.35) + ( 0.8) ( 0.15) Back

FM GL0011 SOA FM 1979 8 FMH0013 SOA FM 1985N 4 FMK0064 SOA FM 1992N 8 FMS DM 0112 SOA FM 1993N 16 FMWL0007 SOA FM 1993N 17

FM GL0011 Consider the following information: Year 1 2 3 4 5 6 Zero-coupon bond price 0.96618 0.92991 0.89157 0.85152 0.81017 0.76790 Solve for the fixed rate in a 3-year interest rate swap with the first settlement in Year 2. A) 4.08% B) 4.29% C) 4.50% D) 5.02% E) 5.55% FRONT

Solution R( 0.92991+ 0.89157 + 0.85152) + 1( 0.85152) = 1( 0.96618) R = 0.96618 0.85152 2.673 = 0.042896 4.29% Back

FMH0013 A two year par value bond with face value of 1,000 pays 10% annual coupons is sold for 980. Find the effective annual yield of this bond. A) 11.0% B) 11.1% C) 11.2% D) 11.3% E) 11.4% FRONT

980 = 100v + 1,100v 2 Solution 1,100v 2 + 100v 980 = 0 Now we can use the quadratic formula: v = b ± b2 4ac 2a a = 1,100 b = 100 c = 980 ( 980) v = 100 ± 1002 4 1,100 = 2 1,100 100 ± 2,078.9420 = 0.8995 2,200 The positive answer is the only one that is relevant in this case. 1 1+ i = 0.8995 i = 11.2% Back

FMK0064 You are an actuary for ABC Insurance Company. You currently have an annuity product that will pay a level $1,000 per year at the end of each year for 20 years. You are considering adding a second annuity that is identical to the first, but it has a cost of living adjustment (COLA) of 4% per year starting in the second year. What is the difference in the cost to your company of the COLA annuity compared to the level annuity assuming an annual effective rate of interest of 5.00%. A) 3,540 B) 4,580 C) 4,960 D) 5,120 E) 5,830 FRONT

Solution 1,000 PV ( increasing,4% ) a 20 0.05 = 1,000 1 ( 1.04 ) 20 1.05 1.05 1.04 a 20 0.05 = 17,418.87 12,462.21 = 4,956.66 4,960 Back

FMS DM 0112 Which one of the following is true of American and European options? A) American options can be exercised before expiration of the option. B) European options can only be purchased through a broker or exchange based in Europe. C) A single American option can be exercised multiple times. D) An example of a European option is an option with 2 years of maturity with quarterly exercise date. E) The strike price of an American option changes over time depending on the price of the underlying. FRONT

Solution A) True by definition B) False - European option purchases are not limited to a certain region. C) False - American options can only be exercised once. D) False - European options can only be exercised at maturity. E) False - The strike price stays the same on a standard American option. Back

FMWL0007 A 5-year $1,000 par value bond has a 7% coupon rate convertible semiannually. It is callable at par anytime between 3 and 5 years inclusive. An investor wishes to buy the bond to yield 6% convertible semiannually. Determine the purchase price of the bond. A) 961.55 B) 982.40 C) 1,006.26 D) 1,027.09 E) 1,050.98 FRONT

Solution yield rate < coupon rate bond sells at premium loss at maturity assume called of earliest possible date because that gives lowest value ( maximum to pay is minimum value) Price to pay = 1,000 ( 0.035)a + 1,000v 6 6 0.03 = 1,027.09 Back

SOA FM 1979 8 An investor buys two 20-year bonds each having semiannual coupons and each maturing at par. For each bond the purchase price produces the same yield rate to maturity. One bond has a par value of $500 and a coupon of $45; the other bond has a par value of $1,000 and a coupon of $30. The dollar amount of premium on the first bond is twice as great as the dollar amount of discount on the second bond. What yield rate, convertible semiannually, does the investor realize? (Answer to the nearest 0.1%) A) 0.042 B) 0.060 C) 0.070 D) 0.084 E) 0.120 FRONT

Solution 500 + P = 500v 40 + 45a 40 1,000 D = 1,000v 40 + 30a 40 where P = 2D Solving, eliminating P and D, 2,500 = 2,500v 40 + 105a 40 From calculator: N = 40 PV = 2,500 PMT = 105 FV = 2,500 Notice that this equivalent to a bond selling for 2,500, having 40 coupons of 105, and the redemption value is 2,500. CPT I / Y = 4.2 6-month effective rate = 0.042 ( i 2 ) = 0.042 ( 2 ) = 0.084 Back

SOA FM 1985N 4 The duration of a bond at interest rate I is defined as: t t ( t C v t ) t ( C v t ) t Where C t represents the net cash flow from the coupons and the maturity value of the bond at time t. You are given a 1,000 par value 20-year bond with 4 percent annual coupons and a maturity value of 1,000. Calculate the duration of this bond at 5 percent interest. A) 5.5 B) 8.9 C) 13.7 D) 20.0 E) 24.0 FRONT

Solution Numerator: ( t C t v t ) t ( 1) ( 40)v + ( 2) ( 40)v 2 +!+ ( 20) ( 40)v 20 + ( 20) ( 1,000 )v 20 = 40 Ia 20 0.05 + 20,000v 20 a = 40!! 20 0.05 20v 20 + 7,537.79 0.05 = 11,975.81 Denominator: ( C t v t ) t ( 40)v + ( 40)v 2 +!+ ( 40)v 20 + ( 1,000 )v 20 Duration = 11,975.81 875.38 = 13.7 = 40a 20 0.05 + 1,000v 20 = 875.38 Back

SOA FM 1992N 8 Amy invests 1,000 at an effective annual rate of 14% for 10 years. Interest is payable annually and is reinvested at an annual effective rate of i. At the end of 10 years the accumulated interest is 2,341.08. Bob invests 150 at the end of each year for 20 years at an annual effective rate of 15%. Interest is payable annually and is reinvested at an annual effective rate of i. Find Bob s accumulated interest at the end of 20 years. A) 9,000 B) 9,010 C) 9,020 D) 9,030 E) 9,040 FRONT

Solution 140s 10 i = 2,341.80 i = 0.11 Recall ( Is) s ( n + 1 ) = n+1 i% n i i For Bob, each reinvestment increases by 0.15 150 : Let n = 19( note: n + 1= 20) and i = 0.11 s 20 150 20 + 0.15 20 11% 0.11 150( 20) 0.15( 150) s 20 20 11% 0.11 = 9,041 Back

SOA FM 1993N 16 Lenny buys a stock for 200 which pays a dividend of 12 at the end of every 6 months. Lenny deposits the dividend payments into a bank account earning a nominal interest rate of 10% convertible semiannually. At the end of 10 years, immediately after receiving the 20th dividend payment of 12, Lenny sells the stock. The sale price assumes a nominal yield of 8% convertible semiannually and that the semiannual dividend of 12 will continue forever. Lenny s annual effective yield over the 10-year period is i. Calculate i. A) 12.80% B) 12.95% C) 13.15% D) 13.30% E) 13.45% FRONT

Solution AV of dividends = 12s 20 5% = 396.79 P = 12 0.04 = 300 200( 1+ i) 10 = 396.79 + 300 ( 1+ i) 10 = 3.4840 i = 13.29% Back

SOA FM 1993N 17 Don and Rob each sell a different security short. Don sells his security short for a price of 960, and Rob sells his security short for 900. Both investors buy back their securities for X at the end of 1 year. In addition, the required margin is 50% for both investors, and both receive 10% annual effective interest on their margin deposits. No dividends are paid on either security. Don s yield rate on his short sale is 50% greater than Rob s yield rate on his short sale. Calculate Don s yield rate. A) 33% B) 35% C) 41% D) 43% E) 54% FRONT

Solution 960 x + 48 480 x = 840 yield = = 1,008 840 480 900 x + 45 450 = 35% ( 1.5 ) Back

MFEC0176 MFENH0012 MFET0228 MFET0256 MFET0257 MFET0267 MFET0280 MFET0292 MFET0331 MFET0347

MFEC0176 Assume the Black-Scholes framework. The price of a stock is $144. The stock pays dividends at a continuously compounded rate of 3% per year. The stock s volatility is 40%. A European call option the stock has a strike of $150 and a premium of $48.70. The call option has a delta of 0.6579, a gamma of 0.0025, and a theta of -5.4293. Calculate the continuously compounded risk-free interest rate. A) 6% B) 9% C) 12% D) 15% E) 18% FRONT

Solution The general form of the Black-Scholes equation is ( r δ )SV + 1 S 2 σ 2 S 2 V +V = rv SS t For a call option, V = C,V S = Δ,V SS = Γ and V t = θ. Therefore, ( r 0.03) ( 144) ( 0.6579) + 1 ( 2 0.4 ) 2 ( 144) 2 ( 0.0025) + ( 5.4293 ) = r ( 48.70) r = 0.0896 0.09 Back

MFENH0012 You are given the following information on the price of a stock: Date Stock Price Jul. 1, 2007 35.30 Aug. 1, 2007 33.90 Sep. 1, 2007 41.20 Oct. 1, 2007 31.95 Nov. 1, 2007 38.25 Dec. 1, 2007 46.18 Estimate the annual volatility of continuously compounded return on the stock. A) 0.20 B) 0.62 C) 0.68 D) 0.69 E) 0.75 FRONT

Solution To estimate volatility from historical data: σ ˆ = N n x 2 t n 1 n x 2, where x = ln t S t S 1, x = t n x t N is the number of periods per year, n is the number one less than the number of observations of stock price. Back In our problem, N = 12 and n = 5. We calculate x = ln t 35.30 S t S t 1 and x 2 : t S x x 2 t t t 33.90-0.0405 0.0016 41.20 0.1950 0.0380 31.95-0.2543 0.0647 38.25 0.1800 0.0324 46.18 0.1884 0.0355 5 = 0.2687 t=1 x t x = 0.2687 5 5 x 2 = 0.1722 t t=1 s 2 = 5 4 = 0.03944 = 0.05373 0.1722 5 0.05373 2 s monthly = 0.03944 = 0.1986 s annual = 0.1986 12 = 0.688 0.69

MFET0228 Assume the Black-Scholes framework. Three months ago, an investor borrowed money at the risk-free rate to purchase a one-year at-the-money European put option on a stock. At that time, the price of the stock was 50. Today, the stock price is 52. The investor decides to close out all positions. You are given: (i) The stock pays dividends continuously at a rate proportional to its price. The dividend yield is 2%. (ii) The volatility of the stock is 30%. (iii) The continuously compounded risk-free interest rate is 6%. Calculate the three-month holding profit of the investor. A) -1.31 B) -1.46 C) 0 D) 1.31 E) 1.46 FRONT

Solution Introduction Three months ago, the value of the put option is: d 1 = 0.06 0.02 + 0.5( 0.3) 2 0.3 = 0.28333 d 2 = 0.28333 0.3 = 0.01667 N( d ) = N( 0.28333 ) = 0.38846 N( d ) = N( 0.01667 ) = 0.50665 1 2 P = 50e 0.06 ( 0.50665 ) 50e 0.02 ( 0.38846 ) = 4.8189 Today (after three months), the value of the put option is: d 1 = ln 52 50 2 + 0.06 0.02 + 0.5 0.3 0.3 0.75 ( 0.75 ) d 2 = 0.39633 0.3 0.75 = 0.13653 N( d ) = N( 0.39633 ) = 0.34593 1 = 0.39633 0.06( 0.75) P = 50e N( d ) = N( 0.13653 ) = 0.44570 2 ( 0.44570 ) 0.02( 0.75) 52e ( 0.34593 ) = 3.5839 Back 0.06( 0.25) The profit to the investor is: 3.5839 4.8189e = 1.308 1.31

MFET0256 Consider four 5-year European options with different strike price. The price of a 50-strike call option is higher than the price of a 60- strike call option by $5.00. The price of a 50-strike put option is lower than the price of a 60- strike put option by $1.70. All options are on the same stock and the stock pays dividends continuously at a rate proportional to its price. Determine the continuously compounded risk-free interest rate. A) 0.06 B) 0.07 C) 0.08 D) 0.09 E) 0.10 FRONT

Solution C( 50) C( 60) = 5 1 P ( 50) P ( 60) = 1.7 2 1-2 yields: C( 50) P ( 50) S e δ ( 5 ) 50e r( 5) 0 C ( 60 ) P ( 60) = 5 ( 1.7 ) ( S e δ ( 5 ) 60e r( 5) ) = 6.70 0 10e 5r = 6.70 r = 0.08 Back

{ Z ( t) } measure. A) 0.31 B) 0.42 C) 0.57 D) 0.68 E) 0.71 MFET0257 is a standard Brownian motion under the true probability Calculate the true probability that Z a 2 Pr a < Z a 2 < a, where a is a constant. is between a and a, FRONT

Solution Under the true probability measure, Z ( t) N 0,1, so Z a 2 N 0, a 2. Pr a < Z a 2 < a = Pr Z a2 = Pr Z a 2 < a < a = 2Pr Z a 2 < a 0 = 2Pr Z < 1 1 < a ( ) < a Pr Z a2 1 Pr Z a2 1 Z a 2 = 2Pr < a 0 1 a 2 a 2 = 2( 0.84134) 1 = 0.68268 Back 0.68

MFET0267 Assume the Black-Scholes framework applies. Consider a European put option on a stock with strike K and time to expiration T. The current price of the stock is S0. An investor wishes to replicate the aforementioned put option with an assetor-nothing option and a cash-or-nothing option. Assume all options in the following answer choices are on the same stock and have the same time to expiration. Which one of the following describes the transactions that the investor should make? A) Buy S0 units of an asset-or-nothing put with trigger price K and sell K units of a cash-or nothing put with trigger price K. B) Sell S0 units of an asset-or-nothing put with trigger price K and buy K units of a cash-or nothing put with trigger price K. C) Buy one unit of an asset-or-nothing put with trigger price K and sell K units of a cash-or nothing put with trigger price K. D) Sell one unit of an asset-or-nothing put with trigger price K and buy K units of a cash-or nothing put with trigger price K. FRONT E) Sell one unit of an asset-or-nothing put with trigger price K and buy one unit of a cash-or nothing put with trigger price K.

Solution The price of one unit of asset-or-nothing put with trigger price K is: AssetPut( K ) = S e δt N( d ) 0 1 The price of one unit of cash-or-nothing put with trigger price K is: CashPut( K ) = e rt N( d ) 2 Using the Black-Scholes formula, the price of put option is: P = Ke rt N( d ) S e δt N( d ) 2 0 1 = K CashPut( K ) AssetPut( K ) So, the put option can be replicated by: buying K units of cash-or-nothing put option with trigger K, and selling one unit of asset-or-nothing put option with trigger K. Back

MFET0280 Assume the Black-Scholes framework. Let S( t) denote the price at time t of a stock that pays no dividends. Consider a European put option with exercise date T, T > 0, and exercise price S( 0)e rt, where r is the continuously compounded risk-free interest rate. You are given: (i) T = 4 (ii) Var lns t = 0.25t, t > 0 Let P ( 0) denote the time-0 price of the put option. Calculate P ( 0) S( 0). A) 0.099 B) 0.125 C) 0.383 D) 0.560 E) 0.800 FRONT

From (ii), we have: σ 2 = 0.25 P 0 S 0 where S ( 0 )e rt = σ = 0.50 e rt N d 2 = N( d ) N( d ) 2 1 = N( d ) N( d ) 1 2 d 1 = = S 0 ln S( 0)e rt σ rt + rt + σ 2 σ T S 0 S 0 + r + σ 2 2 T T 2 T Solution N d 1 d 2 = 0.50 0.50 4 = 0.50 N( d ) = N( 0.50) 1 = 0.69146 N( d ) = N( 0.50) 2 P 0 S 0 = 1 0.69146 = 0.30854 = 0.69146 0.30854 = 0.38292 Back = σ T 2 = 0.50 4 2 = 0.50

FRONT S j MFET0292 Consider a model with two stocks. Each stock pays dividends continuously at a rate proportional to its price. ( t) denotes the price of one share of stock j at time t. Consider a claim maturing at time 4. The payoff of the claim is max S ( 4), S 4 1 2 You are given: A) 83 = S ( 0) = $100 2 (i) S 0 1 (ii) Stock 1 pays dividends of amount ( 0.1)S ( t) dt between time 1 t and time t + dt. (iii) Stock 2 pays dividends of amount ( 0.1)S ( t) dt between time 2 t and time t + dt. (iv) The volatility of Stock 1 is 20%. (v) The volatility of Stock 2 is 30%. (vi) The correlation between the continuously compounded returns on Stock 1 and Stock 2 is 1/3. Calculate the price of the claim. B) 90 C) 116. D) 124 E) 130

Solution Introduction The payoff of the claim at time 4 is: max S ( 1 4),S 2 4 = S 2 4 Accordingly, the price of the claim at time 0 is: P F 0,4 P F 0,4 + max S ( 1 4) S 2 4 S 2 4 S 2 4 + Call S,S 1 2 = S ( 2 0)e δ 2 T = 100e 0.1 ( 4 ) = 67.0320, 0 Call( S 1,S ) = F P ( 2 S 1 ) N( d ) 1 F P ( S 2 ) N( d ) 2 F P ( S ) 1 = F P ( S ) 2 = 100e 0.1 ( 4 ) where = 67.0320 d 1 = ln F P S 1 F P S 2 σ N( d ) 1 = 0.61791 + 0.5σ 2 T T d 2 = 0.30 0.30 4 = 0.30 N( d ) 2 = 0.38209 Call( S 1,S ) 2 = 0.38292 = 67.032( 0.61791) 67.032( 0.38209) = 15.8075 Hence, the price of the claim is: 67.0320 + 15.8075 = 82.84 σ = σ 2 S1 + σ 2 S2 2ρσ S1 σ S2 = 0.20 2 + 0.30 2 ( 2 )( 0.20) ( 0.30) = 0.30 3 Back

MFET0331 X ( t) follows an Ito's process defined by dx t X t = αdt + βdz ( t) where Z ( t) { } is a standard Brownian motion. Let Y ( t) = X ( t) 2 You are given that dy t Y t = adt + bdz ( t) Find a b. A) α + β 2 β B) α + 2β 2 β C) 2a + β 2 β D) α + 2β 2 2β E) 2α + β 2 2β FRONT

Solution Using Ito's Lemma, dy = Y X dx + 0.5Y XX Y = X 2 Y X = 2X 2 1 = 2X Y XX = 2X 1 1 = 2 Y t = 0 dx = α Xdt + β XdZ ( dx) 2 +Y t dt, where ( dx ) 2 = 0 + ( β XdZ ) 2 = β 2 X 2 ( dz ) 2 = β 2 X 2 dt Hence, dy = 2X α Xdt + β XdZ + 0.5 2 dy = ( 2α + β 2 ) X 2 dt + 2β X 2 dz dy Y = 2α + β 2 dt + 2βdZ Thus, a = 2α + β 2 and b = 2β, β 2 X 2 dt + 0dt So, a b = 2α + β 2 β Back

MFET0347 Consider the following Itô processes for two nondividend paying stocks, Stock X and Stock Y: dx t X t dy t Y t where Z t = 0.10dt + 0.20Z ( t) = 0.1125dt + 0.25dZ ( t) { } is a standard Brownian motion. You are also given that X ( 0) = 1 and Y ( 0) = 2. Find Pr X ( 300) > Y ( 300) A) 0.09 B) 0.11. C) 0.13 D) 0.15 E) 0.17 FRONT

For a stock: = α δ ds t S t Hence, dx t X t dt + σ dz t S( t) = S 0 Solution e = 0.10dt + 0.20dZ ( t) X ( t) = X 0 e ( 2 α δ 0.5σ )t+σ Z t 2 0.10 0.5 0.20 t+0.20z t = ( 1 )e 0.08t+0.20Z ( t ) Back dy t Y t Then, = 1 2 X t Y t = 0.1125dt + 0.25dZ ( t) X ( t) = X 0 e 0.08t+0.20Z ( t ) e 0.08125t+0.25Z t α δ 0.5σ 2 = 0.00125 σ = 0.05 Pr X 300 > Y ( 300) e = 0.5e 0.00125t 0.05Z ( t ) X t Y t > 1 = Pr X 300 Y 300 = N = N d! 2 2 0.1125 0.5 0.25 0.5 ln + ( 0.00125 )( 300) 1 0.05 300 t+0.25z t = ( 2 )e 0.08125t+0.25Z ( t ) follows geometric Brownian motion = N 1.23339 = 0.11

What is coaching actuaries? FRONT

The includes a study manual, video instruction/lecture and Adapt with video solutions. It also has a Pass Guarantee which allows you a free 180-day extension if you don t pass your exam. In addition, you will get support help from Coaching Actuaries All-Star team when you purchase the. Back

Coaching Actuaries Blog Click here to go!! FRONT

We recently started a blog for our students and we d love for you to check it out! It includes students stories, Coach Kester s thoughts and a lot of helpful hints and entertaining posts. If you would like to blog, click here. Back

@coachingactuary /coachingactuaries

The Team!