MATH20180: Foundations of Financial Mathematics

Transcription

1 MATH20180: Foundations of Financial Mathematics Vincent Astier office: room S1.72 (Science South) Lecture 1 Vincent Astier MATH / 35

2 Our goal: the Black-Scholes Formula Black-Scholes Formula for pricing a call option: ( log(x0 /k) + rt C T = X 0 Φ σ + 1 ) ( T 2 σ T ke rt log(x0 /k) + rt Φ σ 1 ) T 2 σ T Vincent Astier MATH / 35

3 To reach this goal we need to learn... Mathematics: some set and measure theory, probability, random variables, expected values, Brownian motion. Finance: interest rates, present value, discounted value, hedging, risk, bonds,shares, call (and put) options, arbitrage. Text: Sean Dineen "Probability Theory in Finance: A Mathematical Guide to the Black-Scholes Formula" (in small doses). Vincent Astier MATH / 35

4 Interest Rates Definition The interest represents the money paid to the lender for the use of the borrowed money. It also represents the money earned when a given amount of capital is invested. The principal P is the total amount of money that is either borrowed or invested. The rate of interest r (usually given in percentage) is the amount charged for the use of principal for a given length of time, usually on yearly basis. The future value S is the principal P plus the accumulated interest. Vincent Astier MATH / 35

5 Simple interest Definition The simple interest is the interest computed once on the principal for the entire period of time when it is borrowed. Formula: If a principal P is borrowed (invested) at a simple interest rate of r per year for a period of t years, then 1 the interest is 2 the future value S is given by I = Prt S = P + I = P + Prt = P (1 + rt). (t does not have to be an integer.) Vincent Astier MATH / 35

6 Simple interest: Examples Example 1. John wants to open a special savings account earning 6.25% simple interest per annum with an initial deposit of e500. How much will be in the account at the end of 8 months? Answer: /12. Example 2. At what annual rate of simple interest will a principal of e960 accumulate to a future value of e1,000 in 10 months? Answer: Solve r 10/12 = Vincent Astier MATH / 35

7 Simple interest Remark: Simple interest is rarely used by banks. If e1,000 is deposited for 2 years at a rate of 10% simple interest, then the amount accumulated at the end of two years would be e1,200. If, however, at the end of year one the amount accumulated at that time e1,100 is withdrawn and immediately deposited for a further year at the same interest rate, then, after one more year the amount accumulated would be e1,210, that is, a gain of e10 on the previous amount. If simple interest would be the norm, people would be in and out of banks regularly withdrawing and immediately re-depositing their savings. Vincent Astier MATH / 35

8 Compound interest Definition The compound interest is obtained by applying simple interest at regular intervals to the amount accumulated at the start of each period. Details: If a principal P is borrowed (invested) at a compounded interest rate r for n periods in a year, then 1 The future value after one period is P + P r n = P(1 + r n ), i.e. simple interest at the rate of r/n. The new value is obtained by multiplying the previous one by 1 + r n. Therefore: 2 the future value after k periods is P(1 + r n )k, 3 the future value S after t years is P(1 + r n )nt, 4 the interest I after t years is I = S P. Vincent Astier MATH / 35

9 Compounded interest: examples Example 1. A deposit of e400 is placed in an account earning interest at 10% per year compounded quarterly. Find the future value after three years. How much compounded interest was earned? Answer: 400( /4) 12. Example 2. At what interest rate r per year compounded quarterly will e250 invested today amount to e1,900 in ten years? Answer: Solve 250(1 + r/4) 40 = Example 3. Which of the following yields more profit for e50 invested for a period of one year at the interest rate of 10%: (a) compounded annually; (b) compounded monthly; (c) compounded daily. Vincent Astier MATH / 35

10 Compounded interest: Solution to Example 3 (a) The compounding period is 1 year. Therefore n = 1 so ( S = ) 1 = (b) The compounding period is 1/12 year. Therefore n = 12 and ( S = ) 12 = (c) The compounding period is 1/365 year. ( S = ) 365 = Note that the more often the 10% is compounded, the higher the return. What if we were to take the 10% per annum and compound it continuously? What would the future value S be? Vincent Astier MATH / 35

11 Continuously compounded interest Lemma Let a be a real number. Then lim n ( 1 + a n ) n = e a. Remark. In our case lim n ( 1 + r n) nt = e rt. Definition If a principal P is invested (borrowed) at an interest rate r per year compounded continuously, after t years the future value S is given by S = Pe rt Vincent Astier MATH / 35

12 Continuously compounded interest Exercise: Suppose you invest e100 at interest rate of 5% per annum compounded continuously. How much will you have in 3 years? Solution: S = Pe rt = 100e = Remark: We can reverse the process and say that the present value (present worth, principal) of e in three years s time is e100. In this way we can determine, for a given fixed rate of interest, the present value of any amount from any future time. Vincent Astier MATH / 35

13 Continuously compounded interest Example 2: If P is the present value of e5,000 in six years s time at 7% interest rate compounded continuously, then Pe = 5, 000 therefore P = 5, 000 e 0.42 = 3, The procedure of finding the principal value from the future amount is called discounting back to the present. The present worth of a future amount is called its discounted value. In the above example, e3, is the discounted value of e5,000 in six years s time at 7% interest rate compounded continuously. Vincent Astier MATH / 35

14 Continuously compounded interest Formula: The discounted value of an amount S at a future time t assuming a constant continuously compounded interest rate r is given by Se rt. Vincent Astier MATH / 35

15 Other financial Instruments-Bonds A bond is a contract under which the issuer borrows money from the holder of the bond. The money (called the face value) will be repaid at the end of the contract (called the maturity date). The bond may include the payment of interest (the coupon) at regular intervals. Remark 1: Bonds can be traded between investors. Remark 2: Bond is related to interest rate because of its coupon. For example, if we buy a one-year bond with face value e100 and coupon e6, then we will get e106 after one year, independent of the interest rate. Vincent Astier MATH / 35

16 Other financial Instruments-Bonds: examples Example: A bond has a face value of e1,000 and a coupon worth e50. The (coupon) interest rate is % = 5 100% = 5%. 100 If the interest rate increases to 7%, then the actual value of the bond will decrease. An investor would never pay e1,000 for a bond with a coupon rate of 5% on the secondary market when new issues of similar quality are paying 7%. In order to remain competitive with new issues, the bond would sell at a discount to its face value. If x is the new value of the bond, then 0.07x = 50 and so x = = e Vincent Astier MATH / 35

17 Fair Games Definition A game is said to be fair if it is equally favorable to all players. In Game Theory, by winning we mean net winning, and thus we subtract off any losses and treat a loss as a win of a negative amount. Vincent Astier MATH / 35

18 Fair Games: examples Example: Consider a simple betting game (e.g. tossing a coin) between two players, John and Mark. If the game is favorable to John, then his expected winnings, E[W J ], is at least as large as those of Mark, E[W M ]; that is E[W J ] E[W M ]. Similarly, if the game is favorable to Mark, then E[W M ] E[W J ]. The game is called fair if E[W J ] = E[W M ] (1) If the bets placed by John and Mark are B J and B M, respectively, then the total input will be B J + B M. If the total output, that is the sum of all the players winnings, is also B J + B M, we call the game a zero-sum game. Vincent Astier MATH / 35

19 Fair Games If the game is a zero-sum game we have E[W J ] + E[W M ] = Output Input = 0 (2) and, combining the above two equalities (1) and (2), we obtain The converse is also true. Theorem E[W J ] = E[W M ] = 0. A zero-sum game is a fair game if and only if the expected winnings of each player is zero. Vincent Astier MATH / 35

20 Fair Games Game: John and Mark each bet e5 on the toss of a coin: John wins when a head (H) comes up, Mark wins when a tail (T) comes up and the winner gets e10. In tossing an unbiased coin we expect half of the outcomes, no matter how many games are played, to result in a head, and we interpret this as the probability that a head appears. Similarly, the probability of a tail appearing is 1/2. Thus in 1, 000 games John would expect to win 500 games and also to lose 500. His expected winnings are E[W J ] = 500 (10 5) (0 5) = 0. Since Mark s expected winnings are also 0, we have verified the fair game criterion. Vincent Astier MATH / 35

21 Fair Games Question: What happens if we change these parameters in a zero-sum game? Consider the following variations: (A) John bets e3 and Mark e7, Answer: E[W J ] = 1 2 (10 3) (0 3) = 2, so the game is not fair. (B) on average the coin turns up heads 80% of the time. Answer: E[W J ] = 8 10 (10 5) (0 5) = 3, the game is not fair. Vincent Astier MATH / 35

22 Fair Games We conclude: Theorem Probabilities (or risks) and rewards (or winnings/losses) are both used in calculating the expected return. In a zero-sum game if one of these is given, then the other can be chosen to make the game fair. Vincent Astier MATH / 35

23 Hedging Motivating example: Consider another game in which John and Mark place a bet with a bookmaker on a two-horse race. Terminology: A person who accepts bets on horse races (and other sporting events) is called a bookmaker. Members of the public who place bets are called punters. Remark: In contrast to the coin tossing game, John and Mark do not negotiate the odds on each horse. These are set by the bookmaker, who does not care what odds are given as long as he gets his percentage of the total amount wagered. In fact the bookmaker s strategy is to eliminate (or minimize) any risk to his percentage share. This is called hedging. Vincent Astier MATH / 35

24 Hedging Simple case: John bets e400 on horse A while Mark bets e100 on horse B. The bookmaker knows that they are the only ones playing, and wants 10% profit from the sum of all bets (e500). So the winner takes e450. What odds should the bookmaker give on horses A and B? If horse A wins, John wins e50, so a bet of e400 brings a win of e50, i.e. the odds are 1 to 8 for horse A. If horse B wins, Mark wins e350, so a bet of e100 brings a win of e350, i.e. the odds are 3.5 to 1, i.e. 7 to 2 for horse B. Vincent Astier MATH / 35

25 Hedging What about the point of view of John and Mark? Let p be what John thinks is the probability that horse A wins, and q what Mark thinks is the probability that horse B wins. If they both think the game is fair, then E[W J ] = p50 + (1 p)( 400) = 0, so p = 8 9. E[W M ] = q350 + (1 q)( 100) = 0, so q = 2 9. Vincent Astier MATH / 35

26 Hedging Motivating example (continued): Consider another game in which John and Mark place a bet with a bookmaker on a two-horse race. John places a bet of e400 on horse A while Mark bets e100 on horse B. The bookmaker wants to make a profit of 10%. Suppose now that a further bet for e300 is placed on horse B at the odds 7 : 2. Question: What is the bookmaker s position in this case? Let W B be the winnings of the bookmaker. If horse A wins then W B = = 350, if horse B wins then W B = = To avoid or lower the risk, the bookmaker can: refuse the bet, change the odds on horse A to increase the bets on it, place a limit on how much can be bet on a horse, etc. Vincent Astier MATH / 35

27 Hedging Another option is to bet himself on horse B (with another bookmaker) to make up for the losses in case horse B wins. For instance if another bookmaker offers odds of 3 to 1 for horse B, and our bookmaker bet ex on horse B. Then: If horse A wins, we have W B = 350 x If horse B wins, we have W B = x. Both outcomes will be positive, and thus guarantee a gain for the bookmaker, if < x < 350. If he wants to know in advance how much he will gain, we must have 350 x = x, i.e. x = and the bookmaker will win e12.5, regardless of who wins the race. Vincent Astier MATH / 35

28 Hedging Conclusions: (1) To remove the uncertainty associated with unpredictable future events, identify the desired rewards (or penalties) and develop a hedging strategy by working backwards. (2) To reduce the potential loss due to an unfavorable event occurring, place a bet in favour of the event happening. Most people follow (2) in their daily lives by taking out insurance, and insurance companies also lay off bets but call it re-insuring. In the financial world this form of playing safe is called hedging. Vincent Astier MATH / 35

29 Arbitrage Definition Arbitrage means any situation, opportunity or price which allows a guaranteed profit without risk. Vincent Astier MATH / 35

30 Arbitrage: examples Example 1: Suppose we have two banks A and B, operating side by side. Bank A offers customers a 10% interest per annum on savings, while bank B offers loans to customers at an 8% interest rate per annum. We can take advantage of this situation: Borrow as much as possible from B, and immediately place it on deposit in Bank A. If, for example, one obtains a loan for e1,000,000 for a year, then at the end of the year the principal in Bank A amounts to 1, 000, 000e 0.1 = 1, 105, 171 while the loan repayment to Bank B amounts to 1, 000, 000e 0.08 = 1, 083, 287. This gives a risk-free guaranteed profit of e21,884 at the end of the year. Vincent Astier MATH / 35

31 Arbitrage In reality, arbitrage opportunities close down very rapidly. In the above example, the demand on Bank B would increase rapidly, and as a result, interest rates would quickly be adjusted until equilibrium was established. Vincent Astier MATH / 35

32 Arbitrage: examples Example 2: Two bookmakers offer different odds on a race between two horses A and B. The first bookmaker offers the odds 7 : 2 on the horse B while the second bookmaker offers odds 5 : 2 on the horse A. Mary has e160. How much should she bet on each horse in order to make a no-risk guaranteed profit? Solution. Assume Mary bets x on horse A, so she bets 160 x on horse B. After this she has no money left. Mary s position if horse A wins : x + 5x 2 Mary s position if horse B wins : (160 x) + To run no risk, we must have 7(160 x) 2 x + 5x 7(160 x) = (160 x) which gives x = 90 and so Mary gets 315. Thus her profit is e155 regardless of who wins the race. Vincent Astier MATH / 35

33 Arbitrage In any game we have: either (a) all bets are fair; or (b) there exists a betting strategy which produces a positive return independent of the outcome of the game. Vincent Astier MATH / 35

34 More examples: example 3 A bookmaker knows there will be only three bets of e40, e60 and e100 on three different horses in the same race. Determine the odds he should give on each horse so that he will make a profit of 10%. If, after the odds are fixed, a further bet of e150 is placed on the third horse and the bookmaker responds by placing a bet of ex at odds of 2 to 1 on the same horse in order to run no risk, find x and the bookmaker s profit/loss. Answer: The odds are 7 to 2 on the first horse, 2 to 1 on the second horse, and 4 to 5 on the third horse. After his bet of ex on horse 3 at 2 to 1, we have E[W B ] = x if horse 3 wins, and E[W B ] = 170 x if horse 3 loses. To run no risk, we must have x = 170 x so x = 90 and E[W B ] = 80. Vincent Astier MATH / 35

35 Exercise At a horserace a bookmaker has a guaranteed profit of e150. At a last minute, a new bet of e50 is placed on a certain horse and the bookmaker is able to place a bet on the same horse at odds of a to 1. If he bets e100 it may increase his profits by e50 and if he bets e50 it may reduce his profits by e50. (a) Find a and the odds on the horse offered by the bookmaker (b) How much should he bet in order to run no risk? What will his profit/loss be in this case? Vincent Astier MATH / 35