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THE BLACK-SCHOLES-MERTON MODEL 指導老師:王詩韻老師 學生:曾雅琪 (69936017) ,藍婉綺 (69936011)

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Presentation on theme: "THE BLACK-SCHOLES-MERTON MODEL 指導老師:王詩韻老師 學生:曾雅琪 (69936017) ,藍婉綺 (69936011)"— Presentation transcript:

1 THE BLACK-SCHOLES-MERTON MODEL 指導老師:王詩韻老師 學生:曾雅琪 (69936017) ,藍婉綺 (69936011)

2 Contents Lognormal property of stock prices The distribution of the rate of return The expected return Volatility Concept underlying the Black-Scholes-Merton differential equation Derivation of the Black-Scholes-Merton differential equation Risk-neutral valuation Black-Scholes pricing formulas Cumulative normal distribution function Implied volatilities Dividends

3 Assume : in a short period of time(Δt) : normal distribution Define μ : expected return on stock per year σ : volatility of the stock price per year W iener process

4 ItÔ’s Process ItÔ’s Lemma

5 Example 13.1 A stock with an initial price of $40 An expected return of 16% per annum A volatility of 20% per annum Ask : the probability distribution of the stock price in 6 months ? Thus, there is a 95% probability that the stock price in 6 months will lie between $32.55~$56.56

6 Lognormal distribution A variable that has s lognormal distribution can take any value between zero and infinity.

7 The nth moment of V

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10 Example 13.2 A stock where the current price is $20 An expected return of 20% per annum A volatility of 40% per annum Ask : the expected stock price, and the variance of the stock price, in 1year ?

11 The distribution of the rate of return

12 Information Define The probability distribution of the continuously compounded rate of return earned on a stock between times 0 and T. X : the continuously compounded rate of return per annum realized between times 0 and T.

13 Example 13.3 A stock with an expected return 17% per annum A volatility of 20% per annum Ask : the average rate of return(continuously compounded) realized over 3 years ? Thus, we can be 95% confident that the average return realized over 3 years will be between -7.6%~37.6%

14 The expected return

15 Expected return=E(R)= μExpected return=E(x)= Assume In fact Arithmetic mean Geometric mean

16 Example 13.4 Initial investment of a mutual fund is $100 The returns per annum report over the last five years : 15%, 20%, 30%, -20%, 25% Arithmetic mean>geometric mean Arithmetic meanGeometric mean

17 a. Estimating volatility from historical data b. Trading days vs. calendar days Volatility

18 σ : a measure of our uncertainty about the returns provided by the stock Between 15%~60% The standard deviation of the return The standard deviation of the percentage change in the stock price With the square root of how far ahead we are looking The standard deviation of the stock price in 4 weeks is approximately twice the standard deviation in 1 week.

19 Estimating volatility from historical data Define n+1 : number of observations Si : stock price at end of ith interval, with i=0,1,2……,n τ : length of time interval in years Standard error( )

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21 Trading days vs. calendar days 1. The variance of stock price returns between the close of trading on one day and the close of trading on the next day when there are no intervening nontrading days. 2. The variance of the stock price returns between the close of trading on Friday and close of trading on Monday. Research Reasonably expect The first is a variance over a 1-day period. We might reasonably expect the second variance to be three times as great as the first variance. The second variance to be, respectively, 22%, 19%, and 10.7% higher than the first variance. In fact ‧ Volatility is much higher when the exchange is open for trading than when it is closed. ‧ Practitioners tend to ignore days when the exchange is closed.

22 The number of trading days in a years is usually assumed to be 252 for stocks.

23 Standard error( )

24 Concepts derivation stock A riskless portfolio Return(portfolio)=risk-free interest rate(r) No arbitrage opportunities Example p.290 △ c=0.4 △ S →1. a long position in 0.4 shares 2. a short position in 1 call option △ c=o.5 △ S →1. an extra 0.1 share be purchased 2. for each call option sold

25 The stock price follows the process developed in CH12 with μ and σ constant The short selling of securities with full use of proceeds if permitted. There are no transactions costs or taxes. All securities are perfectly divisible. There are no dividends during the life of the derivative. There are no riskless arbitrage opportunities. Security trading is continuous. The risk-free rate of interest, r, is constant and the same for all maturities. Assumptions

26 Define f : the price of the call option or other derivative contingent on S →f must be come function of S and t. π : the value of the portfolio equation portfolio -1 : derivative : shares

27 process This equation does not involve △ z

28 △ π=rπ △ t The portfolio must instantaneously earn the same rate of return as other short-term risk-free securities. (Black-Schiles-Merton differential equation)

29 r Example 13.5 P.108_equation(5.5)

30 Application to Forward Contracts on a Stock Risk-neutral valuation

31 It is the single most important tool for the analysis of derivatives. Risk-neutral valuation They all are independent of risk preferences t: time S:the current stock price : stock price volatility rf: the risk-free rate of interest If the expected return,u,involved in the above eqution. Independent of risk preferences

32 Risk-neutral valuation All investors are risk neutral Assumption: The expected return of all investment asset is the risk-free rate of interest, r. the expected return = r Why?

33 The risk-neutral investors do not require a premium to induce them to take risks. 1. r=rf + s Risk-neutral valuation Reason: present valueexpected value Discount by r 2.2. Any cash flow

34 How to use risk-neutral valuation A derivative provides a payoff at one particular time. step1 the expected return from the underlying asset is the risk-free interest rate, r. Assume: Step2 Calculate the expected payoff from the derivative Step3 Discount the expected payoff at the risk-free interest rate.

35 Step3 Application to Forward Contracts on a Stock Verify Equation(5. 5) f = S 0 – Ke -rt A long forward contract Maturity: time T Delivery price: K A long forward contract Maturity: time T Delivery price: K step1 Calculate: the value at maturity S T - K S T : stock price at time T Step2 Calculate: the value at time 0 f = e -rT E(ST – K) = e -rT E(S T ) –Ke -rT (13.17) Equation13.4 E(S T ) = S 0 e rT (13.18) Substitute equation(13.18) into (13.17) f = S 0 – Ke -rt step3

36 European Option Pricing c= European call Price P=European put Price S 0 =the stock price at time zero K=the strike price r=risk-free rate =stock price volatility T=time to maturity of the option N(X)= the cumulative probability distribution function for a standardized normal distribution c= European call Price P=European put Price S 0 =the stock price at time zero K=the strike price r=risk-free rate =stock price volatility T=time to maturity of the option N(X)= the cumulative probability distribution function for a standardized normal distribution Formula

37 Black-Scholes pricing formulas S T >K, the expected value of a variable is equal to S(T) S T <K,0 The probability that the option will be exercised in a risk- neutral world a. solve the differential equation(13.16 ) Deriving the Black-Scholes formulas,we can use b. use risk-neutral valuation An European call option

38 Derivation If ln V~ N(m, w) E [max(V − K, 0)] = E(V )N(d1) − KN(d2) where E : expected value Key result

39 Derivation Define g(V) as the probability density function of V. It follows that ln V ~ N(m, w) m=ln[E(V)] - w 2 /2 (13A.3) Step1 ( 13A.2)

40 Derivation Define a new variable (13A.4) Q ~ N(0,1) Denote the density function for Q, h(Q) Step2

41 Derivation Step3 Using equation(13A.4) to convert the expression on the right-hand side of equation(13A.2)from an integral over V to an integral over Q (13A.5)

42 Derivation step4 (13A.6)

43 Derivation N(X): probability that a variable with a mean of zero and a standard deviation of 1 is less than x. = = Substituting for m from equation (13A.3) Step5

44 Properties of the Black-Scholes Formulas When stock price becomes very large N(d1)  1 N(d2 )  1 N(-d1)  0 N(-d2 )  0 d1.d2 become very large N(-d1)=1-N(d1) N(-d2)=1-N(d2 )

45 price p approaches zero So- Ke -rt very similar to a forward contract with delivery price K.

46 When volatility approaches zero the stock is riskless and the price grows at rate r to. S 0 > S 0 < ln(S 0 /K)+rT >0 d1,d2  + ln(S0/K)+rT<0 d1,d2  Call price = max put price=max When  0

47 Cumulative normal distribution function

48 Example Example 13.6 The stock price 6months from the expiration of an option is $ 42 The exercise price of the option is $40 The risk-free interest rate is 10% The volatility of 20% per annum Ke -rt =40e -0.1*0.5 = 38.049 c= 42N(0.7693)- 38.049N(0.6278) p= 38.049 N(-0.6278)- 38.049N(-0.7693 ) N(0.7693)=0.7791 N(-0.7693)=0.2209 N(0.6278)=0.7349 N(0.-6278)=0.2651

49 Implied volatilities In the Black-Scholes pricing formulas the volatility of the stock price. Functions : a.Monitor the market’s opinion about the volatility of particular stock. b. From actively traded option on a certain asset,Traders use to calculate the appropriate volatility for pricing a less actively traded option on the same stock. calculated by option price observed in the market. We can’t find

50 σ=0.2c=1.76too low σ=0.3c=2.1too high σ=0.25too high European call option (No dividend) C=1.875 So= 21 K= 20 r=10% (per annum) T=0.25 Ask: σ =??  Implied volatility How to do?? Iterative search C is an increasing function of σ σ lies between 0.2 and 0.25. In this example, the implied volatilities is 0.235

51 Dividends Assumption : a.The amount and timing of the dividends during the life of an option can be predicted with certainty. b.The date on which the dividend is paid should be assumed to be the ex-dividend date. c.On this date the stock price declines by the amount of the dividend.

52 European options Assumption Stock price riskless componentrisky component 1. the present value of all the dividends during the life of the option discounted from the ex-dividend to the present at the risk-free rate. 2.By the time the option matures, the dividends will have been paid and the this part will no longer exist. 1. S 0 is equal to the risky component of the stock price. 2. is the volatility of the process followed by the risky component

53 Example 13.8 European call option on a stock Ex-dividend dates in two months and five months The dividend is expected to be $0.5 The current share price is$40 The exercise price is $40 The risk-free interest rate is 9% The volatility of 30% per annum Time maturity is six months ASK:European call option price??

54 Calculate the present value of the dividends 0.5e -0.1667*0.09 + 0.5e -0.4167*0.09 = 0.9741 2/12 5/12 So minus the present value of the dividends 40 - 0.9741=39.0259 Use Black-Scholes pricing formulas d1=0.2017 N(d1)=0.58 d2=-0.0104 N(d2)=0.4959 Call price: 39.0259 × 0.58 - 40e -0.5*0.09 ×0.4959 =3.67

55 American Option The dividends corresponding to these times will be denoted by D1,D2 …,Dn,respectively. Assumption When there are dividends, it is optimal to exercise only at a time immediately before the stock goes ex-dividend. n ex-dividend dates are anticipated and they are at times t1,t2 …tn, with(t1<t2< …<tn)

56 American Option (9.5) Final ex-dividend date(tn ) The option is exercised S(tn)-K The option is not exercised S(tn)-Dn

57 American Option If then It cannot be optimal to exercise at time tn It is not optimal to exercise immediately prior to time ti It can be optimal to exercise at time tn If Very large T-tn is small


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