Introduction Random Walk One-Period Option Pricing Binomial Option Pricing Nice Math. Binomial Models. Christopher Ting.

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1 Binomial Models Christopher Ting Christopher Ting : christopherting@smu.edu.sg : : LKCSB 5036 October 14, 2016 Christopher Ting QF 101 Week 9 October 14, /42

2 Table of Contents 1 Introduction 2 Random Walk 3 One-Period Option Pricing 4 Binomial Option Pricing 5 Nice Math Christopher Ting QF 101 Week 9 October 14, /42

3 Applications of Randomness? Casting of lot I Ching (Yi Jing) Source: CARM Source: brainpickings.org Christopher Ting QF 101 Week 9 October 14, /42

4 Binary Stochastic Processes Blank coin with blue dot on one side and red dot on the other side. Equally likely for either blue or red to turn up Random variable C(ω) is a mapping into numbers. { 1, if ω is blue; C(ω) = 1, if ω is red. Christopher Ting QF 101 Week 9 October 14, /42

5 3-Period Binomial Tree D 1 D 2 D 3 U 1 D 2 D 3 D 1 U 2 D 3 U 1 U 2 D 3 D 1 D 2 U 3 U 1 D 2 U 3 D 1 U 2 U 3 U 1 U 2 U 3 0 t Christopher Ting QF 101 Week 9 October 14, /42

6 One-Dimensional Random Walk A model of random walk S t S t = S t 1 + C t. Sum of up-down moves: S t = S 0 + t i=1 C i Mean of C t E ( C t ) = ( 1) = 0 Variance of C t V ( C t ) = 1 2 (1 0) ( 1 0)2 = 1 Christopher Ting QF 101 Week 9 October 14, /42

7 Mean-Variance Analysis of 1-D Random Walk Unconditional mean of random walk E ( ) ) t t S t = E (S 0 + C i = S 0 + E ( ) C i = S0 i=1 i=1 Unconditional variance of random walk V ( ) ( ) t t ) S t = V (S 0 + C i = V C i i=1 i=1 Since C i and C j are independent for i j, ( t ) t V C i = V ( ) t C i = 1 = t. i=1 i=1 i=1 Christopher Ting QF 101 Week 9 October 14, /42

8 Conditional Mean of 1-D Random Walk At time t, S t 1 is known. Given S t 1, what is the expected value of S t? E ( S t S t 1 ) = E ( St 1 + C t S t 1 ) = E ( S t 1 S t 1 ) + E ( Ct S t 1 ) = S t 1. What is the interpretation of this conditional mean? Answer: Christopher Ting QF 101 Week 9 October 14, /42

9 Martingale Every coin toss C i is independent of every other. Intuitively, the drunken man has no memory of where he has been before. Even if all information of the past is provided, still E ( S t S t 1, S t 2,..., S 0 ) = St 1. A fundamental theorem of financial mathematics A financial market is viable (i.e., no risk-free arbitrage opportunity) if and only if there exists a probability measure under which the prices are martingales. Christopher Ting QF 101 Week 9 October 14, /42

10 Conditional Variance of 1-D Random Walk Given S t 1, what is the variance of S t? V ( ) ( ) S t S t 1 = V St 1 + C t S t 1 = V ( ) ( ) S t 1 S t 1 + V Ct S t 1 = = 1 What is the interpretation of this conditional variance? Answer: Christopher Ting QF 101 Week 9 October 14, /42

11 Where is the Random Walker After n Steps? Let n + be the number of upward moves, and n the number of downward moves. By definition, the total must add up: n + + n = n. Also, S n = S 0 + n + i=1 U i + n j=1 D 1 = S 0 + n + n. Christopher Ting QF 101 Week 9 October 14, /42

12 Multiplicative Random Walk Positive random variable Λ t { u > 1, if ω is blue; Λ t (ω) = d < 1, if ω is red. Multiplicative random walk S t = S t 1 Λ t Up to time t from time 0, the price of the asset is obtained as t S t = S 0 Λ i. (1) i=1 Christopher Ting QF 101 Week 9 October 14, /42

13 Recombinant Binomial Tree The order of movements does not matter. S t = uds t 2 = dus t 2 = S t 2. To make the binomial tree recombinant, you must have u = 1 d. Otherwise, for 100 periods, the number of end nodes S T is At each node, you compute S T K for a call option. How long does it take on the fastest supercomputer? Given Pflop/sec, which is , the time taken is = seconds 1,186,339 years!. Christopher Ting QF 101 Week 9 October 14, /42

14 One-Period Case At time 0, the underlying asset price is S 0, At time 1, { us0, if ω is blue; S 1 (ω) = ds 0, if ω is red. At time 0, construct a portfolio consisting a long position x 0 units of risky asset at S 0 per unit, and a short position in a call option at c 0 struck at K, for which ds 0 < K < us 0. The value (opposite of cash flow) of the portfolio is V 0 = x 0 S 0 c 0. At time 1, the value of the portfolio is If ω is blue, V 1 + = x 0uS 0 (us 0 K). If ω is red, V 1 = x 0dS 0. Christopher Ting QF 101 Week 9 October 14, /42

15 Toy-Model Pricing To remove uncertainty and therefore risk, set V 1 + = V 1 V + 1 = x 0uS 0 (us 0 K) = x 0 ds 0 = V 1. Solving for x 0, x 0 = us 0 K (u d)s 0. (2) Therefore, the portfolio value at time 1 is also known at time 0, and we have V 1 = V 1 + = V 1 = d(us 0 K) = S 0 dk u d u d. Christopher Ting QF 101 Week 9 October 14, /42

16 Toy-Model Pricing (Cont d) Since there is no uncertainty, by the first Principle of QF, the present value of the portfolio at time 0 must be discounted by the risk-free rate r 0 as follows: V 0 = e r 0 V 1 = e r 0 d(us 0 K). u d We shall call e r 0 the risk-free discount factor. But the portfolio value at time 0 is V 0 = x 0 S 0 c 0 Substituting in x 0 from (2), we find that the value c 0 of the call option should be c 0 = x 0 S 0 e r 0 d(us 0 K) u d = 1 e r 0 d u d (us 0 K). (3) Christopher Ting QF 101 Week 9 October 14, /42

17 Risk Neutrality In toy-model pricing, the probability p of upward movement by the underlying asset is not needed. By definition, the risk-free rate r 0 is indifferent to the up-or-down outcome. If the dollar value corresponding to the risky asset S 0 is invested in the risk-free bond, the value of this bond will become e r 0 S 0 at time 1 for sure. On the other hand, with probability p, the upward outcome is us 0, whereas the downward outcome ds 0 occurs with a probability of 1 p. Being risk neutral means that the expected return of the risky asset is the risk-free r 0. E Q 0 (S 1) S 0 = e r 0 = us 0 p + ds 0 (1 p) = e r 0 S 0. (4) Christopher Ting QF 101 Week 9 October 14, /42

18 Risk-Neutral Probability Solving Equation (4) for p, we obtain the probability p of upward movement: p = er 0 d u d. (5) The probability q of downward movement is q = 1 p = u er 0 u d. (6) Since the probability is positive, it must be that d < e r 0 < u. (7) Christopher Ting QF 101 Week 9 October 14, /42

19 Numerical Illustration (Thanks for the Mid-Term Feedback!) Suppose the up factor u is That is 25% return. For the binomial tree to be recombinant, the down factor is d = 1/u = 1/1.25 = 0.80, i.e., a decline of 20%. Let r 0 be 1%, hence e 0.01 = The risk-neutral probability of upward movement is p = = 46.67% The risk-neutral probability of downward movement is q = 1 p = = 53.33%. Christopher Ting QF 101 Week 9 October 14, /42

20 Risk-Neutral Probability in Call and Put Prices Applying the put-call parity, you can compute from (3) to obtain p 0 = ue r 0 1 u d (K ds 0). (8) The risk-neutral probability (5) is embodied in the pricing formulas for call and put options, (3) and (8), respectively. To see it, we rewrite these pricing formulas as c 0 = er 0 d u d e r 0 (us 0 K), p 0 = u er0 u d e r 0 (K ds 0 ). It is easy to find that c 0 = p e r 0 (us 0 K), p 0 = q e r 0 (K ds 0 ). Christopher Ting QF 101 Week 9 October 14, /42

21 Numerical Examples From the earlier example, we have p = 46.67%, q = 53.33%, u = 1.25, and d = Also the risk-free rate is r 0 = 1%. Suppose S 0 = $5, and the strike price K is $5.5. The call price is c 0 = e 0.01 ( 1.25 $5 $5.25 ) = $0.46. The put price is p 0 = e 0.01 ( $ $5 ) = $0.66. Christopher Ting QF 101 Week 9 October 14, /42

22 Revisiting the First Principle of QF We denote the value of the call option at time 1 as c + 1 := us 0 K for the up state, and that for the down state as c 1 = 0. In general c 1 may not be zero when the options are deep in the money. With the risk-neutral probability p, we invoke the notion of expected future value of c 1, which is denoted by E Q ( ) 0 c1, and we write c 0 = e r 0 ( pc (1 p)c 1 p 0 = e r 0 ( pp (1 p)p 1 ) = e r 0 E Q ( ) 0 c1. (9) ) = e r 0 E Q ( ) 0 p1. (10) Under the risk-neutral probability p, the fair price of option today is the expected value of its future payoff discounted by the risk free rate. Christopher Ting QF 101 Week 9 October 14, /42

23 Delta Hedging To make the portfolio value invariant to both outcomes, i.e., remove the uncertainty arising from the randomness of coin tossing, the long position in the stock is required to hedge against a short position in the call option. Delta-hedging ratio x 0 = c+ 1 c 1 S + 1 S 1, (11) where S 1 + is the value S 1 of the underlying asset at time 1 in the up state and S1 is the value for the down state. Since S 1 + = us 0, S1 = ds 0, c + 1 = us 0 K, and c 1 = 0, the delta-hedging ratio (11) indeed yields the same result as (2): x 0 = us 0 K. (u d)s 0 Christopher Ting QF 101 Week 9 October 14, /42

24 One-Period Change of Wealth The initial cash amount or wealth is denoted by W 0. You buy x 0 shares of the underlying asset of the call option at the known price of S 0. The left-over cash is M 0 := W 0 x 0 S 0, (12) which is invested in the risk-free money market. One period later, the wealth W 1 will become { W + 1 = x 0 S er 0 M 0, blue outcome; W 1 = W 1 = x 0S 1 + er 0 M 0, red outcome. It is important to note that the two outcomes are anticipated at time 0. Christopher Ting QF 101 Week 9 October 14, /42

25 Replication of Call s Payoff The replication approach is to make the cash flow W 1 at time 1 equal the option s payoff c 1 : W 1 = x 0 S 1 + e r 0 M 0 = c 1. To achieve this replication, using definition (12), we first rewrite the cash flow W 1 of the portfolio as W 1 = e r 0 W 0 + x 0 ( S1 e r 0 S 0 ) = e r 0 ( W 0 + x 0 (e r 0 S 1 S 0 ) ). We express the replication by matching each of the possible outcome: W 0 + x 0 ( e r 0 S + 1 S 0) = e r 0 c + 1, W 0 + x 0 ( e r 0 S 1 S 0) = e r 0 c 1. Christopher Ting QF 101 Week 9 October 14, /42

26 Replication of Call s Payoff (Cont d) To arrive at this equalization, we need to find the values of x 0 and W 0. Multiplying the upward outcome by p and the downward outcome by 1 p, and after adding them together, we obtain ( W 0 +x 0 e r 0 ( ps (1 p)s1 ) ) S0 = e r 0 ( pc + 1 +(1 p)c ) 1. Because of (4), the sum of the two terms, ps (1 p)s 1, is equal to e r 0 S 0. Consequently, W 0 = e r ( 0 pc (1 ) p)c 1. In view of (9), W 0 is in fact the value of the option c 0 at time 0. Christopher Ting QF 101 Week 9 October 14, /42

27 Multi-Period Generalization For each node on the binomial tree that is not an ending node, the cash flow W t at time t is { W + W t = t = x t 1 S t + + e r 0 M t 1, blue outcome; Wt = x t 1 St + e r 0 M t 1, red outcome. The money market account M t maturing at time t + 1 is M t = W t x t S t, for t = 0, 1, 2,..., T 1. It is the fund left (or needed if M t is negative) after taking a long position of x t in the risky underlying asset at the price of S t. Christopher Ting QF 101 Week 9 October 14, /42

28 Multi-Period Generalization (Cont d) The delta-hedging ratio at time t for t + 1 is x t = c+ t+1 c t+1 S + t+1 S t+1 Moreover, the risk-neutral pricing model is = c+ t+1 c t+1 (u d)s t. (13) c t = e r 0 E t ( ct+1 ). (14) Proposition W t = c t from time 0 up to time T 1. For each t of the binomial tree, the risk-neutral valuation of a pair of future payoffs is c t = e r0( pc + t+1 + (1 p)c t+1) = e r 0 E ( c t+1 ). (15) Christopher Ting QF 101 Week 9 October 14, /42

29 Proof of Proposition by Induction Assume that W t = c t is true and show that W t+1 = c t+1 also holds. First, replication means that for the up state, i.e., S + t+1 = us t, W t+1 + = x ts t ( ) er 0 W t x t S t = e r 0 W t + x t S t (u e r 0 ), (16) Substituting the delta-hedging ratio x t (13) into (16), we obtain ( c + t+1 c ) t+1)( u e r 0 W t+1 + = er 0 W t + u d = e r 0 W t + (1 p)c + t+1 (1 p)c t+1. Christopher Ting QF 101 Week 9 October 14, /42

30 Proof of Proposition by Induction (Cont d) In view of (14) and the forward induction assumption that W t = c t, we have e r 0 W t = e r 0 c t = E Q ( ) t ct+1 = pc + t+1 + (1 p)c t+1. Hence, W + t+1 = pc+ t+1 + (1 p)c+ t+1 = c+ t+1. Second, using the same method, you can also show that W t+1 = pc t+1 + (1 p)c t+1 = c t+1. Accordingly, if W t ± = c ± t, then W t+1 ± = c± t+1. We have already shown that t = 0 is true, i.e., W 0 = c 0. At time t = 1, W 1 = c 1 must also be true, and so on. Thus, the proof by forward induction is complete. Christopher Ting QF 101 Week 9 October 14, /42

31 Connection of Volatility to the Up and Down Factors The up and down factors depend on the rate of variance σ 2 of the underlying asset s return. The rate of variance σ 2 quantifies the degree of fluctuation exhibited by the return on the underlying asset. The variance for a time period t is σ 2 t, and the volatility is its square root σ t. We set u = e σ t, and d = e σ t. Note from (7) that the risk-free factor e r 0t must be smaller than the up factor u, i.e., e r 0t < e σ t. It follows that the time interval t of each period must satisfy t < σ r 0. Christopher Ting QF 101 Week 9 October 14, /42

32 A Numerical Example of Binomial Option Pricing Asset prices for all nodes S 0 = Put option s days to maturity = 15 days Since N = 3, each period is 15/3 = 5 days 5 days is t = 5/365 = 1/73 years risk-free rate r 0 = 0.25% σ = 73% u = d = $30.00 $32.68 $27.54 $35.59 $30.00 $25.29 $38.76 $32.68 $27.54 $23.22 Christopher Ting QF 101 Week 9 October 14, /42

33 Put Option Prices $0.00 $0.00 Strike price = $28 Upward probability p = 47.89% $0.86 $0.12 $1.53 $0.24 $0.00 $0.46 $2.71 $4.78 Christopher Ting QF 101 Week 9 October 14, /42

34 Random Walk Once More The multiplicative random walk (1) in the context of multi-period binomial model takes the following form: S t = u k d t k S 0. What is the probability of reaching S t = u 2 ds 0? Answer: The binomial probability for the random number Ñ of obtaining the blue dot on top when tossing the blue-red coins T times. In this case, Ñ = 2 and T = 3. Therefore the probability is P ( ( ) 3 3; Ñ = 2) = p 2 (1 p) 1 = 3p 2 (1 p). 2 Christopher Ting QF 101 Week 9 October 14, /42

35 Recall Your Pre-U Math In general, the binomial probability of the number of successes in flipping the blue-red coin, which turns up the blue dot is P ( ( ) t t; Ñ = k) = p k (1 p) t k, (17) k where the binomial coefficient is ( ) t t! := k k!(t k)!. Here Ñ is a random variable, as the number of successes is uncertain before the t tosses are completed. Christopher Ting QF 101 Week 9 October 14, /42

36 Applying Your Pre-U s Binomial Theorem With the probability mass function (17) of the random variable Ñ for the process of flipping the blue-red coin t times, we can compute the expected value at time 0 of S t given S 0 as follows: ( ) t E 0 St = u k d t k S 0 P ( t; Ñ = k). k=0 Interestingly, we find that ( ) t ( ) t E 0 St = u k d t k S 0 p k (1 p) t k k k=0 t ( ) t = S 0 (up) k( d(1 p) ) t k k k=0 = S 0 ( up + d(1 p) ) t. Christopher Ting QF 101 Week 9 October 14, /42

37 First Principle of QF Once Again Moreover, using the one-period risk-neutral probability p, i.e., (5), we obtain E Q 0 ( St ) = S0 ( (u d)p + d ) t = S0 e r 0t. The reason for using (5) is that each time period we consider here is one unit of time. To gain further insight, notice that up + d(1 p) = (u d)p + d = e r 0t Accordingly, under the single-period risk-neutral probability p in (5), the average gross return over one period is simply the forward factor e r 0. Multi-period generalization S 0 = e r 0t E 0 ( St ). (18) Christopher Ting QF 101 Week 9 October 14, /42

38 Large-Scale Binomial Probability Define a probability mass function B(x) of a discrete variable x: B(x) := P ( T ; Ñ = x) = T! x!(t x)! px (1 p) T x. (19) It is the binomial probability of x number of successes in getting the blue dot on top out of T tosses. The large number T is the result of slicing the time period t into many tiny pieces of size δt, which is a very short duration. We write T = t δt. It is noteworthy that T can be made arbitrarily large when δt is set at an arbitrarily small number. Even so, their product, i.e., T δt is a non-zero finite number t. Christopher Ting QF 101 Week 9 October 14, /42

39 Mean and Variance Under the risk-neutral probability p, the mean of the random variable Λ i in (1) is, as computed before, E Q 0 ( Λi ) = up + d(1 p) = e r 0 δt, for each i The variance of Λ i is V Q( ) ( Λ i = u e r 0 δt ) 2 ( p + d e r 0 δt ) 2 (1 p) ( (u = p e r 0 δt ) 2 ( d e r 0 δt ) ) 2 + ( d e r 0δt ) 2 = ( e r 0δt d )( u + d 2e r 0δt ) + ( e r 0δt d ) 2 = p(1 p)(u d) 2. Given that δt is small, the variance V ( Λ i ) is well approximated by V Q( Λ i ) 4p(1 p)σ 2 δt. Christopher Ting QF 101 Week 9 October 14, /42

40 Takeaways Random walk, though simple, is a great model with which to think about randomness and probability. Binomial trees are useful models for pricing options. up factor, down factor, risk-free rate, and risk neutral probability are closely related. Up and down factors are dependent on the volatility of the underlying asset. In the asymptotic limit, the binomial model converges to the Black-Scholes formula for pricing European options. Christopher Ting QF 101 Week 9 October 14, /42

41 Week 9 Assignment 1 Given the parametrization u = e σt and risk-free rate r 0 for a recombinant binomial tree, show that the risk-neutral upward probability p can be well approximated as p = er 0t d u d 1 2 ( ( r σ2) ) t. σ 2 Given t = 1/73, r 0 = 0.25%, and σ = 73% as in Slide 32, how good is the above approximation of p compared to the exact value in Slide 33? 3 Suppose δt = t, i.e., no splicing of the period. Using the same parameter values in Problem (2), compute the variance of Λ i under the risk-neutral probability p. 4 Given the binomial price tree in Slide 32, price the ATM call option. Christopher Ting QF 101 Week 9 October 14, /42

42 Week 9 Additional Exercises 1 For the one-dimensional random walk with p being the upward move by +1 and q being the downward move by 1, what is the probability for the drunken man to be at S n = x > 0 after n steps? (Hint: Let f(m) be the probability that x = m is ever reached. Then f(m + 1) = f(m)f(1).) 2 For the one-dimensional random walk in Problem 1, Let m > 0 and n > 0. What is the probability g(m, n) of reaching the point x = +m before x = n? 3 A stock analyst with some special powers is able to guess correctly the flip of a coin with 60% probability. He starts with two million dollars, and plays a game of guessing the toss with a sovereign wealth fund, which has almost infinite amount of money. What is the probability that the stock analyst will ultimately lose all the money? Christopher Ting QF 101 Week 9 October 14, /42