Numerical Methods for Option Pricing

Chapter 9 Numerical Methods for Option Pricing Equation (8.26) provides a way to evaluate option prices. For some simple options, such as the European call and put options, one can integrate (8.26) directly and obtain a closed-form solution as in (8.18) or (8.24). However, for more complicated options, it may not be easy to do the integration and one has to resort to numerical means in finding the option prices. There are three common methods for evaluating option prices numerically: the binomial method, the Monte Carlo method, and the finite difference method. We begin with the binomial method. 1 One-step Binomial Model Consider a very simple situation where a stock price is currently $20 and it is known that at the end of three months the stock price will be either $22 or $18. We suppose that the stock pays no dividends and that we are interested in valuing a European call option with exercise price $21 expiring in three months. This option will have one of the two values at the end of the three months. If the stock price turns out to be $22, the value of the option will be $1; if the stock price turns out to be $18, the value of the option will be zero, see Figure 1. p Stock price = $22 Option price = $1 Stock price = $20 1 p 3 months Figure 1. One step binomial method. Stock price = $18 Option price = $0 Consider a portfolio consisting of a long position in shares of the stock and a short position in one call option, i.e. Π(t) = V + S, where V is the option price. If the stock price moves up from 20 to 22, the value of the shares is 22 and the value of the option is 1, so that the total value of the portfolio is 22 1. If the stock price moves down from 20 to 18, the value of the shares is 18 and the value of the option

98 MAT4210 Notes by R. Chan is zero, so that the total value of the portfolio is 18. The portfolio is riskless if the value of is chosen so that the final value of the portfolio is the same for both of the alternative stock prices. This means that 22 1 = 18, or = 0.25. So the value of the portfolio at expiry is Π(T ) = 22 0.25 1 = 4.5 = 18 0.25. Riskless portfolio must, in the absence of arbitrage opportunities, earn the riskfree rate of interest. Suppose that in this case the risk-free rate is 12% per annum. It follows that the value of the portfolio today (where t = 0) must be the present value of 4.5, or Π(0) = 4.5e 0.12 0.25 = 4.367. The value of the stock price today is known to be 20. If the option price is denoted by V, the value of the portfolio today is also given by Π(0) = 20 0.25 V = 5 V. It follows that 5 V = 4.367, or V = 0.633. In general, if the stock price S moves up to u S or down to d S with u > 1 and d < 1, and the corresponding payoffs for the option are V u and V d respectively, then we have Π(T ) = Su V u = Sd V d, or see Figure 2. = V u V d Su Sd, (1) p Stock price = us Option price = V u Stock price = S e rt pv u + (1 p)v d 1 p Figure 2. One step binomial method. T Stock price = ds Option price = V d Denoting the risk-free interest rate by r, the present value of the portfolio must be Π(0) = (Su V u )e rt = (Sd V d )e rt. Since the cost of setting up the portfolio at t = 0 is (S V ), where V is the option price now, it follows that S V = Π(0) = (Su V u )e rt = (Sd V d )e rt.

Numerical Methods for Option Pricing 99 Solving for V and using (1), we obtain V = e rt (pv u + (1 p)v d ), (2) where Notice that from (2), we have p = ert d u d. (3) pv u + (1 p)v d = V e rt which is the expected option price at time T when there is no arbitrage. Since V u is what the option will be worth when the stock price goes to us, and V p is what the option will be worth when the stock price goes to ds, the variable p here can be interpreted as the probability of an up movement in the stock price, where as the quantity 1 p is then the probability of a down movement in the stock price. More precisely, if the probability of S going up to us is assumed to be p and there are no arbitrage, then the expected option price at time T is precisely given by the fair value pv u +(1 p)v d. Moreover, in this case, the expected stock price at time T is also given by E(S T ) = psu + (1 p)sd = ps(u d) + Sd. Using (3), it simplifies to E(S T ) = Se rt, (4) which means that the stock price grows on average at the risk-free rate (and hence is the option). Setting the probability of the up movement equal to p is therefore equivalent to assuming that the return on the stock equals the risk-free rate. This is the same idea as the risk-neutrality we mentioned in 8.6, cf. (4) and (8.28). This risk-neutrality interpretation can be used to find the value of the option faster. In the previous example, p must satisfy the equation 22p + 18(1 p) = E(S T ) = 20e 0.12 0.25, or p = 1 4 ( 20e 0.12 0.25 ) = 0.6523. (5) Thus, the expected payoff for the option is 0.6523 1 + 0.3477 0 = 0.6523. Discounting back to today at the risk-free rate, the value of the option today is 0.6523e 0.12 0.25 = 0.633. 2 Two-step Binomial Model We can extend the analysis above one step further to a two-step binomial model. Consider a call option with a strike price $21 and expiring half year from now. Assume that the current stock price is $20, and the risk-free interest rate is 12% per annum. We break the time interval of half a year into two time-steps, each of length 3 months. In each of the two time-steps, we assume the stock price can only go up 10% or down

100 MAT4210 Notes by R. Chan 10%. In Figure 3, the upper number of each node shows the stock price, which we can construct easily from the given information. For example, if the stock price is $22 at the 3-month time, then it can only be $24.2 or $19.8 at the 6-month time. Note that because $22 d = $19.8 = $18 u, we only have three nodes at the second time-step, and not four. Next we compute the probability of p for each time-step according to (5). Using the data, u = 1.1, d = 0.9, r = 0.12, time-step=0.25, we get p = 0.6523. Since we have the stock prices for the nodes D, E and F at the expiry date, we can obtain the payoffs of the option at these nodes. They are given by the lower numbers at the nodes. With the payoffs of the option given at nodes D and E, we can use (2) to compute the value of the option at node B: e 0.12 0.25 (0.6523 3.2 + 0.3477 0) = 2.0257. Similarly, using the value of the option at nodes E and F, we can find the values of the option at the node C (which obviously is equal to 0). Now we repeat the computation once again, using the value of the options at nodes B and C, we get the option price at node A: e 0.12 0.25 (0.6523 2.0257 + 0.3477 0) = 1.2823. 20 1.2823 A 0.6523 0.3477 22 2.0257 18 0.0 B C 0.6523 0.3477 0.6523 0.3477 3 months 3 months Figure 3. Two-step binomial method. 24.2 D 3.2 19.8 E 0.0 16.2 F 0.0 In general, if the stock prices move up or down by a factor u or d respectively in each time-step δt, then we have the binomial tree of Figure 4. As in the above example, we find V u = e rδt (pv uu + (1 p)v ud ), V d = e rδt (pv ud + (1 p)v dd ), where δt is the length of one time-step. Then V = e rδt (pv u + (1 p)v d ), or V = e 2rδt ( p 2 V uu + 2p(1 p)v ud + (1 p) 2 V dd ).

Numerical Methods for Option Pricing 101 S V p 1 p Su V u Sd V d p 1 p p 1 p Su 2 V uu Sud V ud Sd 2 δt δt V dd Figure 4. Two-step binomial method in the general case. 3 The Binomial Tree The binomial models described above can be generalized to an M-step binomial model. We first construct a binomial tree of possible asset prices, and then evaluate the option prices backward one step at a time as we did in the 2-step binomial method. Assume that at time t = 0 we know the asset price S0 0 = S. Then at the next time-step δt there are two possible asset prices, S0 1 = ds0 0 and S1 1 = us0 0. Given u, d, r and δt, we can compute p as in (3): p = erδt d u d. (6) At the following time, 2δt, there are three possible asset prices, S0 2 = d 2 S0, 0 S1 2 = uds0 0 and S2 2 = u2 S0 0, and so on, see Figure 5. In general, at the m-th time mδt there are (m + 1) possible asset prices, S m n = d m n u n S 0 0, n = 0,..., m. Thus, at the final time Mδt there are M + 1 possible asset prices. Example 1. If we set u = 1/d, then the possible stock prices can be simplified further to: Sn M = dm 2n S0 0 = u2n M S0 0, n = 0,..., M. Next we apply the payoff function for the option on S M n. For a put we have, V M n = max(e S M n, 0), n = 0,..., M, where E is the exercise price and Vn M denotes the n-th possible value of the put at time-step M where the asset value is Sn M. For a call we have V M n = max(s M n E, 0), n = 0,..., M. We can find the expected value of the option at the time-step prior to expiry, (M 1)δt, and for possible asset price Sn M 1, n = 0,..., M 1, since we know that

102 MAT4210 Notes by R. Chan S0 0 V 0 0 p 1 p δt S1 1 V 1 1 S0 1 V 1 0 p 1 p p 1 p δt S2 2 V 2 2 S1 2 V 2 1 S0 2 V 2 0 Figure 5. m-step binomial method. the probability of an asset priced at Sn M 1 moving to Sn+1 M during a time-step is p, and the probability of it moving to Sn M is (1 p). Using the risk-neutral argument we can calculate the value of the option at each possible asset price for time-step (M 1). Similarly, this allows us to find the value of the option at time-step (M 2), and so on, until t = 0. This gives us the value of the option at the current time. For a European option e rδt Vn m = pvn+1 m+1 m+1 + (1 p)vn, n = 0,..., m, or V m n = e rδt ( pv m+1 n+1 ) m+1 + (1 p)vn, n = 0,..., m. 4 Other Options by Binomial Methods To evaluate the option prices of other options, say American options, we use the same idea. First, similar to the European options, we construct a binomial tree of possible asset prices. Consider the situation at time-step m and at asset price Sn m+1. The option can be exercised prior to expiry to yield a profit determined by the payoff function. For a put Vn m = max(e Sm n, 0), n = 0,..., m, and for a call V m n = max(s m n E, 0), n = 0,..., m. If the option is retained, its value is, as in the European case, Vn m = e rδt ( pvn+1 m+1 ) m+1 + (1 p)vn, n = 0,..., m. The value of the American option is the maximum of two choices: the choice of exercising the option and the choice of keeping the option, i.e. for a put and for a call, n = 0,..., m. V m n = max ( max(e S m n, 0), e rδt (pv m+1 n+1 V m n = max ( max(s m n E, 0), e rδt (pv m+1 n+1 m+1 + (1 p)vn ) ) m+1 + (1 p)vn ) )

Numerical Methods for Option Pricing 103 Example 2. Let us compute an American put by a 2-step binomial method. We will use the same data as given as in 1, i.e. S 0 = 20, E = 21, T = 0.5, δt = 0.25, and u = 1.1 and d = 0.9. Thus by (3), p = 0.6523. The binomial tree is given in Figure 6. The option prices at the expiry nodes D, E, and F are clear. Next consider node B. If we exercise, we get 0. If we do not exercise, then the option price is the same as the European one: e 0.12 0.25 (0.6523 0 + 0.3477 1.2) = 0.4160. Clearly, in this case, we do not exercise, and the option price is 0.4160. We illustrate this in Figure 6 by underlying the maximum of the two values: 0 and 0.4160. It is similar for node C. If we exercise, we get 3. If we do not exercise, then the option price is the same as the European one: e 0.12 0.25 (0.6523 1.2 + 0.3477 4.8) = 2.4444. Clearly, in this case, we exercise, and the option price is 3. We illustrate this in Figure 6 by underlying the maximum of the two values: 3 and 2.4444. Finally at t = 0, the option price is given by e 0.12 0.25 (0.6523 0.4160 + 0.3477 3) = 1.1179. 20 1.1179 A 0.6523 0.3477 22 {0,0.4160} 18 B C 0.6523 0.3477 0.6523 24.2 D 0 19.8 E 1.2 3 months {3,2.4444} 0.3477 3 months 16.2 F 4.8 Figure 6. Two-step binomial method for an American option. What if the underlying stock pays dividend? We can use the same idea to compute the option price. Just remember that the stock price will drop across the ex-dividend date t d, but the option price will be continuous across t d, see Proposition 4.7. Example 3. Let us compute the same American put as in Example 2 by a 2-step binomial method, but we assume that the stock pays dividend at the end of the first 3 months and the dividend yield is 5%. Since all other parameters are unchanged, we still have p = 0.6523. The binomial tree is given in Figure 7. Since the stock pays dividend at the end of 3 months, we break this particular ex-dividend date into two parts, one is t d, with

104 MAT4210 Notes by R. Chan nodes B and C, and the other part is t + d, with nodes D and E. The stock price at nodes D and E are obtained by reducing the stock prices at nodes B and C by 5%, e.g. at Node D, the stock price is 20.9 = 22 0.95. After the ex-dividend date, the stock prices will move according to a normal binomial tree, i.e. it goes up or down according to u or d. With this, we construct the stock prices at nodes F, G and H. Once we have the stock prices at the terminal nodes, then the remaining computation is pretty much the same as in previous examples. We first compute the option prices at the expiry nodes. Next consider node D. If we exercise, we get 0.1. If we do not exercise, then the option price is the same as the European one: e 0.12 0.25 (0.6523 0 + 0.3477 2.19) = 0.7390. Clearly, in this case, we do exercise, and the option price is 0.7390. It is similar for node E where we would exercise to get 3.9. At t d, the option price will be the same as t+ d. Of course we can do the same thing asking if we should hold or exercise the option at t d. For American put, we can easily see that we will not exercise at t d, for if we do, we get higher return if we exercise at t + d, see nodes C and E. Once we get the option prices at nodes at t d (in our case, they are nodes B and C), we can continue to back-compute the option price until we get back to t = 0. In this example, at t = 0, the option price is given by e 0.12 0.25 (0.6523 0.7390 + 0.3477 3.9) = 1.7839. 20 1.7838 A 0.6523 0.3477 B D 22 20.9 {0,0.7390} {0.1,0.7390} C E 18 17.1 {3,3.9} {3.9,3.2793} 0.6523 0.3477 0.6523 0.3477 3 months t d t + d 3 months Figure 7. Two-step binomial method for an American option. 22.99 F 0 18.81 G 2.19 15.39 H 5.61 As expected, the put option price is higher at t = 0 than the one where the stock does not pay dividend (see Example 2 where V = 1.1179). The price is higher at t = 0 in anticipation that the stock will pay dividend before the expiration time T which will drive down the stock price. 5 Determination of the Parameters In 1 4, we assume we know u and d, and then we calculate p using (6) and then V accordingly. In general, the parameters p, u and d are unknown, and have to be

Numerical Methods for Option Pricing 105 chosen so that they give the correct values for the mean and variance of stock price changes during a time interval of length δt. Under the risk-neutral assumption, the expected return from a stock is the risk-free interest rate, r. Hence at the end of the time interval δt, the expected value of the stock E(S δt ) is Se rδt, see (4) or Corollary 8.5. It follows that Se rδt = E(S δt ) = psu + (1 p)sd, (7) or It is nothing but just (3). By Corollary 8.5 again, Since Var(S δt ) = E(S 2 δt ) (E(S δt)) 2, and it follows that or e rδt = pu + (1 p)d. (8) Var(S δt ) = S 2 e 2rδt (e σ2 δt 1). E(S 2 δt) = p(us) 2 + (1 p)(ds) 2, S 2 e 2rδt (e σ2 δt 1) = pu 2 S 2 + (1 p)d 2 S 2 S 2 (pu + (1 p)d) 2, e 2rδt+σ2 δt = pu 2 + (1 p)d 2. (9) Equations (8) and (9) impose two conditions on p, u and d. We still need one more equation to determine them. One can equate the third moment of S δt, but it will lead to extremely difficult expressions. A popular choice for the third condition is Solving (8), (9) and (10), we obtain u = 1 d. (10) where d = A A 2 1, u = A + A 2 1, p = erδt d u d, (11) A = 1 2 ( e rδt + e (r+σ2 )δt ). (12) Another popular choice is to set p = 1/2. (13) This together with (8) and (9) leads to ( ) ( ) d = e rδt 1 e σ2 δt 1, u = e rδt 1 + e σ2 δt 1. Assumptions (10) and (13) may give different pricing of the options, but one can show that if δt 0, then the two pricings should be the same.

106 MAT4210 Notes by R. Chan M = (T t)/δt T t P DE 16 32 64 128 256 0.25 4.8511 4.8511 4.8511 4.8511 4.8511 4.8511 0.50 4.7048 4.7046 4.7047 4.7047 4.7048 4.7048 0.75 4.5636 4.5626 4.5632 4.5634 4.5634 4.5636 1.00 4.4304 4.4292 4.4300 4.4300 4.4300 4.4304 Table 1. Binomial Method (u = 1/d) for European put with E = 10, S = 5, r = 0.06, σ = 0.3. Example 4. Consider a European put with E = 10, S = 5, r = 0.06 and σ = 0.3. Table 1 shows the comparison of the binomial method with u = 1/d and the PDE solution of the Black-Scholes equation for various expiry time with different time-steps. We see that the results are more accurate if δt is smaller, i.e. M is getting larger. We should note that this is the same with solving PDE using numerical methods. Example 5. Consider an American put with E = 10, S = 9, r = 0.06 and σ = 0.3. Table 2 shows the comparison of the binomial method with u = 1/d and the PDE solution to the Black-Scholes equation for various expiry time with different time-steps. Again we see that the results are more accurate if δt is smaller. M = (T t)/δt T t P DE 16 32 64 128 256 0.25 1.1264 1.1311 1.1238 1.1259 1.1252 1.1260 0.50 1.2569 1.2494 1.2591 1.2535 1.2551 1.2545 0.75 1.3551 1.3550 1.3555 1.3562 1.3551 1.3545 1.00 1.4339 1.4399 1.4310 1.4373 1.4336 1.4347 Table 2. Binomial Method (u = 1/d) for American put with E = 10, S = 9, r = 0.06, σ = 0.3. The binomial method is extremely memory efficient for options with only one underlying asset: although there are O(M 2 ) nodes, the memory required grows only linearly with the number of time-steps, i.e. it is of order O(M) only. The reason is that we can re-use the memory in the (m + 1)-st time-step for the m-th time-step. Thus the largest memory requirement is at the last time-step M, where we require 2(M +1) memory for the stock prices and the option prices. Since there are O(M 2 ) nodes, and each node requires O(1) calculations, the execution time grows quadratically with the number of time-steps, i.e. of order O(M 2 ). For an option whose price depends on d stock prices, the binomial method will require O(M d ) memory locations and O(M d+1 ) calculations. Thus if d is large, it is not an efficient method or one needs a very fast computer to compute such options. It is interesting to note that about 10% of the world top 500 fastest computers are installed in financial institutions across the world to compute option prices. 6 Monte Carlo Method Monte Carlo method is another name for simulation method. Here we try to simulate the stock price at the expiry date T according to the log-normal process (7.7). If we know the stock price S(T ) at T, then we know the option price c(s, T ) at T. By discounting that back to the current time t, we know the option price c(s, t) at t.

Numerical Methods for Option Pricing 107 Clearly, if we only simulate one price at T, the option price will not be accurate at all. We need to simulate many many stock prices at T to get enough samples of c(s, T ) at T. Then we take the average of these samples to get a good estimate of the expected value of c(s, T ). By discounting back to t, then we get the expected value of c(s, t). By (7.7) and the fact that X T t N (0, T t), we have S T = S t e (µ σ2 /2)(T t)+σε T t, where S t is the current asset price and ε is a random variable drawn from a standardized normal distribution N (0, 1). However, in a risk-neutral world (or when we are only interested in computing the option prices), µ is taken as the interest rate r, i.e., S T = S t e (r σ2 /2)(T t)+σε T t, (14) see (8.40). Using this the stock prices can be simulated between t and T. In general, we fix a time-step δt = (T t)/m and the number of paths N, which usually goes from 10,000 to 1,000,000, depending on the accuracy we want. Then we generate the j-th path from t to T : S j i = Sj i 1 e(r σ2 /2)δt+σε j i δt, i = 1,..., M, j = 1,..., N, (15) where ε j i are random numbers distributed as N (0, 1). In Figure 8, we show ten simulated stock price paths using (15), with S t = 100, t = 0, T = 50, and δt = 1. Note that for each j = 1,..., N, S j M is one sample of the stock price at time T, from which one can calculate the payoff of the option. 150 140 130 120 110 100 90 80 70 0 5 10 15 20 25 30 35 40 45 50 Figure 8. Ten simulated stock prices. Note that if we are only interested in evaluating European options, there is no need to know what the stock prices are in between t and T ; we only need to know the stock price at T. We can modify (15) to get it directly: S j T = Sj t e (r σ2 /2)(T t)+σε j T t, j = 1,..., N,

108 MAT4210 Notes by R. Chan i.e. for each ε j sampled, we can generate one sample of S j T. From each of them, a payoff for the option can be calculated at expiry. For example, if it is a European call option, we can compute V j = max(s j T E, 0), j = 1,..., N. Then the expected value of the payoff can be estimated as the arithmetic average of these payoffs, i.e. E(V T ) = 1 N N V j. j=1 The current value of the option can be calculated by discounting the expected payoff value, i.e., V = e r(t t) E(V T ). Example 6. Consider a European put with E = 10, S 0 = 5, r = 0.1 and σ = 0.4. For computing European options, one can use (14) to generate S T directly from S 0, and there is no need to partition the time interval [0, T ]. Table 3 shows the comparison of the Monte Carlo method and the PDE solution of the Black-Scholes equation for different expiry time and numbers of paths. We see that the results are more accurate when we increase the number of paths N. T t P DE N = 10 N = 10 2 N = 10 3 N = 10 4 N = 10 5 0.25 4.7533 4.6187 4.7683 4.7748 4.7568 4.7504 0.50 4.5199 4.1824 4.4687 4.4644 4.5356 4.5221 0.75 4.3121 3.6242 4.0504 4.3016 4.2964 4.3077 1.00 4.1292 4.7011 4.2652 4.2282 4.1485 4.1231 Table 3. Monte Carlo method for European put with E = 10, S 0 = 5, r = 0.1 and σ = 0.4. The number of simulation trials carried out depends on the accuracy required. If K independent trials are carried out as described above, it is usual to calculate the standard deviation as well as the mean of the option prices given by the simulation trials. Denote their mean by µ and the standard deviation by σ. The Central Limit Theorem states that the option price V should be distributed as N ( µ, σ2 K ). Hence a 95% confidence interval for V is given by µ 1.96 σ K < V < µ + 1.96 σ K, or { Prob µ V < 1.96 σ } = 0.95, K see (6.7). This shows that our uncertainty about the value of the option is inversely proportional to the square root of the number of trials K. That would mean that to double the accuracy of the simulation, we must quadruple the number of trials; and to increase the accuracy by a factor of 10, the number of trials must be increased by a factor of 100. Example 7. Consider the European put in Example 6. Now we study how the number of trials K affects the accuracy of simulation. Here we focus on the case with N = 1000

Numerical Methods for Option Pricing 109 and T t = 0.25. First we perform K = 10 independent trials. The mean and the standard deviation of the 10 discounted payoffs are given by µ = 4.7554 and σ = 0.0393, with a 95% confidence interval for the option price V : 4.7311 < V < 4.7798. Note that the simulation s estimate of the option value is accurate to 2 decimal places, compared to the PDE solution V = 4.7533 in Table 3. To increase the accuracy by a factor of 10, i.e. to increase the significant digit by 1, the number of trials must be increased by a factor of 100, i.e., K = 1000. Thus, we then carry out K = 1000 independent trials and obtain the mean and the standard deviation of the discounted payoffs: µ = 4.7535 and σ = 0.0323, with a 95% confidence interval for the option price V : 4.7515 < V < 4.7555. Monte Carlo simulation tends to be numerically more efficient than other procedures when there are three or more underlying assets that the option price is dependent on. This is because the time taken to carry out a Monte Carlo simulation increases approximately linearly with the number of unknown variables, i.e. it is of order O(MN). In contrast, the time taken for most other procedures increases exponentially with the number of unknown variables. For example, the cost of the binomial method and the finite difference method both increase like O(M d+1 ) where d is number of underlying assets. Monte Carlo simulation also has the advantage that it provides a standard error for the estimates that are made. It is an approach that can accommodate complex payoffs and complex stochastic processes. It can be used when the payoff depends on some function of the whole path followed by a variable, not just its terminal value. For example, we can use it to compute barrier options easily. The limitations of the Monte Carlo simulation are that it converges slowly, and it is used normally for European options, for it is very difficult to use for American options. In fact, in computing American options, one may have to generate and remember all the intermediate stock prices in all the paths, i.e. S j i for 1 i M and 1 j N in (15). For these prices are to be used later to compute the price of exercising the options pre-maturely and the price to hold onto the options till the next time-step. The price to hold onto the options till the next time-step is called the continuation price, which is quite difficult to estimate accurately from the sampled prices. 7 Finite Difference Approximation Besides the Black-Scholes equation, in practice, there are many other options that can be modeled by partial differential equations too. Unfortunately not many of them have closed-form solutions as in (8.18) or (8.24). In this section, we discuss how to solve partial differential equations by finite difference approach. The main idea is to approximate the differential operators in the differential equation by difference operators.

110 MAT4210 Notes by R. Chan Given a function u(x), using Taylor s expansion, we have u(x + δx) = u(x) + δx u (x) + δx2 2 u (x) + δx3 6 u (x) + O(δx 4 ), (16) u(x δx) = u(x) δx u (x) + δx2 2 u (x) δx3 6 u (x) + O(δx 4 ). (17) If δx is small, we can approximate the derivative u (x) at x by the backward difference : du(x) u(x) u(x δx) = + O(δx), (18) dx δx or more accurately, by the first order central difference : du(x) dx u(x + δx) u(x δx) = + O(δx 2 ). (19) 2δx From (16) and (17), we can also approximate the second order derivative u (x) at x by the second order central difference : d 2 u(x) dx 2 = u(x + δx) 2u(x) + u(x δx) (δx) 2 + O(δx 2 ). (20) Suppose we are to solve the Black-Scholes equation for a European option V (S, t), V (S, t) t + 1 2 σ2 S 2 2 V (S, t) V (S, t) S 2 + rs rv (S, t) = 0, (21) S with given final condition V (S, T ) (for 0 < S < ) and boundary conditions V (0, t) and V (, t) (for 0 < t < T ). Since it is impossible to work with S = numerically, we first replace the boundary S = by S = S max where S max is a sufficiently large stock price. Then we replace the condition V (, t) by V (S max, t), see Figure 9. Next we partition the solution domain [0, S max ] [0, T ] by grid lines. More precisely, we partition [0, S max ] into N equal sub-intervals, each of length δs; and [0, T ] into M equal sub-intervals, each of length δt, see Figure 9. Let S j = jδs and t i = iδt for 0 j N and 0 i M. Note that for 0 i M and 0 j N, V (S 0, t i ), V (S N, t i ) and V (S j, t M ) are given boundary and final conditions. They are known numbers. Our problem is to find V (S j, t i ) for 0 < j < N and 0 i < M, which are the option prices inside the solution domain. Our idea of finding them is to replace the derivatives in (21) by the finite differences in (18) (20). This reduces the partial differential equation to a difference equation. We then solve the difference equation in a time-marching manner, marching back from time T to 0, at one δt a time. Let us illustrate that by computing V (S j, t M 1 ) for 1 j N. Let us consider (21) at the point (S j, t M ) and apply a backward difference (18) at t M and central differences (19) and (20) at S j. Then we have V (S j, t M ) V (S j, t M 1 ) + 1 δt 2 σ2 Sj 2 V (S j+1, t M ) V (S j 1, t M ) + rs j 2δS V (S j+1, t M ) 2V (S j, t M ) + V (S j 1, t M ) δs 2 rv (S j, t M ) = 0, 0 < j < N. (22) Notice that in (22), all the quantities are known from the boundary and final conditions except V (S j, t M 1 ). Hence we can compute V (S j, t M 1 ) for all 0 < j < N.

Numerical Methods for Option Pricing 111 Figure 9. Solution domain of European options and the grids. Once we have obtained all the option prices at the time step t M 1, we can march one δt backward in time to obtain V (S j, t M 2 ) for all 0 < j < N. For general 0 < i M, applying differencing on (21) at (S j, t i ) will lead to V (S j, t i ) V (S j, t i 1 ) + 1 δt 2 σ2 Sj 2 V (S j+1, t i ) V (S j 1, t i ) + rs j 2δS V (S j+1, t i ) 2V (S j, t i ) + V (S j 1, t i ) δs 2 rv (S j, t i ) = 0, 0 < j < N. (23) With V (S j, t i ) known for all S j, we can compute V (S j, t i 1 ) by (23). By repeating the process M times, we can obtain the option prices V (S j, t 0 ) = V (S j, 0) at the current time for all S j. Obviously, the accuracy of the solution depends on the size of δt and δs. From (18) (20), we see that the error in approximating the partial differential equation (21) by the finite difference equation (23) is O(δt) and O(δS 2 ). In order to balance the two errors, one would choose δt = O(δS 2 ). That would imply that δt will be a very small number and hence M will be very big (of O(N 2 )). That would mean that we will have to do many time-marching steps. Unfortunately, there is no way to overcome this. In fact, one can show that α δt/(δs) 2 has to be less than a certain constant α 0 in order that V (S j, t i ) of (23) converges to the true solution of (21) when δt and δs go to 0. If α α 0, then V (S j, t i ) actually diverges. For some problems, the condition α < α 0 will restrict δt to a very small number, and hence the method may be too slow to find the solution. There are better finite difference schemes that can circumvent this problem but they will require more computational costs. Example 8. Consider a European put option with the exercise price E = 10 and T = 1/2 year. Assume r = 0.05 and σ = 0.2. Using the method, we can obtain the

112 MAT4210 Notes by R. Chan approximate option prices for different asset values S 0, which are shown in Table 7. Here we choose S max = 25, and the number of intervals for [0, T ] is M = 60, i.e., every three days. The step-size δs is determined by α. The exact solution is obtained by the closed-form formula (8.24). We see that the method diverges when α > 0.1, and the solution becomes oscillatory, see Figure 10. S 0 α = 0.02 α = 0.1 α = 0.13 Exact 0.00 9.7531 9.7531 9.7531 9.7531 2.00 7.7530 7.7530 7.7530 7.7531 4.00 5.7530 5.7530 5.7530 5.7531 6.00 3.7532 3.7531 3.7531 3.7532 8.00 1.7996 1.7985 1.7986 1.7987 10.00 0.4427 0.4420 0.4430 0.4420 12.00 0.0492 0.0483 0.0486 0.0483 14.00 0.0029 0.0027 0.0027 0.0028 16.00 0.0001 0.0001 0.0113 0.0001 17.00 0.0000 0.0000 0.0280 0.0000 18.00 0.0000 0.0000 0.4041 0.0000 19.00 0.0000 0.0000 2.2761 0.0000 20.00 0.0000 0.0000 3.5420 0.0000 Table 4. E = 10, r = 0.05, σ = 0.2, M = 60, 6 months to expiry 10 5 Option Price 0 5 0 5 10 15 20 25 S 0 Figure 10. Option price is oscillatory when α > α 0. The finite difference method is extremely efficient for options with only one underlying asset: its computational cost is of O(NM). However, it also has the curse of dimension in that the cost is O(MN d ) for an option whose price depends on d stocks. The time-marching scheme (23) for computing European option prices can be adapted easily to compute the American option prices. The main idea is that once V (S j, t i ) is computed by (23), we have to compare it with the exercise price at (S j, t i )

Numerical Methods for Option Pricing 113 to determine the true value of the option at (S j, t i ). For example, if we are computing the call option price, and c h (S j, t i ) is the price computed by (23), then c(s j, t i ) = max{c h (S j, t i ), S j E}. Example 9. Consider the American put option with parameters the same as in Example 8. Using the above method, we compute the values of the put option for asset prices from 0 to 16. The results are shown in Table 5. S 0 American (α = 0.1) European (α = 0.1) 0.00 10.0000 9.7531 2.00 8.0000 7.7530 4.00 6.0000 5.7530 6.00 4.0000 3.7531 8.00 2.0000 1.7985 9.00 1.0652 0.9878 10.00 0.4651 0.4420 12.00 0.0497 0.0483 14.00 0.0027 0.0027 16.00 0.0001 0.0001 Table 5. E = 10, r = 0.05, σ = 0.2, M = 60, 6 months to expiry

114 MAT4210 Notes by R. Chan