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Basket options 125 Now, instead of sampling from N0 I, an importance sampling distribution N I is used. Thus c =e −r T−t x 0 E Z∼NI n i=1 w i e i √ T−t bZ i − K x 0 + e 1 2 − Z (6.38) A good and stratification variable is now chosen. Observe that if n i=1 w i i √ T −t bZ i is small, then n i=1 w i e i √ T−t bZ i ≈ exp n i=1 w i i √ T −t bZ i since n i=1 w i = 1. Therefore, the aim is to find a good and stratification variable for an option with price c g , where c g = e −r T−t x 0 E Z∼NI exp n i=1 w i i √ T −t bZ i − K x 0 + e 1 2 − Z Now define c i = w i i . Then c g = e −r T−t x 0 E Z∼NI exp √ T −t c bZ − K x 0 + e 1 2 − Z (6.39) Following the same approach as for the Asian option, a good choice of is ∗ where ∗ = ln exp √ T −t c b ∗ − K x 0 Therefore, ∗ = b c √ T −t exp √ T −t c b ∗ exp √ T −t c b ∗ −K/x 0 where exp n i=1 √ T −t c b ∗ −K/x 0 > 0. Therefore ∗ = b c (6.40) where = √ T −t exp √ T −t c c exp √ T −t c c −K/x 0 (6.41) and where >ln K/x 0 / √ T −tc c. With this choice of , and therefore ∗ ,we obtain from Equation (6.39) c g = e −r T−t x 0 E Z∼NI exp −1 √ T −t ∗ Z − K x 0 + e 1 2 ∗ ∗ − ∗ Z 126 Simulation and finance Table 6.3 Results for basket option, using naive Monte Carlo (basket) and importance sampling with post stratification (basketimppostratv2) basket a basketimppostratv2 b K c Var c c Var c v.r.r. c 1 600 8403 0881 8518 00645 (96,306) 1 660 4721 0707 4803 00492 (107,338) 1 720 2349 0514 2404 00282 (171,544) 2 600 7168 0390 7225 00390 (51,164) 2 660 2274 0283 2309 00096 (447,1420) 2 720 274 0100 287 000297 (585,1869) a 10 000 paths. b 25 replications, each consisting of 400 paths over 20 equiprobable strata. c Approximate 95 % confidence interval for the variance reduction ratio. Since this is the expectation of a function of ∗ Z only, the ideal stratification variable for the option with price c g is X = ∗ Z − ∗ ∗ √ ∗ ∗ ∼ N 0 1 (6.42) From Equation (6.38), for the original option with price c the estimator e −r T−t x 0 n i=1 w i e i √ T−t bZ i − K x 0 + e 1 2 ∗ ∗ − ∗ Z (6.43) is used, where Z ∼ N ∗ I ∗ is determined from Equations (6.40) and (6.41), and Equation (6.42) defines the stratification variable. The procedure ‘basketimppoststratv2’ in Appendix 6.7.2 implements this using post stratified sampling. Table 6.3 compares results using this and the naive method for a call option on an underlying basket of four assets. The data are r = 004 x = 5 25 4 3 q = 20 80 60 40 T = 05t = 0, and as given in Equation (6.37). Two sets of cases were considered, one with = 1 =03 02 03 04 , the other with = 2 = 005 01 015 005 . The spot price is q x = 660. 6.6 Stochastic volatility Although the Black–Scholes model is remarkably good, one of its shortcomings is that it assumes a constant volatility. What happens if the parameter is replaced by a known function of time t ? Then dX X = dt + t dB 1 t Stochastic volatility 127 so using Itô’s lemma d ln X = dX X − 2 t X 2 2X 2 dt = dt + t dB t − 1 2 2 t dt = − 1 2 2 t dt + t dB 1 t (6.44) Now define an average squared volatility, V t =1/t t 0 2 u /2du. Given that X 0 = x 0 , Equation (6.44) can be integrated to give X t = x 0 exp − 1 2 V t t + t 0 u dB 1 u = x 0 exp − 1 2 V t t + V t B 1 t Using the principle that the price at time zero of a European call with exercise time T and strike K is the discounted (present) value of the payoff in a risk-neutral world, the price is given by c = e −rT E Z∼N 01 x 0 e r−V T /2 T+ √ TV T Z −K + (6.45) Therefore, the usual Black–Scholes equation may be used by replacing the constant volatility with the average squared volatility. A more realistic model is one that models the variable t as a function of a stochastic process Y t . An example is given in Figure 6.1. For example, Fouque and Tullie (2002) suggested using an Ornstein–Uhlenbeck process (see, for example, Cox and Miller, 1965, pp. 225–9), dY = m −Y dt + dB 2 t (6.46) where the correlation between the two standard Brownian motions B 1 t and B 2 t is , and where > 0. A possible choice ensuring that t > 0is t = e Y t A rationale for Equation (6.46) is that the further Y strays from m the larger the drift towards m. For this reason, Yt is an example of a mean reverting random walk. To solve Equation (6.46), Itô’s lemma is used to give d m −Y e t =−e t dY +e t m −Y dt =−e t m −Y dt + dB 2 t +e t m −Y dt =−e t dB 2 t 128 Simulation and finance A faster mean reverting volatility process (alpha = 5) with the expected volatility 0.08 0.1 0.12 0.14 0.16 0.18 0.2 sigma(t) 012345 t Figure 6.1 An exponential Ornstein–Uhlenbeck volatility process Integrating between s and t >s gives m −Y t e t − m −Y s e s =− t s e u dB 2 u =− t s e 2u du t −s B 2 t −B 2 s =− 2 e 2t −e 2s 2 t −s B 2 t −B 2 s Now define 2 = 2 /2. Then m −Y t e t − m −Y s e s =− e 2t −e 2s t −s B 2 t −B 2 s (6.47) or Y t = e − t−s Y s + 1 −e − t−s m + √ 1 −e −2 t−s Z st where Z st are independent N 0 1 random variables for disjoint intervals s t . Putting s = 0 and letting t →, it is apparent that the stationary distribution of the process is N m 2 ,soif t = e Y t , then for large t, E t ∼ exp m + 2 /2 .To simulate an Ornstein–Uhlenbeck (OU) process in 0T put T = nh and Y j = Y jh . Then Y j = e −h Y j−1 + 1 −e −h m + √ 1 −e −2h Z j Stochastic volatility 129 where Z j are independent N0 1 random variables for j = 1n. Note that there is no Euler approximation here. The generated discrete time process is an exact copy at times 0hnhof a randomly generated continuous time OU process. The procedure ‘meanreverting’ in Appendix 6.8 simulates the volatility process e Y t . Additional plots in the Appendix show the effect of changing .As increases the process reverts to the mean more quickly. Now we turn to the pricing of a European call option subject to a stochastic volatility. The dynamics of this are dX X = dt + Y dB 1 (6.48) dY = m −Y dt + dB 1 + 1 − 2 dB 2 (6.49) Y = exp Y (6.50) where B 1 and B 2 are independent standard Brownian motions. Note the correlation of between the instantaneous return on the asset and dY that drives the volatility Y . There are now two sources of randomness and perfect hedging would be impossible unless there were another traded asset that is driven by B 2 . As it stands, in order to price a derivative of X and Y , theory shows that the drift in Equations (6.48) and (6.49) should be reduced by the corresponding market price of risk multiplied by the volatility of X and Y respectively. The resulting drift is called the risk-neutral drift. Call the two market prices X and Y . A market price of risk can be thought of in the following way. For the process X, say, there are an infinite variety of derivatives. Suppose d i and s i are the instantaneous drift and volatility respectively of the ith one. To compensate for the risk, an investor demands that d i = r + X s i for all i, where X is a function of the process X only, and not of any derivative of it. Remembering that a derivative is a tradeable asset, we notice that one derivative of X is the asset itself, so = r + X Y . Therefore the risk-neutral drift for Equation (6.48) is − X Y = r, which is consistent with what has been used previously. In the case of Y , volatility is not a tradeable asset so we cannot reason similarly; and can only say that the risk-neutral drift for Equation (6.49) is m −Y − Y . It turns out that Y = X + 1 − 2 Y = −r Y + 1 − 2 Y (Hobson, 1998) where X and 1 − 2 Y are the components arising from randomness in B 1 and B 2 . The fact that both X and Y are unknown is unfortunate and accounts for the fact that there is no unique pricing formula for stochastic volatility. Given some view on what Y should be, a derivative is priced by solving dX X = r dt + Y dB 1 dY = m −Y − Y dt + dB 1 + 1 − 2 dB 2 (6.51) Y = exp Y (6.52) 130 Simulation and finance The call option price is the present value of the expected payoff in a risk-neutral world. It is now a simple matter to modify the procedure ‘meanreverting’ to find the payoff, XT −K + , for a realized path X t Y t 0 ≤ t ≤ T (see Problem 10). If the X and Y processes are independent then the valuation is much simplified. Let c denote the call price at time zero for such an option expiring at time T. Then c = e −rT E B 2 B 1 X T −K + where X is sampled in a risk-neutral world as described above. Therefore c = e −rT E B 2 E B 1 B 2 X T −K + Since B 1 is independent of B 2 , it follows that E B 1 B 2 X T −K + is simply the Black– Scholes price for a call option, with average squared volatility V B 2 t = 1 t t 0 1 2 2 B 2 u du where 2 B 2 u is a realization of the volatility path. Therefore, an unbiased estimate of c is obtained by sampling such a volatility path using Equations (6.51) and (6.52) with = 0. This is an example of conditional Monte Carlo. If T = nh, there are usually 2n variables in the integration. However, with independence, = 0. This design integrates out n of the variables analytically. The remaining n variables are integrated using Monte Carlo. 6.7 Problems 1. Show that the delta for a European put, at time t, on an asset earning interest continuously at rate r f ,is−e −r f T−t −d r f where d r f is as given in Equation (6.22). 2. Use the procedure ‘bscurrency’ in Appendix 6.5 to price European put options on a block of 1000 shares offering no dividends, where the current price is 345 pence per share, the volatility is 30 % per annum, the risk-free interest rate is 4.5 % per annum, and the strike prices are (a) 330, (b) 345, and (c) 360 pence respectively. The options expire in 3 months time. If you have just sold these puts to a client and you wish to hedge the risk in each case, how many shares should you ‘short’ (i.e. borrow and sell) initially in each case? 3. A bank offers investors a bond with a life of 4 years on the following terms. At maturity the bond is guaranteed to return £1. In addition if the FTSE index at maturity is higher than it was when the bond was purchased, interest on £1 equal to one-half of Problems 131 the % rise in the index is added. However, this interest is capped at £030 . The risk-free interest rate is 4 % per annum and the volatility of the FTSE is 0.2 per annum. The aim is to find a fair (arbitrage-free) price for the bond V xt t for 0 ≤ t ≤4 where xt is the index value at time t, using the principle that the price of any derivative of the FTSE is the present value of the expected payoff at t = 4 years, in a risk neutral world. (a) Deduce a definite integral whose value equals V xt t . The integrand should contain the standard normal density z. (b) Since integration is over one variable only, Monte Carlo is not justified providing numerical integration is convenient. Therefore, use numerical integration with Maple to find Vx0 0. (c) After 2 years the index is standing at 18x0. What is the value of the bond now? 4. The holder of a call option has the right to buy a share at time T for price K. However, the holder of a forward contract has the obligation to do so. The derivation of the price of such a derivative is easier than that for a call option: (a) Consider a portfolio A consisting at time zero of one forward contract on the share and an amount of cash Ke −rT . Consider another portfolio B comprising one share. Show that the two portfolios always have equal values in 0T. Hence show that if a forward contract is made at time zero, its value at time t when the price of the share is x t is given by V x t t = x t −Ke −rT−t (b) Show that V X t t satisfies the Black–Scholes differential equation (6.14). (c) What is the hedging strategy for a forward contract that results in a riskless portfolio? (d) What happens if K = x 0 e rT ? 5. A bank has sold a European call option to a customer, with exercise time T from now, for a (divisible) share. The price of one share (which does not yield a dividend) is determined by dX X = dt + dB The risk-free interest rate is r and the volatility is . The following policy is used by the bank to hedge its exposure to risk (recall that it will have a payoff of X T −K + at time T): at time t = 0 it borrows 0 X 0 to purchase 0 shares, while at time t =T it purchases an additional 1 − 0 shares. Therefore, it is employing an extreme form of discrete hedging, changing its position in the shares only at the beginning and end of the option’s life. At these times the bank has decided it will use 132 Simulation and finance the deltas calculated for continuous hedging. Let C denote the total cost of writing the option and hedging it. Show that E C =c +X 0 e −r T d − d +Ke −rT d − √ T − d − √ T where c is the Black–Scholes price of the option at time zero, d = + 2 /2 T +ln x0/K √ T and d = r + 2 /2 T +ln x 0 /K √ T Plot E C −c when r = 005 = 01X 0 = 680K= 700, and T = 05 for ∈ −03 03. 6. Consider an average price Asian call with expiry time T. The average is a geometric one sampled at discrete times h 2hnh=T. Let c g denote the price of the option at time zero. Show that c g = e −rT E Z∼N 01 X 0 exp r − 1 2 2 n +1 2 h + √ ahZ n −K + where a = n n +1 2n +1 /6. By taking the limit as n →, show that the price for the continuously sampled geometric average is the same as the price of a European call where the volatility is / √ 3 and where the asset earns a continuous income at rate r/2 + 2 /12. 7. When stratifying a standard normal variate, as, for example, in Section 6.4.2, an algorithm for sampling from N 0 1 subject to X ∈ −1 j −1/m −1 j/m for j = 1m is required. Write a Maple procedure for this. Use a uniform envelope for j =2m−1 and one proportional to x exp −x 2 /2 for j =1 and m. Derive expressions for the probability of acceptance for each of the m strata. Such a procedure can be used for sampling from N 0 1 by sampling from stratum j, j = 1m, with probability 1/m. What is the overall probability of acceptance in that case? For an alternative method for sampling from intervals j = 1 and m (the tails of a normal) see Dagpunar (1988b). Problems 133 8. Using the result in Problem 6, obtain the prices of continuous time geometric average price Asian call options. Use the parameter values given in Table 6.2. Obtain results for the corresponding arithmetic average price Asian call options when the number of sampling points in the average are 50, 200, and 500 respectively. Compare the last of these with the continuous time geometric average price Asian call option prices. 9. Let c denote the price at t = 0 of a European arithmetic average price call with expiry time T . The risk-free interest rate is r, the strike price is K, and the asset price at time t is X t . The average is computed at times h 2hnhwhere nh = T . (a) Make the substitutions x j = x 0 exp r jh −T and j = √ jh/T for j = 1n. Hence show that c is also the price of a basket option on n assets, where there are 1/n units of asset j which has a price of x j at time zero, j = 1n, and where the correlation between the returns on assets j and m is √ j/m for 1 ≤ j ≤m ≤ n. (b) Refer to the results in Table 6.2. Verify any of these by estimating the price of the equivalent basket option. 10. (a) Modify the procedure ‘meanreverting’ in Appendix 6.8 so that it prices a European call option on an asset subject to stochastic volatility, assuming Y = 0. (b) Suggest how the precision may be improved by variance reduction techniques. 11. Use the modified procedure referred to in Problem 10(a) to evaluate options when the Brownian motions driving the asset price and volatility processes are independent. Then repeat using conditional Monte Carlo. Estimate the variance reduction achieved. [...]... 2 T = 35 m =70 [0.4683920028, 0. 858 4469999, 1.3248481 95, 1 .56 4463 956 , 2.34210 358 9, 2. 753 604 757 , 3.161013 255 , 3. 355 203918] [0.31 054 258 25, 0.6 851 910142, 1.0 255 06 152 , 1.036301499, 1.247404803, 1.370810129, 2.376811 957 , 2.377386193, 2 .56 4390192, 3.436339776] [0.8816330302, 0.99 951 87699, 1.733006037, 1.92 655 7 959 , 1.926642493, 2.803064014] [1 .59 6872439, 2.036243709, 2.04299 955 2, 2.3414 453 60, 2 .51 3 656 874,... Microsaint, and Extend From a historical viewpoint the book by Tocher (1963) is interesting Other books emphasizing practical aspects of building discrete event simulation models include those by Banks et al (20 05) , Fishman (1978), Law and Kelton (2000), and Pidd (1998) Simulation and Monte Carlo: With applications in finance and MCMC © 2007 John Wiley & Sons, Ltd J S Dagpunar 136 Discrete event simulation. .. Let P 1 of before observing the values of certain random variables in the model Suppose we now observe those values and let D denote these data Let P D denote the likelihood d Using Bayes’ theorem, the conditional of the data and let P D = S P D P probability distribution of given the observed data is P D = P D P P D Simulation and Monte Carlo: With applications in finance and MCMC © 2007 John Wiley... is relaxed Here, n=20 = 2 5 per day, = 0 15 per day, and = 2 5, giving a mean and standard deviation of length of stay of 5. 92 and 2 .53 days respectively Early departures are not possible =0 Appendix 7.2.4 shows some point estimates for a 20 bedded ward when = 2 5 per day and the mean and standard deviation of a patient’s length of stay are 5. 92 days and 2 .53 days respectively Early departures are... processes in the plane Consider a two-dimensional domain, D Let C ⊆ D and E ⊆ D where C ∩ E = and is the empty set Let N C and N E denote the number of randomly occuring points in C and E respectively Suppose there exists a positive constant , such that, for all such C and E N C and N E are independent Poisson random variables with means C dx dy dx dy respectively Then the point process N H H ⊆ D is defined... number infected increases by one is nk p t + o t The probability that the number infected increases by more than one is o t Once infected the time to death for the individual is negative exponential, mean 1/ 1 , and this is independent of corresponding times for other individuals During the interval t t + t , for each noninfected individual there is an independent probability, t , that the individual... 0 15 0 03 0 02 0 ⎜0 07 02 01 0 ⎟ ⎜ ⎟ ⎜ ⎟ 0 06 0 35 0 05 P = ⎜0 ⎜ ⎟ ⎝0 0 0 09 01 ⎠ 0 0 0 0 1 Starting in state 1, simulate the times at which the machine changes state until the Markov chain is absorbed into state 5 Perform 1000 replications of this and find a 95 % confidence interval for the mean lifetime of the equipment Compare this with the theoretical result 8 Suppose there are k infected and. ..7 Discrete event simulation A discrete event system will be defined as one in which the time variable is discrete and the state variables may be discrete or continuous Correspondingly, a continuous event system is one in which the time variable is continuous and all state variables are continuous This leaves one remaining type of system, one where the time variable is continuous and the state variables... at least one of the quantities in (a) to (c) for arbitrary f (not exponential), , and M 8 Markov chain Monte Carlo Markov chain Monte Carlo (MCMC) refers to a class of methods for sampling random (generally, they are not independent) from a multivariate distribution vectors X 0 X 1 f Note that in this chapter the convention will be dropped of using bold face type to indicate that a quantity is a vector... patient, and (v) referral elsewhere of arriving patient The following variables are used: simtim = duration of simulation n = number of beds in unit t0 j = ranked (ascending) scheduled departure times at start of simulation, with t0 j = infinity indicating bed is currently unoccupied, j = 1 n = threshold early departure parameter seeda = seed for random number stream generating arrivals seedb = seed for random . sortd: print(arrival_times); end do: lambda = 2 T = 3 5 m = 70 [0.4683920028, 0. 858 4469999, 1.3248481 95, 1 .56 4463 956 , 2.34210 358 9, 2. 753 604 757 , 3.161013 255 , 3. 355 203918] [0.31 054 258 25, 0.6 851 910142,. simulation models include those by Banks et al. (20 05) , Fishman (1978), Law and Kelton (2000), and Pidd (1998). Simulation and Monte Carlo: With applications in finance and MCMC J. S. Dagpunar ©. 2.036243709, 2.04299 955 2, 2.3414 453 60, 2 .51 3 656 874, 2.9879 858 32, 3.1 857 27007, 3.370120432] [0. 956 6889486, 1. 358 244739, 2.99849 657 6] Another way to obtain the uniform order statistics avoids sorting. Let
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