Monte Carlo Methods. School of Mathematics, University of Edinburgh 2016/17

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1 Monte Carlo Methods David Šiška School of Mathematics, University of Edinburgh 2016/17 Contents 1 Introduction Numerical integration Derivative pricing Some useful identities Convergence Random variables, their distribution, density and characteristic functions Convergence modes Law of large numbers and Central limit theorem Generating random samples Pseudorandom numbers, uniform distribution Inversion method Acceptance rejection method Box Muller method for generating normally distributed samples Generating correlated normally distributed samples Summary Variance reduction Antithetic variates Control variates Multiple control variates Summary Further reading This is the first draft of the notes. There will be mistakes that need to be corrected and material will be added and removed as will be appropriate for the lectures. Last updated 1st June

2 5 Some applications Asian options Options on several risky assets Approximating sensitivity on parameters Background General Setting Finite Difference Approach Calculating Sensitivities Pointwise in Ω The Log-likelihood Method Solutions to some exercises 40 A Appendix 43 A.1 Multi-dimensional Itô s formula Introduction We are interested in Monte Carlo methods as a general simulation technique. However many most of our examples will come from financial mathematics. 1.1 Numerical integration We start with examples that are not directly related to derivative pricing. This is to let us understand the main idea behind Monte Carlo methods without getting confused by general derivate pricing issues. Example 1.1 Numerical integration in one dimension. Let f : [a, b] R be given and say that we want to approximate I = b a fxdx. Assume that f 0 on [a, b] and that f is bounded on the interval [a, b] and let M := sup x [a,b] fx. Assume that we know how to generate samples from U0, 1 that is the uniform distribution on the interval 0 to 1. Let u i N and v i N be two collections of N samples each from U0, 1. Let x i := a + b au i and y i := Mv i. Let 1 A be equal to 1 if A is true and 0 otherwise. Then we can approximate I with I N := b am 1 N N 1 {fxi y i }. That is, we count the number of times when y i is equal to or less than fx i and then we divide by N. Finally, we scale this by the area of the rectangle inside which we are sampling our random points. One would hope that I N converges to I, in some sense, as N. 2

3 Example 1.2 Multidimensional numerical integration. Let Ω R d be bounded inside the hypercube [a 1, b 1 ] [a 2, b 2 ] [a d, b d ]. Let f : Ω R + be measurable, integrable and bounded. We wish to approximate I = fxdx. Let Ω M := sup fx. x Ω For j = 1,..., d + 1 and i = 1,..., N sample u ij independently from U0, 1. Let x i1 := a 1 + b 1 a 1 u i1, x i2 := a 2 + b 2 a 2 u i2,. x id := a d + b d a d u id, y i := m + M mu id+1. Let x i := x i1, x i2,..., x id T. First we approximate the volume of Ω, denoted by V. Let V N denote the approximation of the volume. V N := b 1 a 1 b 2 a 2 b d a d 1 N N 1 {xi Ω}. We can now approximate I with I N := V N M 1 Ñ N 1 {xi Ω}1 {fxi y i }, Ñ = N 1 {xi Ω}. That is, we first calculate Ñ, that is the number of x i that lie inside Ω. Then we count the number of times y i is equal to or less than fx i and we divide by Ñ. Finally we scale by the volume of Ω [0, M]. Again we hope that I N converges to I, in some sense, as N. 1.2 Derivative pricing We now give some examples of pricing derivatives with Monte Carlo methods. Let Ω, F, P be a probability space and F t t [0,T ] a given filtration to which the traded assets are adapted. It can shown that for any option whose payoff is given by a F T -measurable random variable h has the value at time t < T given by V t = E Q Dt, T h F t, where Dt, T is the discounting factor for the time period t to T, which in the simplest case can be Dt, T = exp rt t for some risk free rate r 0 and where E Q denotes the expectation under the risk neutral measure Q. This is the measure under which the discounted traded assets are martingales. 3

4 We have shown that in the particular case of European call and put options in the Black Scholes framework we have vt, S = E Q e rt t gs T S t = S, where g is the function giving the option payoff at exercise time T. Of course in this case we have the well known Black Scholes formula giving the option price. Example 1.3 Classical Black Scholes. Say that we have derived the Black Scholes formula ourselves but we are not sure whether we have performed all the calculations correctly. One way for us to check would be to use Monte Carlo methods to approximate the option payoff by simulating the behaviour of the risky asset. Recall that the model for the risky asset in the real-world measure P is ds t = µs t dt + σs t dw t, where W t t [0,T ] is a P-Wiener process with respect to F t t [0,T ], µ R and σ > 0. Say that gs := [S K] +, that is, the option is an European call option. We have shown that in the risk-neutral measure Q the evolution of the risky asset is given by ds t = rs t dt + σs t d W t, 1 where W t t [0,T ] is a Q-Wiener process with respect to F t t [0,T ]. We further know that S T = S t exp r 12 σ2 T t + σ WT W t. The option price is thus given by vt, S = E Q e [S rt t exp r 12 σ2 T t + σ WT W ] t K. + By definition W T W t is normally distributed with mean 0 and variance T t. If Z N0, 1 then T tz has the same distribution as W T W t. So vt, S = E Q e [S rt t exp r 12 σ2 T t + σ ] T tz K. 2 Now we will use a Monte Carlo method to evaluate 2. Assume for now that we know how to draw samples from standard normal distribution. Let us take N independent samples z i N from N0, 1. The approximation is given by v N t, S := 1 N N e [S rt t exp r 12 σ2 T t + σ ] T tz i K. + We would hope that for fixed t and S we can say that v N t, S converges in some sense to vt, S as N. Now we can compare v N t, S to the option price given by the Black Scholes formula. If the numbers are close and on average decreasing as N increases then we would have every reason to believe we are using the correct Black Scholes formula. + 4

5 1.3 Some useful identities Let X, Y be random variables and recall that [ Var[X] := E X E[X] 2] = E [ X 2] E[X] 2, [ ] Cov[X, Y ] := E X E[X] Y E[Y ] = E[XY ] E[X]E[Y ]. If λ, µ are constants, then E[µ + X] = E[X] + µ, Var[µ + X] = Var[X], E[λ X] = λ E[X], Var[λ X] = λ 2 Var[X], E[X + Y ] = E[X] + E[Y ], Var[X + Y ] = Var[X] + Var[Y ] + 2 Cov[X, Y ]. 5

6 2 Convergence So far we have not discussed the convergence of Monte Carlo algorithms. It is clear that the usual notions of convergence are insufficient when analysing Monte Carlo methods. No matter how large sample we take, we can always be extremely unlucky, draw a unrepresentative sample and get a bad estimate for the true solution of our problem. In this section we introduce the appropriate notion of convergence, law of large numbers and the central limit theorem, which provides the convergence of Monte Carlo algorithms. 2.1 Random variables, their distribution, density and characteristic functions Let Ω, F, P be a probability space. If X is an R d -valued random variable then its distribution function sometimes called the cumulative distribution function or CDF is F : R d [0, 1] given for R d x = x 1,..., x n by F x = F x 1,..., x n = PX 1 x 1,..., X d x d. We recall that if for some g : R d C d we have E gx < then E[gX] = gy df y. R d In particular taking gy = 1 B y for some B BR d leads to PX B = E[1 B X] = E[gX] = df y. We say that the distribution function F has density f if f : R d [0, is such that R fydy = 1 and d x1 xd F x = F x 1,..., x d = fy 1,..., y d dy d dy 1. If X is a random variable with a distribution that has a density then we call X continuous 1. Recall that for a continuous random variable X with density f we have EgX = gyfy dy R d whenever E gx <. The characteristic function ϕ of a distribution F or a random variable with distribution F is defined by ϕz := e izx df x, z R d. R d Here zx := d z ix i is the inner dot product. We see that if X has the distribution F then its characteristic function is ϕz = E[e izx ]. Let X k k N be independent random variables and let S n = X X n. Then [ n ] n ϕ Sn t = E e itx k = E [ e itx ] n k = ϕ Xk t. 3 k=1 k=1 1 This is a completely different concept to continuity of functions! B k=1 6

7 Theorem 2.1. Let X be an R-valued random variable with distribution F and characteristic function ϕz = E[e izx ]. Then ϕ satisfies the following: 1. ϕt ϕ0 = t ϕt is uniformly continuous. 3. ϕt equals the complex conjugate of ϕ t. 4. ϕt is real-valued if and only if F is symmetric in the sense that PB = P B, where B := {x : x B}. 5. If E X n < for some n 1 then d r dt r ϕt = ϕr t = R ix r e itx df x exists for all r n and E[X r ] = i r ϕ r 0. Moreover ϕt = n it r E[X r ] + itn E n t, 4 r! n! r=0 with E n t 3E X n and E n t 0 as t 0. Compare the expansion in 4 to the Taylor expansion: ϕt = 1 + tϕ 0 + t2 2! ϕii tn n! ϕn 0 + itn E n t. n! 2.2 Convergence modes We now look at various types of convergence. Let X n n N be a sequence of random variables. Definition 2.2 Pointwise Convergence. We say that the random variables converge pointwise to X if for all ω Ω we have X n ω Xω as n. We say that an event E occurs almost surely or a.s. for short if PΩ\E = PE c = 0. From this follows the definition of almost sure convergence. Definition 2.3 Almost sure Convergence. We say that the random variables converge almost surely to X if there is an event E with PE c = 0 such that for all ω E we have X n ω Xω as n. We can immediately see that pointwise convergence implies almost sure convergence. Definition 2.4 L p Convergence. Let p > 0. We say that the random variables converge in L p to X if E[ X n X p ] 0 as n. Definition 2.5 Convergence in probability. We say that the random variables converge in probability to random variable X if for all ε > 0 we have P [ X n X ε] 0 as n. 7

8 Definition 2.6 Convergence in Distribution. Let X n n N be random variables with distributions F n n N. We say that the random variables converge in distribution to a random variable X with distribution F if F n x F x as n for all real numbers x at which F is continuous. We make the following remarks. 1. We will sometimes use the notation X n d X to denote that Xn n N converges to X is distribution as n. 2. The random variables need not be defined on the same probability space when one considers convergence in distribution. Indeed the statement only involves the distribution functions. The following two theorems give the relations between different types of convergence. Theorem 2.7. We have: i Almost sure convergence implies convergence in probability. ii L p convergence implies convergence in probability. iii Convergence in probability implies convergence in distribution. For proof see Shiryaev [7, Ch. II, 10, Theorem 2]. The following theorem says that there are at least three equivalent ways to see convergence in distribution. Theorem 2.8. Let ϕ Xn and ϕ X be the characteristic functions of X n and X respectively. Then the following are equivalent: i X n X as n in distribution. ii EfX n EfX as n for all bounded and continuous functions f. iii ϕ Xn t ϕ X t as n for all t R. For proofs of a more general result and further reading see Shiryaev [7, Ch. III, 1, Theorem 1 and Ch. III, 3, Theorem 1]. 2.3 Law of large numbers and Central limit theorem We now have all the tools we will need to prove the Law of large numbers and the Central Limit Theorem. The proofs are those given in Shiryaev [7, Ch. 3]. Theorem 2.9 Law of large numbers. Let X k k N be a sequence of independent and identically distributed random variables such that E X 1 = m <. Let Then Sn n m in probability. S n = X X n. 8

9 Proof. Let ϕ and ϕ Sn/n be the distribution functions of the random variables X 1 and S n /n respectively. That is [ ] ϕt = E[e itx 1 ], ϕ Sn/nt = E e it Sn n. Then, since X i are independent, with same calculation as for 3, we have t n ϕ Sn/nt = ϕ. n From 4 we know that we can write ϕ as t ϕ = 1 + itm n n + it n E 1 t. n From Theorem 2.1 we know that E t 1 n 0 as n for each fixed t. So we can write 2 t ϕ = 1 + itm 1 n n + o n and so ϕ Sn/nt = [ 1 + itm n + o 1 n] n e itm for all n > N. The function t e itm is the characteristic function of a random variable Z = m almost surely. From Theorem 2.8 we know that convergence of characteristic functions is equivalent to convergence in distribution. In general convergence in distribution does not imply convergence in probability. However, in the special case when S n /n Z = m as n in distribution, we can conclude that the convergence is in probability too. Theorem 2.10 Central limit theorem. Let X k k N be independent and identically distributed with EX k = µ and VarX k = σ 2. Then X n := S n n := 1 n n k=1 X k satisfies where Z N0, 1. ξ n := n Xn µ σ d Z as n, 5 Proof. Let ϕ be the the characteristic function of X 1 µ i.e. ϕt = E [ e itx 1 µ ] and let ϕ n be the the characteristic function of ξ n. Observe that n Xn µ ξ n = σ = n S n n E S n n σ. n 2 We say that fn = ogn if for any ε > 0 there is N such that fn ε gn for all n > N. 9

10 Hence, due to the same independence type calculation as in 3, we get [ n k=1 ϕ n t := E [exp itξ n ] = E exp it X ] k µ σ = n [ ] t n = ϕ σ. n From Theorem 2.1 and 4 we get Hence ϕ n t = [1 σ2 t 2 2σ 2 n ϕt = 1 σ2 t 2 2 t2 2 E 2t. ] n t2 2σ 2 n E 2t = [1 t2 2n + o n k=1 [ E exp it X ] k µ σ n ] n 1 e t2 2 as n. n The function t e t2 2 is the characteristic function of a N0, 1 random variable and Theorem 2.8 tells us that convergence of characteristic functions is equivalent to convergence in distributions. The proof can also be found in Grimmett and Stirzaker [3, Chapter 5, Section 10]. Proposition Let us take X n n N and X n as in the Central limit theorem. Let Φ denote the distribution of a standard normal random variable. Then for any δ > 0 we have σ P X n z δ/2 n µ X σ n + z δ/2 n 1 δ as n, where z δ/2 is a number such that 1 Φz δ/2 = δ/2. Proof. Since Φx = x φzdz = x 1 2π e z2 /2 dz we see that Φ is continuous. Hence, due to 5 and the definition of convergence in distributions, we know that for all x R n Xn µ P x Φx as n. σ Thus, taking x equal to ϕ and to ψ above, with 0 ϕ, ψ <, we get n Xn µ n Xn µ P ϕ Φϕ and P < ψ Φ ψ as n. σ σ Therefore P X n ϕ σ µ X n + ψ σ = P n n n Xn µ ψ σ ϕ n Xn µ n Xn µ = P ϕ P ψ Φϕ Φ ψ as n. σ σ 10

11 For any δ > 0 we can choose ϕ and ψ such that Φϕ Φ ψ = 1 δ. In particular letting z δ/2 be a number such that 1 Φz δ/2 = δ/2 we see that Φz δ/2 Φ z δ/2 = 1 δ 2 δ 2 = 1 δ. Hence σ P X n z δ/2 n µ X σ n + z δ/2 n 1 δ as n. Roughly speaking this means that the estimator X n is a correct estimate for µ up σ to an error of z δ/2 n, with probability 1 δ. That is, with n sufficiently large, we can halve the error by quadrupling the number n. Another way of looking at this is that to be able to say that X n is correct up to an error of ɛ > 0 we need to take n > zδ/2 2 σ2 ɛ 2. Of course σ would typically be unknown in Monte-Carlo simulations and so this does not give us a usable error estimate. Nevertheless it is a constant and so we can still say that the Monte-Carlo method would converge with order 1/2 we halve the error by quadrupling N. Definition 2.12 Estimator for EX. Let us take X n n N and X n. We will call Xn the estimator for EX n = µ. Note that E Xn = 1 n n EX k = 1 n k=1 n EX = EX. An estimator with the property that the expectation of the estimator recall it is a random variable is equal to the parameter we are estimating is called unbiased. An estimator that does not have this property is called biased. Example Let us now return to the setting of Example 1.3. At time t, when the risky asset is worth S, we have the option price vt, S given by 2. As before, we take N independent samples z i N from N0, 1. The Monte Carlo approximation is given by v N t, S := 1 N N k=1 e [S rt t exp r 12 σ2 T t + σ ] T tz i K. + To use the central limit theorem let us take X i := e [S rt t exp r 12 σ2 T t + σ ] T tz i K where Z i N are independent and identically distributed standard normal variables. Of course in the actual Monte Carlo experiment we will have z i N samples from the standard normal distribution. But to do the mathematical analysis we have to replace those by random variables. The the expectation of X i does not depend on i and the same for the variance and we have µ = vt, S = EX i and σ vn = VarX i. Clearly both these quantities are unknown unless we calculate µ = vt, S using the + 11

12 Black Scholes formula; but that is not the point here. Using central limit theorem we get that NvN t, S µ d Z as n. σ In particular we have the same asymptotic estimate as above. That is, for large N we know that vt, S v N t, S z δ/2 σ vn N 1/2, v N t, S + z δ/2 σ vn N 1/2. As we said before σ vn is an unknown number but it is a constant. The central limit theorem gives us the order of convergence of a Monte Carlo algorithm. Reducing the variance will reduce the error in the approximation for a fixed number of samples. Hence finding ways that reduce the variance of Monte Carlo simulations is an area of active research interest. 12

13 3 Generating random samples To use Monte Carlo methods we need to generate random samples from various distributions. Of course a computer algorithm will never generate truly random numbers, but there are ways of generating sequences of numbers that look random, unless we actually know the algorithm that generated them. We will say that such sequences are pseudorandom. From those we can easily get samples from U0, 1, the uniform distribution on the interval 0 to 1. Now we would like to be able to generate random samples from any distribution efficiently. We will present several methods: inversion, acceptancerejection method and the Box Muller method for generating normally distributed random samples. 3.1 Linear congruential pseudorandom number generators and generating uniformly distributed random samples One of the most commonly used methods for generating pseudorandom numbers is the linear congruentiual pseudorandom number generator. Given a random seed x 0 we generate pseudorandom numbers x 1, x 2,... using the recurrence relation x i+1 = ax i + c mod m, i = 0, 1,... with a a given multiplier, c a given increment and m a given modulus. The period of the generator is the smallest p N such that x i = x i+p for any i = 0, 1,.... The period of the generator will never exceed m i.e. p m. See Knuth [4, Section 3.2.1]. Clearly the smallest value x i can have is 0 and the maximum is m 1. Example 3.1. Take x i+1 = 7x i + 8 mod 15. With seed x 0 = 1 we get x 1 = 15 mod 15 = 0, x 2 = 8 mod 15 = 8 etc. The period is p = 12 because x 12 = 1 = x 0 and for all i = 1, 2,..., 11 we have x i x 0. With seed x 0 = 3 we get x 1 = 29 mod 15 = 14, since = 29. If we use linear congruential pseudorandom number generator then we can generate a sequence u i,...,m of samples from U0, 1 by taking u i = x i /m 1, where x i are the numbers produced by the generator with maximum period m. 3.2 Inversion method From now onwards we will assume that we can generate not just pseudorandom but truly random samples from the uniform distribution. The inversion method is a method for generating samples from distributions of random variables that take values in R. Say we want to generate samples following the distribution F : R [0, 1]. Assume that the inverse F 1 of F exists. Recall that we say that the random variable X : Ω R has the distribution F if P X x = F x for any x R. If F is continuous and strictly increasing then for each u 0, 1 there is F 1 u, given by the usual inverse of the strictly increasing continuous function F. If F is a general distribution function then we define F 1 u := inf{x : F x u} for 0 < u < 1. 13

14 Let U U0, 1. Consider X := F 1 U. Then PX x = PF 1 U x = PF F 1 U F x = PU F x. But U has got uniform distribution and so PU u = u for any u R. Hence Thus X has the distribution F. PX x = F x. This means that if u i i N are samples from U0, 1 then x i i N given by x i = F 1 u i are samples from the distribution F. Example 3.2. Say that we would like to generate N random samples from the exponential distribution with parameter λ > 0. First we would like to invert F. To that end we solve y = 1 e λx for x: and hence x = F 1 y = ln1 y λ ln1 y. λ So we can generate N samples from U0, 1, denote them by u i N and get x i := F 1 u i = ln1 u i. λ Of course if U U0, 1 then 1 U U0, 1 and so we can equally well take x i := F 1 1 u i = lnu i. λ Exercise 3.3. We would like to sample from the double exponential distribution, also known as the Laplace distribution, which has the density given by fx = exp x. 2 a Show that the distribution given by the above density is { 1 F x = 2 ex if x 0, e x if x > 0. b Show that the inverse of F is given by { F 1 ln2x if x 1 x = 2, ln 2x + 2 if x > 1 2. c Say we have generated N random samples u i N distributed uniformly in [0, 1]. How to generate x i N samples from the distribution given by the Laplace density? 14

15 Very often we would like to generate random samples from the normal distribution. We know that for normally distributed random variables we can write their distribution in terms of the density P X x = F x = x φydy = 1 2π x exp y2 dy. 2 Since φx is strictly positive for all x R we see that F is a strictly increasing function of x and hence its inverse F 1 : 0, 1 R exists. Nevertheless there is no closed form formula for F 1. This would suggest that one can not use the inversion method for generating normally distributed random numbers. This is not the case. We can either approximate F 1 or we can use Newton s method to find the inverse of F numerically. 3.3 Acceptance rejection method This is a method for generating random samples from a continuous distribution with density f. To use it we have to assume that we can sample from U0, 1 and also from another distribution with a density g. Finally, we have assume that there is c > 0 such that fx cgx x R. 6 To generate a sample from the distribution with density g we can use the following algorithm: 1. Generate a sample u from U0, Generate a sample y from distribution with density g. 3. If u fy cgy then x = y is a random sample from a distribution with density f and we stop. Otherwise go to step 1. Exercise 3.4. We know how to generate samples from a Laplace or double exponential distribution gx = exp x, 7 2 see Exercise 3.3. We wish to use this to generate random variables with normal distribution, that is with density φx = 1 exp y2. 8 2π 2 To that end we would first have to find c > 0 such that φx cgx for all x R. Only if such c exists can we use the acceptance-rejection algorithm. a Show that the function ξ : R R given by ξx = fx/gx is symmetric about x = 0. b Show that ξ has a maximum at x = 1. c Hence show that the inequality 6 is satisfied with 2e c = π. 9 15

16 From Exercise 3.4 we know that the condition 6 is satisfied for the normal density and double exponential density with c given by 9. To understand what the acceptance rejection algorithm actually does, let us look at Figure 1. In steps one and two we sample u from the uniform distribution and x from the proposal distribution that is, the distribution we already know how to sample from, in this case the double exponential distribution. The value of x gives us the x-coordinate of each point in the plot. To get the the y-coordinate of each point we take u and scale it by cgx Here g is the known density given by 7. Now we check whether y = cugx is smaller or larger than φx, which is given by 8. If y lies on or under φx it is accepted, while if it lies above φx it is rejected Normal dens. Scaled double exp. dens. Accepted Rejected Figure 1: Acceptance-rejection used to generate samples from normal density Looking at the algorithm we see that unless the generated u and x satisfy u fx cgx we will be repeating steps one to three forever. So a natural question is what is the probability that the algorithm terminates at step three? Proposition 3.5. Assume that U U0, 1 and that Y : Ω R d is a random variable with density g. Let f be a density function. Let there be c > 0 such that 6 holds. Then P U fy = 1 cgy c. Proof. As U, Y are independent with known densities, we have their joint density: P U fy = gy du dy cgy {u,y 0,1 R:u fy/cgy} fy/cgy 1 = gy du dy = fy dy. c R 0 R Thus the sample generated in step two is accepted with probability 1/c. So the algorithm will need to generate random samples u and x exactly K times with the 16

17 probability P K = k = 1 1 k 1 1 c c, k N. Clearly the algorithm will be the most efficient if c is very close to 1. So far we have only given an algorithm without justifying why the generated random sample has the desired distribution. Proposition 3.6. Assume that U U0, 1 and that Y : Ω R d is a random variable with density g. Let f be a density function. Let there be c > 0 such that 6 holds. Let X be the random variable with distribution given by the distribution of Y conditional on U fy cgy 1. That is, for A R d, PX A := P Y A U fy. cgy Then X has the density f. Proof. Let A be a measurable subset of R d. To prove the proposition we need to show that P Y A U fy = fydy. 10 cgy First we note that P Y A U fy = cgy P A { } {Y A} U fy cgy. P U fy cgy This means that, due to Proposition 3.5, P Y A U fy { = cp {Y A} U fy } cgy cgy fy/cgy = c gy du dy = fy dy. A 0 A But this is exactly 10, which concludes the proof. 3.4 Box Muller method for generating normally distributed samples Very often we need to sample from the standard normal distribution. We have seen that we can use the acceptance-rejection method to that end or even the inversion method if we either approximate the normal density or use a numerical method for finding the inverse of the distribution function. The Box Muller method is a method designed to produce samples from standard normal distribution efficiently. It is based on the following observation. Proposition 3.7. The random variables X and Y are normally distributed and independent with mean 0 and variance 1 if and only if the random variables R := Y X 2 + Y 2 and Θ := arctan 11 X 17

18 are such that R 2 is exponentially distributed with parameter 1/2 and Θ is uniformly distributed over the interval [0, 2π] and R and Θ are independent. Proof. Assume that X and Y are independent standard normal random variables. The joint density of the random variables X and Y is then given by f X,Y x, y := 1 2π exp x2 + y 2, 2 since for independent continuous random variables their joint density is just the product of densities. We wish to calculate in the joint density of R and Θ. Recall that those are given by 11. We will now carry out essentially just the change of integration variables from cartesian to polar. Notice that with gx, y := y x 2 + y 2 and hx, y = arctan x we have R = gx, Y and Θ = hx, Y. Furthermore, letting J denote the Jacobian of the transformation, det J = det = g h x y g h y x. Now g x = g x h x x x 2 + y 2 g y h y and Recall that d dx arctanx = 1 + x2 1. Hence g y = h x = y 1 x 2 = y 1 + y2 x 2 + y 2 and h x 2 Altogether, letting r = gx, y, det J = x r y = 1 x x r 2 y y r r 2 = 1 r. y x 2 + y 2. 1 x = 1 + y2 x 2 + y 2. x 2 Then, letting θ = hx, y, the joint density of R and Θ is f R,Θ r, θ = f X,Y x, ydet J 1 = 1 2π exp r2 r. 2 Note that this is a standard calculation for the joint density of a pair of random variables that are given as functions of another pair of random variables. See e.g. Ross [6, Chapter 6, Section 7]. Let f R r := re r2 /2 and f Θ θ = 2π 1. We see that f R,Θ r, θ = f R rf Θ θ. Hence the random variables are independent. The random variable Θ already has the required distribution. The random variable R has the Raleigh distribution but we are more interested in the distribution of R 2. We see that for x 0 we immediately have PR 2 x = 0. For x > 0: PR 2 x = PR x = x Thus R 2 has exponential density with parameter 1/2. 0 re 1 2 r2 dr = 1 e 1 2 x. To prove the implication in the other direction we could start with the joint density f R,Θ r, θ, carry out a change of variables, and derive the joint density f X,Y. 18

19 Armed with this knowledge we can give the Box Muller algorithm for generating a pair of independent samples from the joint density of two independent standard normal random variables X and Y. 1. Sample d from the exponential distribution 1 e x/2. 2. Sample θ from the uniform distribution on 0, 2π. 3. Let r = d and x = r cos θ and y = r sin θ. Then x and y are the required samples. Note that we can use the inversion method to sample from the exponential distribution. 3.5 Generating correlated normally distributed samples We will define a multivariate normal distribution as follows. Let µ R d be given and let Σ be a given symmetric, invertible, positive definite d d matrix it is also possible to consider positive semi-definite matrix Σ but for simplicity we ignore that situation here. A matrix is positive definite if, for any x R d such that x 0, the inequality x T Σx > 0 holds and positive semi-definite if we only have x T Σx 0. From linear algebra we know that this is equivalent to: 1. There are d eigenvalues of a positive definite matrix Σ are all strictly positive for positive semi-definite matrix they are non-negative and the d corresponding eigenvectors are orthonormal. 2. There is a unique up to multiplication by 1 lower-triangular matrix B such that BB T = Σ. This is given by Cholesky decomposition. For our purposes the matrix B s.t. BB T = Σ doesn t need to be lower triangular and we can use another method 3 to find it: let u i, λ i d be the eigenvectors and eigenvalues of Σ. Let Λ := diagλ 1,..., λ d and U be the matrix of the eigenvectors i.e. U := u 1,..., u n. Since the eigenvectors are orthonormal UU T = I. Moreover, we have ΣU = UΛ. Hence Σ = UΛU T. Define Λ 1/2 := diag λ 1,..., λ d. Then Σ = UΛ 1/2 Λ 1/2 U T = BB T with B = UΛ 1/2. Let B be a d d matrix such that BB T = Σ. Let X i d be independent random variables with N0, 1 distribution. Let X = X 1,..., X d T and Z := µ + BX. We then say Z Nµ, Σ and call Σ the covariance matrix of Z. Exercise 3.8. Show that CovZ i, Z j = EZ i EZ i Z j EZ j = Σ ij. This justifies the name covariance matrix for Σ. It is possible to show that the density function of Nµ, Σ is 1 fx = 2π d/2 detσ exp 1 2 x µt Σ 1 x µ. 12 Note that if Σ is symmetric and invertible then Σ 1 is also symmetric. 3 This is sometimes referred to as Principal Component Analysis PCA. 19

20 Exercise 3.9. You will show that Z = BX defined above has the density f given by 12 if µ = 0. i Show that the characteristic function of Y N0, 1 is t exp t 2 /2. In other words, show that Ee ity = exp t 2 /2. Hint. complete the squares. ii Show that the characteristic function of a random variable Y with density f given by 12 is E e iσ 1 ξ T Y = exp 1 2 ξt Σ 1 ξ. By taking y = Σ 1 ξ conclude that E e iyt Y = exp 1 2 yt Σy. Hint. use a similar trick to completing squares. You can use the fact that since Σ 1 is symmetric ξ T Σ 1 x = Σ 1 ξ T x. iii Recall that two distributions are identiacal if and only if their characteristic functions are identical. Compute E e iyt Z for Z = BX and X = X 1,..., X d T with X i d independent random variables such that X i N0, 1. Hence conclude that Z has density given by 12 with µ = 0. You can now also try to show that all this works with µ 0. To generate N independent samples from Nµ, Σ with µ R d and Σ a d d matrix we propose the following algorithm: 1. Use PCA or Cholesky decomposition to find B such that Σ = BB T. 2. Generate N d samples from N0, 1 and collect them in N vectors each with d components, labelled x i N, with xi R d for each i = 1,..., N. 3. For each i = 1,..., N let z i := µ+bx i. Now z i N samples from Nµ, Σ. are independent 3.6 Summary We have seen that linear congruential generators can be used to give sequences of pseudorandom natural numbers. These can be used to generate samples from the uniform distribution. We can then use the inversion method or the acceptance-rejection method to generate samples from other distributions. For generating samples from the normal density the Box Muller algorithm is generally sufficiently efficient. The Ziggurat algorithm which is based on acceptance rejection method optimized for efficient implementation is what is used by state of the art numerical libraries and Matlab. 20

21 4 Variance reduction We will discuss variance reduction techniques in this section. These are techniques which allow us to get a better estimate, on average, without increasing the sample size. 4.1 Antithetic variates The idea is to reduce variance by introducing negative dependence in pairs of replications. Intuitively, an extremely large draw from a distribution can be compensated by an extremely low one and so the variance of the average will be reduced. Example 4.1. We can use antithetic variates when sampling from the following distributions: a If U U0, 1 then 1 U U0, 1. b If Z N0, 1 then Z N0, 1. c If U U0, 1 then F 1 U and F 1 1 U both have distribution F. The method then consists of considering N pairs X 1, X 1,..., X N, X N that are independent and identically distributed but such that for each i the random variables X i and X i are identically distributed but not independent. Assume that there are random variables X and X with the same distribution as X i and as X i respectively and such that EX i = EX and VarX i = VarX and E X i = E X and Var X i = Var X for i = 1,..., N. Definition 4.2 Antithetic variates estimator for EX. Let X AV N := N N X i + 1 N be the antithetic variates estimator for EX. It is easy to check that this estimator is unbiased. We would like to apply central limit theorem to XN AV. Of course we can not use the sequence X 1, X 1, X 2, X 2,..., X N, X N since those random variables are not independent. But the random variables are independent and X AV N = 1 2 X 1 + X N N X i, X 2 + X 2,..., X N + X N 2 2 N X i + 1 N N X i = 1 N N X i + X i. 2 Hence due to central limit theorem N XN AV E X+ X 2 d N0, 1 as N. σ AV 21

22 Of course X and X are identically distributed and so E X+ X 2 = EX. Here σ AV = X + X Var. 2 Now we calculate σ AV. The question of course is how it compares to VarX. First VarX + X = VarX + Var X + 2CovX, X = 2VarX + 2CovX, X. Thus σ 2 AV = 1 2 VarX + CovX, X. So the method will decrease variance provided that CovX, X < 0. The problem is that, typically, EX is unknown and so VarX is unknown and also CovX, X is unknown. One way to overcome this is to test experimentally, that is estimate CovX, X itself using Monte Carlo. There is also a theoretical result that may help in some situations. Say that for example Z i d are independent and distributed according to N0, 1. Let X := fz 1, Z 2,..., Z d for some increasing f which is an increasing function of all its arguments. Then with X := f Z 1, Z 2,..., Z d we have EX X EXE X and hence CovX, X 0. The same is also true if we replace Z i with U i and Z i with 1 U i. Example 4.3. We will employ antithetic variates in a simple situation. Imagine that we would like to use Monte Carlo method to estimate vt, S given by 2. We would then use X i := e [S rt t exp r 12 σ2 T t + σ ] T tz i K + where Z i N are independent and identically distributed standard normal variables together with X i := e [S rt t exp r 12 σ2 T t σ ] T tz i K. + An estimator for vt, S is then v N t, S := 1 N N X i + X i Control variates This is another variance reduction technique. Recall that our aim is to reduce the variance of our Monte Carlo estimate and thus the improve the estimate, while keeping the number of samples fixed. The number of random samples used by a Monte Carlo method is a good proxy for the computational effort required. Thus improving accuracy while keeping the number of samples fixed mean we are improving accuracy while keeping the computational effort fixed. 22

23 The main idea behind control variates is to use a random variable with a known expectation that is highly correlated with the random variable whose expectation we seek to correct our estimate. For example while Ee rt [S T K] + is unknown unless we use the Black Scholes formula the quantity Ee rt S T = S, since the evolution of the discounted risky asset is a Martingale. In what follows we will use X to denote the random variable whose expectation is known. We will use Y to denote the random variable whose expectation we wish to estimate. Assume we have X i, Y i N independent and identically distributed with the same distribution as X, Y. As always, these will be used in the analysis instead of specific samples x i N and y i N. Definition 4.4 Control variates estimator with parameter b R for EY. Let Y i b := Y i bx i EX and let Ȳ N b := 1 N N Y i b be the control variate estimator with parameter b for EY. Let ȲN and X N be the estimators for EY and EX respectively. Recall that EX is assumed to be known. Note that E Ȳ N b = E Ȳ N b X N EX = ȲN = EY and so the control variates estimator with parameter b is for b R is unbiased. Let σ b := VarY i b. From the central limit theorem we know that N ȲN b E Ȳ N b σ b d N0, 1 as N. Of course, as E Ȳ N b = EY we get the same asymptotic error bounds as for the ordinary estimator, see Proposition 2.11 but with σ replaced by σ b. Proposition 4.5. Let σ Y := VarY and σ X := VarX. Let ρ XY := CovX, Y σ X σ Y. Then there is b R such that σ 2 b = σ 2 Y 1 ρ2 XY. Before we proceed to prove this result let us make some observations about what this implies. Remark 4.6. We can conclude the following. a The higher the correlation between X and Y the higher variance reduction can be achieved. b The sign of the correlation is not important. 23

24 c With an ordinary estimator we would need N 1 ρ 2 XY samples to achieve the same asymptotic error bound i.e. accuracy as the control variates estimator. This means that if we can get X i without increasing the computational effort then control variate estimator always performs better than ordinary estimator. In practice producing X i and the slightly more complicated calculation costs something in terms of computing time and so one would not use control variates unless ρ XY is reasonably high. What this means must almost always be determined experimentally. d If we assume that we get X i for free then a correlation of 0.95 produces a speedup of factor 10 we can use ten times smaller sample size while maintaining accuracy. Correlation of 2 1/2 0.7 produces only a speedup of factor 2. Proof of Proposition 4.5. Recall that, for a general random variable Z we have Further recall that Hence VarZ = E Z EZ 2 = EZ 2 EZ 2. CovX, Y = E X EXY EY. σ X σ Y ρ XY = CovX, Y VarXVarY = E X EXY EY VarXVarY = EXY EXEY. Now recall that Y i b = Y i bx i EX and so EY i b = EY i as EX i = EX. So, if we use the fact that X i and Y i have the same distribution as X and Y respectively we obtain VarY i b = E Y i b 2 EY i 2 = E Y 2 i 2bY i X i EX + b 2 X i EX 2 EY i 2 = EY 2 2bEY X + 2bEY EX + b 2 E X EX 2 EY 2 = VarY 2bCovX, Y + b 2 VarX = σ 2 Y 2bσ X σ Y ρ XY + b 2 σ 2 X. Our aim is to minimise the variance. So we must choose b that minimises the above expression. Hence we seek b such that 0 = d db σ 2 Y 2bσ X σ Y ρ XY + b 2 σx 2 = 2σX σ Y ρ XY + 2bσX. 2 So b = σ Y ρ XY σ 1 X. Then σ 2 b = VarY ib = σ 2 Y σ 2 Y ρ 2 XY. 24

25 There is one small problem remaining. Remember that we are trying to estimate EY. But if we do not know this then it is rather unlikely that we will actually know σ Y = VarY and CovX, Y and hence ρ XY. So while the above result of variance reduction of the control variates method is correct it is not usually usable in practice. What one can do though, is to take an estimate for b, using the samples generated during the Monte Carlo method and use ˆb N := N x i x N y i ȳ N N x i x N Let ˆB N denote the random variable we obtain if in the above equation we use X i and Y i in place of x i and y i and X N and ȲN in place of x N and ȳ N. Then EȲN ˆB N = E ȲN ˆB N XN EX = EY E ˆB N X + E ˆB N EX = EY + E ˆB N EX E ˆB N X. We see that EȲN ˆB N is no longer an unbiased estimator for EY. We have bias equal to E ˆB N EX E ˆB N X. It can be shown, though we do not do it here, that the bias is of order 1/N. Since for a Monte Carlo method the error is of order 1/ N we can say that for large N this is not significant. Hence in practice one would use control variates with the estimate given by 13. Example 4.7. We consider a call option price in the Black Scholes framework so we know the exact price as it is given by the Black Scholes formula. In the risk neutral measure the evolution of the risky asset is given by ds u = rs u du + σs u dw u, S t = S. Here W t t [0,T ] is a Wiener process in the risk neutral measure. We have shown before that in the risk neutral measure the process e rt u S u u [t,t ] is a martingale and hence E e rt t S T = S t = S. Now we would like to use control variates to estimate vt, S = E e rt t [S T K] +. We take Y = e rt t [S T K] + and so we are estimating EY. We take X = e rt t S T as our control since we know that EX = S t = S. Now we generate N samples from standard normal distribution and denote them by z i N. We then get an estimate x N = 1 N e rt t S exp r 12 N σ2 T t + σ T tz i for EX = Ee rt t S T. We also calculate ȳ N = 1 N e [S rt t exp r 12 N σ2 T t + σ ] T tz i K. + This is, by itself, an estimate for EY. But we would like to use control variates and so we use 13 as an estimate for ˆb N. Then our control variates estimate for EY is given by ȳˆb N = ȳ N ˆb N x N S. 25

26 This is a number we know how to calculate. Notice that we did not have to generate another set of random samples to use control variates in this case. This means that whatever reduction in variance we are achieving, it is achieved at the cost of evaluating exp and few additions N times. 4.3 Multiple control variates It is also possible to generalize the control variate method to multiple controls. Imagine you have a random variable Y and you wish to estimate EY. Assume that you have X k with EX k known for k = 1,..., m. Let Σ X be the m m covariance matrix for X, and let Σ XY be the m 1 covariance matrix for X k, Y, i.e. Σ X jk := CovX j, X k, Σ XY j1 := CovX j, Y and as before σ 2 Y = VarY is a scalar. Hence we have the correlation matrix ΣX Σ XY Σ T XY σ 2 Y for the R m+1 valued r.v. X 1,..., X m, Y. Define, for b R m, Proposition 4.8. For b R m we have Y b := Y b T X EX. VarY b = σ 2 Y 2b T Σ XY + b T Σ X b. The b R m which minimizes Var[Y b] is given by 4.4 Summary b = Σ 1 X Σ XY. We have shown that the appropriate convergence concept for Monte Carlo methods is convergence in distributions. We have used central limit theorem to derive an asymptotic error bound for a Monte Carlo approximation in terms of number of samples and variance. We have seen that convergence is always of order 1/2. We have seen that reducing variance improves the estimate. We have discussed two techniques for variance reduction: antithetic variates and control variates. We have seen that both provide a tangible improvement only in specific situation and hence one must analyse the problem before deciding whether to use a specific variance reduction technique. There are other variance reduction techniques that we have not discussed. 26

27 4.5 Further reading This material is based in particular on Glasserman [1]. See in particular Chapter 2 and Chapter 4, Section 1 and 2. For more details on variance reduction techniques and various applications see again Glasserman [1]. 27

28 5 Some applications 5.1 Asian options Asian options differ sligtly from the European options: the payoff is not based only on the price of the risky asset at the exercise date T but on the average price of the risky asset over several dates before the exercise date Geometric Asian option Let S t t [0,T ] denote the price of the risky asset at time t. Let n 1/n S G := ST i, where T i N are some dates that are fixed in the option contract and are such that t < T 1 < T 2 <... < T N = T. The option contract also specifies the strike K > 0. The option payoff at the expiry time T is given by [ S G K] +. Assume we work in the Black Scholes framework. Then the price, denoted by v AA, is given by v AG t, S = E e rt t [ S G K] +, where the expectation is, as always, in the risk neutral measure. Exercise 5.1. Show that the Black Scholes formula can be used with expiry time T t, where, T := n 1 n T i, risk free rate r, strike K, volatility σ given by 1 σ σ = n 2i 1T n+1 i t. T t n and spot price Se γ T t, where γ := 1/2 σ 2 σ Arithmetic average option Let S t t [0,T ] denote the price of the risky asset at time t. Let n S A := 1 ST i, n where T i N are some dates that are fixed in the option contract and are such that t < T 1 < T 2 <... < T N = T. The option contract also specifies the strike K > 0. The option payoff at the expiry time T is given by [ S A K] +. Assume we work in the Black Scholes framework. Then the price, in the risk neutral measure, denoted by v AA is given by v AA t, S = E e rt t [ S A K] +. 28

29 There is no known formula that would tell us the price of this option. We can use Monte Carlo to estimate the value of the option. In order to do that we need to simulate the prices of the risky asset at times T i with i = 1,..., n. Since we are working in the Black Scholes framework we know that if S t = S then Su = S exp r 12 σ2 u t + σw u W t u [t, T ]. Hence for any T i we know that Su = S Ti exp r 12 σ2 u T i + σw u W Ti u [T i, T ]. So, setting T 0 := t, we get ST i = S Ti 1 exp r 12 σ2 T i T i 1 + σw Ti W Ti 1 d = S Ti 1 exp r 12 σ2 T i T i 1 + σ T i T i 1 Z i, i = 1,..., N. 14 where Z i N0, 1 are independent standard normal random variables and where d = is used to denote that two random variables have the same distribution. For each i = 1, 2,... n we can take N samples from N0, 1 and thus in total we have n N samples and denote them zj i, i = 1,..., n and j = 1,..., N. Let us define s j T 0 := S and define, for j = 1,..., N, s j T i := s j T i 1 exp r 12 σ2 T i T i 1 + σ T i T i 1 zj i, i = 1,..., N. Let x j := e rt t [ 1 n Our Monte Carlo approximation is then ] n s j T i K +. v AA,Nt, S = 1 N N x j. j=1 If we wanted an approximation for the error we would need to use Proposition To that end let for each i = 1,..., n let there be N independent standard normal random variables denoted Zj in j=1. Let S jt 0 := S and S j T i := S j T i 1 exp r 12 σ2 T i T i 1 + σ T i T i 1 Zj i, i = 1,..., N. Let X j = e rt t [ 1 n ] n S j T i K Note that EX j = v AA t, S is the quantity we wish to estimate. Let us use σ v := VarX j. Then X N := 1 N X j N 29 j=1 +.

30 is an unbiased estimator for v AA t, S and δ > 0 σ v P X N z δ/2 v AA t, S X σ v N + z δ/2 1 δ N N as N. Remark 5.2. Note that in Section the option price depends on the path of the of the process used to model the risky asset at times T 1, T 2,..., T n and not just T. Since we know the solution to the equation ds u = µs u du + σs u dw u, S t = S, we know exactly the distributions the random variables ST 1, ST 2,..., ST n have. It is given by 14. If we use a general stochastic differential equation to model the risky asset we have dx u = bu, X u du + σu, X u dw u, X t = x. Under appropriate assumptions on b and σ we would know that a solution of such equation exists but we would not necessarily know what it is. In this case we could approximate XT 1 X 1, XT 2 X 2,..., XT n X n using, for example the Explicit Euler scheme X i = X i 1 + bt i 1, X i 1 T i T i 1 + σt i 1, X i 1 W Ti W Ti 1. If we now proceed as before there would be two sources of error in our approximation. One would arise, as always, from the use of a Monte Carlo method and can be estimated using the Central limit theorem. The other error would arise from the approximation of X u u [t,t ] by X i n and is a type of discretization error. In practice one would subdivide [t, T ] into more subintervals than just those required for the arithmetic average option in order to decrease the discretization error. Example 5.3 Control variates for arithmetic average option. We can use the price of the geometric average option as a control variate in a Monte Carlo method when estimating the arithmetic average option price. Of course we could also use the discounted evolution of the risky asset as in Example 4.7 but it can be shown at least experimentally the the correlation between the payoff of the geometric average option and the payoff of the arithmetic average option are higher. This method is very useful because it works in many situations where a simple model leads to a price given by a formula that we can then use to improve our Monte Carlo method in a more realistic model. Other examples of use are e.g. option pricing with stochastic volatility with the price given by Black Scholes formula given as a control. 30

31 5.2 Options on several risky assets We can use Monte Carlo methods to price options on more than one risky asset. In order to do that we need some models for the evolution of the risky assets. The basic one extends the idea of Geometric brownian motion to several dimensions using a d-dimensional Wiener process Wiener process in R d It will occasionally be more convenient to write Xt instead of X t for some stochastic process {X t } t 0 = {Xt} t 0. This is just a matter of notation. A process {W t} t 0 is a Wiener process on R d if W 0 = 0 almost surely, it has independent increments, the function t W t is almost surely continuous and where I is a d d identity matrix. W t W s N0, t si, Note that if {W 1 t} t 0, {W 2 t} t 0,..., W d t t 0 are independent, 1-dimensional Wiener processes, then the process given by W t = W 1 t, W 2 t,..., W d t T satisfies the above definition and so is a Wiener process on R d Multi-dimensional geometric Brownian motion Let {W t} t 0 be a Wiener process on R k. Then is a process with covariance Σ. In fact Xt := BW t, Xt Xs N 0, t sσ. If, for a vector z we write z = z 1, z 2,..., z d T then then X i t = B i1 W 1 t + B i2 W 2 t + + B ik W k t, i = 1, 2,... d. 15 Let µ, σ R d. We can model d correlated risky assets using dsu = diagsu µdu + diagσdxu, St = S. Here diagz denotes a d d matrix with the diagonal equal to z and all off-diagonal elements equal to zero. The above equation is equivalent to ds i u = S i u µ i du + σ i dx i u, S i t = S, i = 1, 2,..., d. which is in turn k ds i u = S i u µ i du + σ i B ij dw j u, S i t = S, i = 1, 2,..., d. j=1 We would like to obtain an explicit solution to the stochastic differential equaiton just like in the 1-dimensional case. For this we need the multi-dimensionla Itô formula, 31