Elements of Financial Engineering Course

Transcription

1 Elements of Financial Engineering Course NSD Baruch MFE Summer Camp 206 Lecture 6 Tai Ho Wang Agenda Methods of simulation: inverse transformation method, acceptance rejection method Variance reduction techniques Copulas and methods for copula simulation Financial applications: Multiasset option pricing by simulation Portfolio VaR and CVaR calculation Inverse transformation method Theorem X Q X X Q X (U) U U[0, ] Let be a random variable and be its qf. Then, where. Proof Homework Note A pseudo code is like u, u2,, u n U[0, ] Generate sample points from uniform distribution For each i =,, n, let x i = Q X ( u i ) Return x,, x n Example

2 Example X Exp(λ) F X x 0 Let. Note that the cdf is given by, for, p (0, ) Hence, for, F X x (x) = P [X x] = λe λξ dξ = e λx. Q X (p) = F X Therefore,, where. X ln( U) λ 0 (p) = ln( p). U U[0, ] λ

3 In [8]: NumSim < e4 # generate sample points from U[0,] U < runif(numsim) # inverse tranform lambda < 3 Qf < function(p) /lambda*log( p) X < Qf(U) # plot hist(x,breaks=00,prob=t) curve(dexp(x,rate=lambda),from=0,col='blue',add=t) Empiricial cumulative distribution function In practice, where is the qf from? Empirical qf as the inverse of empirical cdf. Definition

4 Definition X j j =,, J F Let,, be a random sample ~ from a common distribution with cdf. The nondecreasing, càdlàg (why?) function F defined by ~ F J(x) = (x) J [ Xj, ) J j= F X j j =,, J is called the empirical cdf of from the sample,. Remark X j j =,, J Y X j P [Y = ] = X j (ω) We can reconcile the empirical cdf from the sample,, as the cdf of the random variable uniformly distributed among the points, i.e., X j. Empirical cdf is random, the randomness is from the samples empirical cdf has actually two arguments: x and ω.. Therefore, an J Glivenko Cantelli lemma The Glivenko Cantelli ~ lemma asserts that, as the total number of samples J approaches infinity, the empirical cdf F J converges almost surely (in ω) and uniformly (in x) to the original behindthe scene cdf F which generates the samples. Theorem X j j =,, J Assume that, ~, are independent random samples drawn from the common distribution with cdf F and F J be the empirical cdf from the random samples X j. Then lim sup (x) F(x) = 0 J x R ~ F J almost surely. Remark The pointwise convergence version of the Glivenko Cantelli lemma can be obtained by applying the strong law of large numbers. Namely, for any fixed, The rate of convergence in the Glivenko Cantelli lemma is have finite third moment, according to the Berry Esseen theorem. x R lim (x) F(x) = 0 almost surely. J ~ F J n if the random variables

5 In [9]: # the code illustrates the convergences of empirical cdf in Glivenko Cantelli lemm NumSim < 00 df < 3 X < rt(numsim,df=df) plot.ecdf(x,xlim=c( 3,3)) curve(pt(x,df=df),col='blue',add=t) hist(x, breaks=20, prob=t) curve(dt(x,df=df), col='blue', add=t)

6 Acceptance rejection method Let X and Y with pdfs f X and f Y respectively. We have a scheme to simulate Y and wish to simulate X through the samples from Y. Assumption: there exists a constant c such that f X for every x, i.e., f X is dominated by cf Y. The procedure goes as follows. (x) c Generate samples from and samples from independently. f For i =,, n, accept y i if it satisfies u X ( y i ) i ; otherwise, reject y. c f Y ( y ) i i Return the accepted 's f Y (x) y,, y n Y u,, u n U[0, ] y i

7 Example As an example, we use the method of acceptance rejection to simulate normal distribution by double exponential distribution. Let X N(0, ) and Y DE(). The pdf of X is x2 f (x) = X e 2 whereas the pdf of Y is f. 2π Y (y) = 2 e y We determine the acceptance threshold c as (x) f X(x) f Y 2 e 2 e x π x 2 π e x2 2 = = =: c ( + x x) 2π x x 2e 2 2

8 In [23]: # the code demonstrate the acceptance rejection method by using Laplace and Gaussi # X ~ normal # Y ~ dobule exponential NumSim < e4 lambda < # Simulate samples from Y and U independently U < runif(numsim) Y < runif(numsim) Qf < function(p) ifelse(p<=0.5,/lambda*log(2*p), /lambda*log(2 2*p)) Y < Qf(Y) # plot histogram of Y fy < function(x) lambda/2*exp( lambda*abs(x)) hist(y,breaks=50,prob=t) curve(fy,add=t,col='blue') # setting the acceptance threshold c c < sqrt(2*exp()/pi) # check acceptance rejection criterion fx < function(x) dnorm(x) AccRej < (U <= fx(y)/fy(y)/c) # fetch out accepted Y as sample points for X and plot X < Y[AccRej == T] hist(x,breaks=50,prob=t) curve(dnorm,add=t,col='blue')

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10 In [24]: length(x) Out[24]: 7570 Why it works? Note that, since the sample points from Y are accepted only when the inequality is satisfied, we can rewrite the acceptance rejection criterion in terms of conditional expectation as

11 We calculate the numerator and denominator as follows. denominator = numerator = It follows that = = = f ) P[Y P[Y x U X(Y x, U f X(Y) ] c f ] = Y c f Y ) P[U ] f X(Y) = = = P[U ] = P[U Y = y] (y)dy c f X(Y ) f X(Y ) f Y ) c ) P[U f X(y) ] f Y (y)dy ( Y and U are independent) c f Y f Y f X(y) (y)dy ( (u) = u) c c f Y f Y F U ( f X (x)dx = ) P[Y x, U ] = P[Y x, U Y = y] (y)dy x x c F X c f X(Y ) f X(Y ) f Y ) c ) f X(y) P[U ] (y)dy ( Y and U are independent) c (x). f X(y) c f Y f Y (y)dy ( (u) = u) f Y f Y F U f P[Y x U X(Y )] = = F (x). c ) X f Y F X c(x) c c f Y f Y f Y f Y Variance reduction Antithetic variate For each simulated sample point ω, we immediately add its reflected point ω. Quote from Glasserman: The method of antithetic variates can take various forms; the most broadly applicable is based on the observation that if U is uniformly distributed over [0, ], then U is too. Hence, if we generate a path using as inputs U,, U n, we can generate a second path using U,, U n without changing the law of the simulated process. The variables U i and U i form an antithetic pair in the sense that a large value of one is accompanied by a small value of the other. This suggests that an unusually large or small output computed from the first path may be balanced by the value computed from the antithetic path, resulting in a reduction in variance. Variance reduction Importance sampling

12 The goal is to estimate, for some function so that is integrable. We rewrite the expectation as where p and q are pdfs for the random variables X and Y respectively. Therefore, we can use g(y)p(y) as an unbiased (why?) estimator for, i.e., sample an iid sequence Y of Y then calculate the sample mean E[g(X)] g g(x) p(x) q(x) p(y ) q(y ) E[g(X)] = g(x)p(x)dx = g(x) q(x)dx = E[g(Y ) ], q(y) E[g(X)],, N N n= p( Y ) g( Y n ) n. q( ) In other words, instead of sampling an iid sequence directly from the distribution of X and calculating the sample mean g( ), we sample an iid sequence N N n= X n Y,, Y N from the distribution of Y, which we have the flexibility to choose, then calculate the sample mean p( g( ). N N Y n ) n= Y n q( ) Y n Y n,, X X N Y N So the question is: why would this help reduce the variance? Let's start with calculating their variances. From Cauchy Schwarz inequality we have g(x) p(x) q(x) q(x) g(x) (x) 2 p 2 g(y ) q(x)dx = 2 p 2 (Y ). q 2 (x) q 2 (Y ) E g(x) = g(x) p(x)dx = E[ ] Therefore, we obtain a lower bound for the variance of the random variable Var[ ] g(y ) 2 p2 (Y ). q 2 [E g(x) ] 2 (E[g(X)]) 2 On the other hand, again by Cauchy Schwarz inequality, we obtain the same lower bound for the variance of : g(x) (Y ) dx g(y)p(y) q(y) Var[g(X)] = E g(x) 2 (E[g(X)]) 2 [E g(x) ] 2 (E[g(X)]) 2. as At first glance, there seems no advantage in either case because they are both greater than or equal to the same right hand side. We can't do anything to the inequality Var[g(X)] [E g(x) ] 2 (E[g(X)]) 2. there is nothing to play with. However, we can try to make the inequality Var[ ] g(y ) 2 p2 (Y ). q 2 [E g(x) ] 2 (E[g(X)]) 2 (Y )

13 q(y) = g(y) p(y) an equality by smartly choosing q! In fact, simply pick. Moreover, if g doesn't E g(x) change sign, i.e., g(x) 0 or g(x) 0 for all x, then the right hand side of the above inequality g(y)p(y) is zero, which means with such a choice of q, the random variable has zero variance! Of q(y) course, this is too good to be true because in order to be in such an ideal situation, we need to know p and which are exactly the answers we are searching for. E[g(X)] From marginal to joint For notational simplicity, we shall denote by the unit interval and by the unit square hereafter. I 2 = {(x, y) : 0 x, y } (X, Y ) Let be a pair of random variables which are (marginally) uniformly distributed on I, i.e., X, Y U[0, ]. How much do we know about the joint distribution F XY (x, y) of (X, Y )? Basically nothing, since we have no information on the "linkage" between X and Y. There is no hope to reconstruct F XY (x, y) because there are infinitely many possibilities. It is in fact an ill posed problem However, the very least we have is the following. Apparently, since the random vector is supported in the unit square, the relevant values o F XY are within the unit square I 2. The above formula simply tells what the values of F XY on the boundaries of I 2 should be. In other words, there is no freedom of picking the values of F XY on the boundary of the unit square. For values of inside, we note that is 2 increasing, i.e., for all x x2 and 0 y y2, the inequality holds since, F XY (x, y) = P [X x, Y y] = y, x, 0, I [0, ] = {x : 0 x } if x, y ; if x, 0 y ; if 0 x, y ; if x < 0 or y < 0. (X, Y ) I 2 F XY I 2 F XY 0 F XY ( x2, y2) F XY ( x2, y) F XY ( x, y2) + F XY ( x, y) 0, F XY ( x2, y2) F XY ( x2, y) F XY ( x, y2) + F XY ( x, y) = P [ x X x2, y Y Y2 Copula in two dimensions In fact, any bivariate function defined on I 2 satisfies the aforementioned properties defines a 2 copula. Definition R A bivariate function C : I 2 is call 2 increasing if for every non empty rectangle u u2 v v2 I 2 [, ] [, ] C(, ) C(, ) C(, ) + C(, ) 0. u2 v2 u2 v u v2 u v

14 We now give a formal definition of copula. Definition A bivariate function C : I 2 I is called a copula if it satisfies C(0, u) = C(u, 0) = 0 for every u I. C(u, ) = u and C(, v) = v for every u, v I. C is 2 increasing in I 2. Remark In other words, a copula is simply the joint distribution function of the random vector (U, V ) with U and V being marginally uniformly distributed in the unit interval I. Recall that, if X and Y are continuous random variables with quantile functions Q X and Q Y respectively, then X Q X (U) and Y Q Y (V ), where U and V are uniformly distributed in I. Therefore, we readily have that, let F XY denote the joint distribution function of X and Y, the function C given by C(u, v) F XY ( Q X (u), Q Y (v)) defines a copula. (Check the definition.) In other words, we can "extract" the copula from a joint distribution. Natural questions to ask are a) is such copula unique? b) are all joint distribution given by copulas? Answers to the questions are given by Sklar's theorem. Sklar's theorem The Sklar's theorem in a sense means the decomposition of the joint distribution of a random vector into its marginal distribution combined with a (margin independent) copula function. In other words, one component comes purely from the marginal distributions, which statistically is more accessible to infer a la Glivenko Cantelli; and the other component (the copula) is unitized so that it accounts solely for "linking" the margins. Theorem (x) Let F and F2 be two one dimensional cdfs. If is any copula then F F2 is a 2 dimensional distribution function, whose marginal distribution functions are F (x) and F2(y). If is a dimensional distribution function with marginals F and F2, then there exists a copula C(v, z) such that F(x, y) = C( F(x), F2 (y)). If F(x) and F2(y) are continuous then C is unique. Thus, we can either (y) C C( (x), (y)) F(x, y) 2 (x) link two marginal distributions by a given copula or extract copula from a given joint distribution (y) Example

15 We mean.x. demonstrate sd.x. how to mean.y. extract the copula, sd.y. referred cor.x..y. to as the normal or Gaussian copula, from a joint normal distribution. A pseudo code will be like (, ),, (, ) u i F X x i = ( ) Simulate joint normal sample points x y x n y n For each i =,, n, let = ( ) and v i F Y y i Return the joint sample points ( u, v),, ( u n, v n ) Note It doesn't matter if the marginals X and Y are standard or not because F X (X) and F Y (Y ) are always uniformly distributed in. [0, ] In [25]: # the code demonstrate how to simulate Gaussian copula NumSim < e4 rho < 0.7 # First simulate indpendent normals X < rnorm(numsim) Y < rnorm(numsim) # correlate the independent normals X < X mean(x) X < X/sd(X) Y < rho*x + sqrt( rho^2)*y Y < Y mean(y) Y < Y/sd(Y) In [26]: Out[26]: data.frame(mean(x),sd(x),mean(y),sd(y),cor(x,y)) mean.x. sd.x. mean.y. sd.y. cor.x..y..98e e

16 In [27]: # copula sample points U < pnorm(x) V < pnorm(y) # plot par(mfrow=c(2,2)) plot(u,v,col='blue') Vh < hist(v,breaks=50,plot=f) barplot(vh$density,horiz=t,space=0,,main='histogram of V') abline(v=,col='blue') hist(u,breaks=50,prob=t) abline(h=,col='blue') Now we can use the simulated Gaussian copula sample points to link two marginal distributions, say, a distribution with degrees of freedom 3 and an exponential distribution with parameter 2. t

17 In [28]: X < qt(u,df=3) Y < qexp(v,rate=2) # plot par(mfrow=c(2,2)) plot(x,y,col='blue') yh < hist(y,breaks=50,plot=f) barplot(yh$density,space=0,horiz=t,main='histogram of Y') #curve(dexp(x,rate=2),col='blue',add=t) hist(x,breaks=50,prob=t) curve(dt(x,df=3),col='blue',add=t) Archimedean copulas Definition An Archimedean copula is a copula of the form

18 C φ (u, v) = φ (φ(u) + φ(v)) (0, ] φ() = 0 for some convex, continuous, decreasing function defined on with. The function φ is called the generator of C φ. (Exercise: check such defined function C is a copula) Commonly used Archimedean copulas Commonly used Archimedean copulas and their generators include Ali Mikhail Haq: Clayton: φ(t) = log( θ( t)) log t, θ [, ) uv C φ (u, v) = θ( u)( v) t φ(t) = θ, θ [, ) {0} θ C φ (u, v) = [( u + ) ] θ v θ + /θ Frank: Gumbel: Note that Joe: θ = φ(t) = log( e θ ) log( e θt ), θ R {0} ( ] C φ (u, v) = log[ + e θu )( e θv ) θ e θ φ(t) = ( log t ) θ, θ C φ (u, v) = e corresponds to the product copula. [( log u +( log v ] )θ ) θ θ φ(t) = ( e t ) θ, θ [, ) C φ (u, v) = [( u ) θ + ( v) θ ( u ) θ ( v ) θ θ ] Simulation of Archimedean copulas Therefore, a pseudo code will be like Simulate the random variable whose moment generating function is φ. U i i =,, n Simulate iid random variables, for, whose common conditional cdf (conditioned on V ) is e vφ(u), by the method of inverse transformation.

19 ,, U n Fetch out samples U. Proof = = = = = = C( u,, u n ) P [ U u,, U n u n ] P [ U u,, U n u n V = v] d F V (v) n i= n i= P [ V = v] d (v) ( 's are conditionally independent) U i u i F V U i e vφ( u ) i F V U i e vφ( ) d (v) ( V has cdf u i ) (v) v e n i= φ( u i ) d F V 0 φ u u n m V φ (φ( ) + + φ( )) ( V has mgf (t) = (t)) Simulation of the Clayton copula

20 In []: # Clayton copula simulation # number of simulations NumSim < e4 # dimension of random vector d < 3 # copula parameter theta < 0.5 # F tilde = inverse of generator Ft < function(t) ( + t)^( /theta) V < rgamma(numsim,shape = /theta,scale=) U < matrix(runif(numsim*d),ncol=d) U < Ft( log(u)/v) plot(u[,2],u[,],col='blue')

21 Simulation of the Gumbel copula

22 In [0]: # Gumbel copula simulation # generator = ( log(u))^theta # for simulating stable distribution, require package: stabledist # requires library(stabledist) library(stabledist) # number of simulations NumSim < e4 # dimension of random vector d < 3 # copula parameter: theta > theta < 5 # Ftile = inverse generator Ft < function(t) exp( t^(/theta)) V < rstable(numsim,alpha = /theta,beta=,gamma=cos(pi/2/theta)^(theta),delta=0) U < matrix(runif(numsim*d),ncol=d) U < Ft( log(u)/v) plot(u[,2],u[,3],col='blue')

23 Simulation of the Marshall Olkin copula

24 In [30]: # Marshall Olkin copula simulation NumSimul < e4 lambda <. lambda2 < 0.2 lambda2 < 0.6 Z < rexp(numsimul,rate=lambda) Z2 < rexp(numsimul,rate=lambda2) Z2 < rexp(numsimul,rate=lambda2) U < exp( (lambda + lambda2)*pmax(z,z2)) U2 < exp( (lambda + lambda2)*pmax(z2,z2)) #plot plot(u,u2) Now we can calculate by simulation the price of multiasset options such as basket, spread, or BestOf. VaR and CVaR of a portfolio For example, assuming each individual asset is lognormally distributed with μ i and σ 2 i, for. We link them by the Clayton copula, say i =, 2, 3 θ =.5

25 In []: # Generate Clayton copula # number of simulations NumSim < e4 # dimension of random vector d < 2 # copula parameter theta < 0.5 # F tilde = inverse of generator Ft < function(t) ( + t)^( /theta) V < rgamma(numsim,shape = /theta,scale=) U < matrix(runif(numsim*d),ncol=d) U < Ft( log(u)/v) # Link log normal margins by Clayton copula # time to expiry t < /52 # volatilites sig < 0.3 sig2 < 0.25 # log price at current time x < log(00) x2 < log(50) # lognormal samples S < qlnorm(u[,], meanlog = x sig^2*t/2, sdlog = sig*sqrt(t)) S2 < qlnorm(u[,2], meanlog = x2 sig2^2*t/2, sdlog = sig2*sqrt(t)) K < 250 payoffbasket < (S + S2 K)*(S + S2 >= K) pricebasket < mean(payoffbasket) In [2]: pricebasket Out[2]: Now we wrap the code up as a function.

26 In [3]: # copula pricemean parameter pricesd theta < 0.5 s < 00 s2 < 50 sig < 0.3 sig2 < 0.25 t < K < 250 pricebasket_clayton < function(k,t,numsim=e4,theta=0.5,m=){ # F tilde = inverse of generator Ft < function(t) ( + t)^( /theta) d < 2 pricemean < numeric(m) pricesd < numeric(m) for (i in :m) { V < rgamma(numsim,shape = /theta,scale=) U < matrix(runif(numsim*d),ncol=d) U < Ft( log(u)/v) # Link log normal margins by Clayton copula S < qlnorm(u[,], meanlog = log(s) sig^2*t/2, sdlog = sig*sqrt(t)) S2 < qlnorm(u[,2], meanlog = log(s2) sig2^2*t/2, sdlog = sig2*sqrt(t)) payoffbasket < (S + S2 K)*(S + S2 >= K) pricemean[i] < mean(payoffbasket) pricesd[i] < sd(payoffbasket) } data.frame(pricemean,pricesd) } In [4]: Out[4]: pricebasket_clayton(k=k,t=t,m=4) pricemean pricesd

27 In [5]: pv < pricebasket_clayton(k=k,t=t,m=e3) hist(pv$pricemean,breaks=50,prob=t) In [6]: mean(pv$pricemean) Out[6]: