Matematik simulation and monte carlo with applications in finance and mcmc phần 9 ppt

At time t the spot price is Pn i¼1qixiðtÞ.. 6.7.1 Naive Monte Carlo The procedure ‘basket’ estimates c using naive Monte Carlo.. asset price at time t; # sigma[i1]=volatility of asset i1

Trang 1

‘K’=55, ‘sigma’=0.1, ‘n’=16

‘# replications’=100, ‘paths per replication’=2500, ‘strata’=100

‘point estimate of price’=0.2023819565

‘estimated standard error’=0.00002354664402

13753

¼ 0:02171666756

‘K’=50, ‘sigma’=0.1, ‘n’=16

13753

¼ 0:01089879722

‘K’=45, ‘sigma’=0.1, ‘n’=16

Consider a basket (or portfolio) consisting of n assets The basket contains a

quantity qi of asset i where i¼ 1; ; n Let r; i; fXiðuÞ; u T ; 0 ug

denote the risk-free interest rate, volatilty, and prices in [0,T ] of one unit of

the ith asset At time t the spot price is Pn

i¼1qixiðtÞ Let denote thecorrelation between the returns on the assets and let the Cholesky decomposi-

tion of this beb bT ¼ r Then the price of a European call option at time t

with strike price K and exercise time T is the discounted expected payoff in a

risk-neutral world, that is

whereW Nð0; rÞ: P ut W ¼ bZ where Z Nð0; IÞ

6.7.1 Naive Monte Carlo

The procedure ‘basket’ estimates c using naive Monte Carlo It performs

npathreplications of the payoff Ensure that the procedure ‘STDNORM’

in Appendix 6.1 and the Linear Algebra and statistics packages are loaded

Trang 2

> basket:=proc(r,x,sigma,q,rho,T,t,n,npath,K) localb,c1,c2,i1,i2,mean,R,spot,stderr,theta,w,xav,xi,z; globali,X2;

#

# Computes call price for basket option using naive Monte Carlo;

# load STDNORM and Linear Algebra package;

#

# r=risk-free interest rate;

# x[i1]= i1th asset price at time t;

# sigma[i1]=volatility of asset i1;

# q[i1]=quantity of asset i1 in basket;

# rho=correlation matrix for returns between assets;

# T=exercise time;

# npath=number of paths

# K=strike price;

#spot:=Transpose(x).q;

for i2 from 1 to npath do;

for i1 from 1 to n do;

print("point estimate of price"=mean);

print("estimated standard error"=stderr);

end proc:

2666666666666666666666666666666666666666666664

Trang 3

‘K’=660, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’¼

0:30:20:30:4

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# paths’ = 10000

‘point estimate of price’ = 47.20505098

‘estimated standard error’ = 0.7072233067

‘K’=600, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

Trang 4

‘sigma’ ¼

0:30:20:30:4

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# paths’ = 10000

266666664

> K:=720;seed:=randomize(9624651);basket(r,x,sigma,q,rho,T,t,n,npath,K);

K:=720seed:=9624651

‘K’= 720, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:30:20:30:4

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# paths’ = 10000

2666666666666666666664

[ Now change the vector of volatilities

> sigma:=Vector([0.05,0.1,0.15,0.05]);

:¼

0:050:10:150:05

264

375

266664

> K:=660;seed:=randomize(9624651);basket(r,x,sigma,q,rho,T,t,n,npath,K);

K:=600;seed:=randomize(9624651);basket(r,x,sigma,q,rho,T,t,n,npath,K);

K:=720;seed:=randomize(9624651);basket(r,x,sigma,q,rho,T,t,n,npath,K);

266666

Trang 5

K :¼ 660seed :¼ 9624651

‘K’=660, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:050:10:150:05

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# paths’ = 10000

K :¼ 600seed :¼ 9624651

‘K’=600, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:050:10:150:05

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# paths’ = 10000

K :¼ 720seed :¼ 9624651

‘K’=720, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:050:10:150:05

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

Trang 6

‘# paths’ = 10000

2664

b0bq

The procedure ‘basketimppoststratv2’ below implements these two iance reduction devices See Table 6.3 in the text for the variance reduc-tion ratios achieved

Trang 7

# Computes call price for basket option using importance

sampling with post stratification

# load STDNORM, Linear Algebra package, and Statistics

package

#

# r=risk-free interest rate;

# x[i1]= i1th asset price at time t;

# sigma[i1]=volatility of asset i1;

# q[i1]:=quantity of asset i1;

# rho[i1, j]:=correlation between returns on assets i1 and j;

# T=exercise time;

# m=number of strata

# npath=number of paths in one replication; should be at

least 20*m for post stratification to be efficient;

# K=strike price;

# p=number of replications;

# upper=an upper bound for lambda;

# f[j]=number of paths falling in stratum j in one

Trang 8

for jj from 1 to p do:

theta:=0;

for j from 1 to m do s[j]:=0;f[j]:=0 end do;

for i2 from 1 to npath do;

print("K"=K,"spot"=spot,"r"=r,"n"=n,"t"=t,"T"=T,"x"=x,

"q"=q,"sigma"=sigma,"rho"=rho);

print("# replications"=p,"paths perreplication"=npath,"strata"=m);

print("point estimate of price"=mean);

print("estimated standard error"=stderr);

2666666666666666666666666666666666666666666666666666

Trang 9

375

‘K’=660, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:30:20:30:4

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# replications’ = 25, ‘paths per replication’ = 400, ‘strata’ = 20

‘approximate 95% confidence interval for reciprocal variance

Trang 10

> seed:=randomize(9624651);K:=600:basketimppoststratv2(r,x,sigma,q,rho,T,t,n,m,npath,K,p,upper);

seed :¼ 9624651

‘lambda’ ¼ 3:589486716; ‘beta’ ¼

0:5897151083850928900:2906095186127429790:3924690008997681610:230185997913606666

264

375

‘K’=600, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:30:20:30:4

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘approximate 95% confidence interval for reciprocal variance of error=’,

124.1839457,

‘to’, 394.1881451

26666666666666666666666666666

seed :¼ 9624651

‘lambda’ ¼ 6:065827969; ‘beta’ ¼

0:9965520647393203420:4910973310477080860:6632283752469805280:388988391569389748

264

375

‘K’=720, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

266666666666666666664

Trang 11

‘sigma’ ¼

0:30:20:30:4

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘approximate 95% confidence interval for reciprocal variance of

264

375

264

375

‘K’=660, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:050:10:150:05

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

Trang 12

‘‘estimated standard error’’= 0.009617279541

‘approximate 95% confidence interval for reciprocal variance of error =’,

5586.582894, ‘‘to’’, 17733.08729

266

seed :¼ 9624651

‘lambda’ ¼ 5:407253416 ‘beta’ ¼

0:3176870924067420880:2193666225964937260:2488741512073879280:0442439933277393033

264

375

‘K’=600, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:050:10:150:05

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘approximate 95% confidence interval for reciprocal variance of error =’,

339.5286687, ‘to’, 1077.741373

266666666666666666666666666

seed :¼ 9624651

‘lambda’ ¼ 20:43485296; ‘beta’ ¼

1:200589009091433290:8290206380612443740:9405341851408761670:167204942779322074

264

375

266666666666664

Trang 13

‘K’=720, ‘spot’=660., ‘r’=0.04, ‘n’=4, ‘t’=0, ‘T’=0.5,

‘x’ ¼

52:543

264

375; ‘q’ ¼

20806040

264

375,

‘sigma’ ¼

0:050:10:150:05

264

375; ‘rho’ ¼

1 0:7 0:5 0:30:7 1 0:6 0:20:5 0:6 1 0:40:3 0:2 0:4 1

264

375

‘# replications’ = 25, ‘paths per replication’ = 400, ‘strata’ = 20

‘approximate 95% confidence interval for reciprocal variance of error=’,

The procedure ‘meanreverting’ generates a volatility processfðtÞg in ½0; T at

t¼ 0; ; nh, where ðtÞ ¼ eYðtÞandYðtÞ is an Ornstein–Uhlenbeck process,

dY ¼ ðm Y Þdt þ dB;

where 0 < ; and fBðtÞg is a standard Brownian motion Remember to

load ‘STDNORM’ in Appendix 6.1 Define 2¼ 2/ð2Þ: There is a

and so

EððtÞÞ ¼ exp½yð0Þe tþ ð1 e tÞm þ 0:5 2ð1 e2 tÞ

> with(stats);

Warning, these names have been redefined: anova, describe,

fit, importdata, random, statevalf, statplots, transform

Trang 14

> meanreverting:=proc(T,n,seed,alpha,m,nu,y0) localh,P,j,z,a,b,d,A,Y;

#

# Procedure generates a list [{[j*h,exp(Y(j*h))],j=0 n}]where Y(j*h) is the position of the OU process (withparameters alpha,m,nu) at time jh

#global i,X2;

i:=false;

randomize(seed);

h:=T/n;

Y:=y0;P[0]:=[0,evalf(exp(y0))];

Trang 15

A slow mean reverting volatility process(alpha=0.5) with the expected volatility

0.130.14

0.110.12

Notice in the realization above that the volatility is taking an exceptionally

long time to reach a value equal to its asymptotic expectation Now increase

from 0.5 to 5 Observe in the plot below that the reversion to the mean is

far more rapid

PLOT(CURVES(v1,v2),TITLE("A faster mean reverting volatility

process\n (alpha=5) with the expected

Trang 16

A faster mean reverting volatility process(alpha=5) with the expected volatility0.3

Trang 17

PLOT(CURVES(v1,v2),TITLE("A fast mean reverting volatility

process \n (alpha=50) with the expected

A fast mean reverting volatility process(alpha=50) with the expected volatility0.3

Trang 19

Appendix 7: Discrete event simulation

7.1 G/G/1 queue simulation using the regenerative technique

For procedure ‘gg1’, the aim is to estimate the long-run average line length.Line length is the number of customers in the queueing system It includes anycustomer currently being served (The queue length is the number of customerswaiting for service.) The program assumes that interarrival times are indepen-dently Weibull distributed with Pðx < XÞ ¼ exp½ðxÞ for 0 < ; 0 < x,and that service durations are independently distributed with complementarycumulative distribution Pðx < DÞ ¼ exp½ðxÞ for 0 < ; 0 < x The expec-tation and standard deviation of these are printed out For stationary beha-viour, the mean interarrival time should be greater than the mean serviceduration Other distributions can be used by suitable modifications to theprogram The three-phase method for discrete event simulation is used Thebound state changes are (i) customer arrival and (ii) customer departure Theonly conditional state change is (iii) start customer service Separate streams areused for the two types of sampling activity This has benefits when performingantithetic replications The regeneration points are those instants at which thesystem enters the empty and idle state and n regeneration cycles are simulated

Trang 20

> gg1:=proc(lambda,mu,gam,beta,n,seed1,seed2) localrho,seeda,seedb,m,clock,L,Lp,a,d,reward,k,tau,prev,delta,u,v,R,tau,TAU,b1,b2,a1,a2,a3,a4,a5,a6,f,stderr,a7,gam1,beta1,gam2,gam3,beta2,beta3,v1;

for m from 1 to n do;

# START OF NEW CYCLEtau:=0;

tau:=min(a,d);

delta:=tau-tauprev;

reward:=reward+delta*L;

# CHECK FOR ’END OF CYCLE’

if tau=a and L=0 then break end if;

# CHECK FOR BOUND ACTIVITY ’ARRIVAL’

if tau=a thenL:=L+1;

Trang 21

d:=infinityend if;

# CHECK FOR CONDITIONAL ACTIVITY ’START SERVICE’

if L >0 and d=infinity then;

# TAU[m]=LENGTH OF m th REGENERATIVE CYCLE

# R[m]=CYCLE ’REWARD’=int(L(t),t=0 TAU[m]), where L(t)IS

NUMBER IN SYSTEM AT TIME t, AND t IS THE ELAPSED TIME SINCE

a7:=a6*(1+(a5/a1/a2-a4/a2/a2)/n); #a7 IS MODIFIED RATIO

ESTIMATOR OF TIN, M.(1965),JASA, 60, 294-307

f:=a3+a4*a7^2-2*a7*a5;

stderr:=sqrt(f/n)/a2;

print("mean cycle length"=a2);

print("estimate of mean line length"=a7);

print("95% Conf Interval for mean line

‘Mean interarrival time’=1 ‘Std dev of interarrival time’=0.5000000000

‘Mean service duration’=0.5000000000, ‘Std dev of service duration’=0.5000000000

‘mean cycle length’=2.003817572

Trang 22

‘estimate of mean line length’=0.9993680613

‘95% Confidence interval for mean line length’=0.9503482328,’to’, 1.048387890

‘ \ _’

‘Mean interarrival time’=1., ‘Std dev of interarrival time’=0.6666666667

‘95%Confidence interval for mean line length’=1.903465437, ‘to’, 2.156929053

‘ \ ’

‘95% Confidence interval for mean line length’=9.087027523, ‘to’, 11.56763834

2

6

> gg1(1,2,1,1,10000,246978,71586);

print(" ");

gg1(1,1.5,1,1,10000,246978,71586);

print(" ");

gg1(1,1.1,1,1,10000,246978,71586);

‘ _\ _’

‘ ’

Tiêu đề	Matematik Simulation and Monte Carlo with Applications in Finance and MCMC Part 9 PPT
Trường học	University of Economics and Finance
Chuyên ngành	Mathematics, Financial Engineering
Thể loại	Slide Presentation
Năm xuất bản	2023
Thành phố	Hanoi

Định dạng
Số trang	35
Dung lượng	710,57 KB