Exercise Sheet 7 - Solutions

Transcription

1 Exercise Sheet 7 - Solutions Alessandro Gnoatto June 10, Exercise 1 1. Let Y t = log X t, then dy t = 1 dx t 1 X t Xt Xt σ dt by Ito s Lemma. Hence This implies that dy t = (µ σ )dt + σdw t. log X t = log x + ) (µ σ t + σw t and We then have σ (µ X t = xe )t+σwt σ (µ E[X t ] = xe )t E[e σwt ]. To compute E[e σwt ] we note that E[σW t ] = E[σW t ] 0 = E[σW t ] E[σW 0 ] = E[σW t σw 0 ] and σ (W t W 0 ) N (0, σ t) therefore E[e σwt ] = e z df(z), where f(z) is the density of an N (0, σ t) r.v. = 1 πσ t = 1 πσ t e z e z σ t dz = e z +zσ t σ t dz We complete the square of z + zσ t as follows: Then 1 z z e σ t dz πσ t z + zσ t σ 4 t + σ 4 t = (z σ t) + σ 4 t. = 1 πσ t e (z σ t) +σ 4 t σ t dz = 1 πσ t e σ t e (z σ t) σ t dz 1

2 Let w = z σ t σ t = 1 πσ t e σ t =e σ t. then dw = dz Therefore E[e σwt ] = e σ t and σ (µ E[X t ] =xe )t e σ t =xe µt. σ t and e (z σ t) σ t dz To compute V ar[x t ] we note that V ar[x t ] = E[X t ] E[X t ]. Computing E[X t ] follows a similar procedure as the computation for E[X t ] since E[Xt σ (µ ] =xe )t E[e σwt ] where σ = σ. Then and Finally =xe (µ σ )t E[e σwt ], E[e σwt ] = e σ t = e σ t xe (µ σ )t E[e σwt ] = xe (µ σ )t e σt = xe µt+σt. V ar[x t ] =E[X t ] E[X t ] =xe µt+σt (xe µt ) =xe µt e σt x e µt =xe µt (e σt x). Now, computing E[Xk ], we observe that for k = 1, E[X 1 ] =E[(1 + µ t 0 )X 0 + σx 0 W 0 ] = (1 + µ t 0 )x. If for some k, then k E[X] = x (1 + µ t l ) (induction hypothesis) E[X k ] =E[(1 + µ t )X + σx W ] =(1 + µ t )E[X ] k =(1 + µ t )x (1 + µ t l ) =x (1 + µ t l ).

3 The induction hypothesis is true for k = therefore we have that E[Xk ] = x (1 + µ t l ) for k = 1,... n with E[X0 ] = x. To compute V ar[x n ] = E[(X n ) ] E[X n ], we first compute E[(X k ) ]. We have that [ ((1 E[(X1 ) ] =E + µ t0 )X0 + σx0 ) ] W 0 =E [ (1 + µ t 1 ) x + σ(1 + µ t 0 )x W 0 + σ x W0 ] =(1 + µ t 0 ) x + σ x E[ W 0 ] =(1 + µ t 0 ) x + σ x t 0 =((1 + µ t 0 ) + σ t 0 )x. If for some k, k E[(X) ] = x ((1 + µ t l ) + σ t l ) (induction hypothesis) then [ ((1 E[(Xk ) ] =E + µ t )X + σx W ) ] =E [ (1 + µ t ) (X) + σ(1 + µ t )(X) W + σ (X) W ] Note that X = (1 + µ t )Xk + σx k W k involves only the random variables ( W l ) k 1 which are all independent from W. Therefore (X ) is independent of both W and ( W ). Therefore E[(X k ) ] =... =(1 + µ t ) E[(X ) ] + σ(1 + µ t )E[(X ) ]E[ W ] + σ E[(X ) ]E[ W ] =(1 + µ t ) E[(X ) ] + σ E[(X ) ] t = ( (1 + µ t ) + σ t ) E[(X ) ] = ( k (1 + µ t ) + σ ) t x ((1 + µ t l ) + σ t l ) =x ((1 + µ t l ) + σ t l ). 1 hint: we can write X as a linear combination of the terms in ( W l) k 3

4 The induction hypothesis is true for k =. Therefore Finally, E[(Xk ) ] =x ((1 + µ t l ) + σ t l ) for k = 1,,... n, E[(X 0 ) ] =x with V ar[xk ] =E[(Xk ) ] E[Xk ] ( =x ((1 + µ t l ) + σ t l ) x (1 + µ t l ) ( ) =x ((1 + µ t l ) + σ t l ) (1 + µ t l ). ). E[X T ] =e = V ar[x T ] =e (e 1 1) = E[Xn ] =( ) 100 = V ar[xn ] =(( ) ) 100 ( ) 00 = The example Java program is reported below package de. math. lmu. exercise07 ; 4 import java. text. DecimalFormat ; import net. finmath. time.*; 6 import net. finmath. montecarlo. BrownianMotion ; import net. finmath. montecarlo. RandomVariable ; 8 import net. finmath. stochastic. RandomVariableInterface ; 10 public class EulerSchemeTests { 1 s t a t i c f i n a l DecimalFormat formattersci6 = new DecimalFormat (" E00 ; E00 "); 14 public s t a t i c void main ( String [] args ) { 16 int numberoftimes = 100; double t0 = 0.0; 18 double T = 1.0; double initialvalue = 1.0; 0 double mu = 0.1; double sigma = 1.0; double[] times = new double[ numberoftimes ]; 4 for ( int countertime =0; countertime < numberoftimes ; countertime ++) { times [ countertime ] = t0 + countertime *( T- t0)/(( double) numberoftimes ); 4

5 6 8 TimeDiscretization mytimes = new TimeDiscretization ( times ) ; 30 int seed = 134; int numberofpaths = 50; 3 BrownianMotion brownianw = new BrownianMotion ( mytimes,1, numberofpaths, seed ); 34 RandomVariableInterface previousvalue = new RandomVariable ( t0, initialvalue ); 36 System. out. println (" time "+ " "+" emp mean X"+ " "+" theor. mean X cont." + " "+" theor. mean X discr."+" "+ " error cont."+ " "+" error discr " 38 + " "+" mean error "); 40 double theoreticalmeandiscrete = initialvalue ; RandomVariableInterface MCerrorContinuous = new RandomVariable (0.0) ; 4 RandomVariableInterface MCerrorDiscrete = new RandomVariable (0.0) ; 44 for ( int countertime =0; countertime < numberoftimes -1; countertime ++) { 46 double dt = mytimes. gettimestep ( countertime ); 48 RandomVariableInterface dw = brownianw. getbrownianincrement ( countertime, 0); RandomVariableInterface nextvalue = previousvalue. mult (1+ mu*dt). add (dw. mult ( previousvalue. mult ( sigma ))); 50 double empiricalmean = nextvalue. getaverage (); 5 //continuous model 54 double theoreticalmeancontinuous = initialvalue * Math. exp (mu* mytimes. gettime ( countertime ));//times[ countertime]); 56 //discrete model i f ( countertime >0) theoreticalmeandiscrete = theoreticalmeandiscrete *(1+ mu* dt); 58 MCerrorContinuous = nextvalue. sub ( theoreticalmeancontinuous ); 60 MCerrorDiscrete = nextvalue. sub ( theoreticalmeandiscrete ) ; double MCmeanError = Math. abs ( empiricalmean - theoreticalmeancontinuous ); 6 System. out. println ( formattersci6. format ( mytimes. gettime ( countertime )) +//times[countertime])+ 64 " " + formattersci6. format ( empiricalmean ) + " " + formattersci6. format ( theoreticalmeancontinuous ) + 66 " " + formattersci6. format ( theoreticalmeandiscrete ) + " " + formattersci6. format ( MCerrorContinuous. getaverage ())+ 68 " " + formattersci6. format ( MCerrorDiscrete. getaverage ())+ 5

6 70 7 " "+ formattersci6. format ( MCmeanError )); previousvalue = new RandomVariable ( nextvalue ); 74 double[] bins = new double [5]; bins [0]= Double. NEGATIVE_INFINITY ; 76 for ( int counterbin = 1; counterbin < bins. length -1; counterbin ++) { 78 bins [ counterbin ] = -1.1+( counterbin -1) *0.1; 80 bins [ bins. length -1] = Double. POSITIVE_INFINITY ; double[] hist = MCerrorContinuous. gethistogram ( bins ); 8 84 System. out. println (); 86 System. out. println (" "); System. out. println (" Histogram of Error :"); 88 System. out. println (" "); 90 System. out. println (" bin right " + " " + " counts "); for ( int counterbin = 0; counterbin < bins. length -1; counterbin ++) { 9 System. out. println ( formattersci6. format ( bins [ counterbin +1]) +" "+ hist [ counterbin ]); 94 System. out. println (); 96 System. out. println (" "); System. out. println (" Sample Paths :"); 98 System. out. println (" "); previousvalue = new RandomVariable ( t0, initialvalue ); 100 for ( int countertime =0; countertime < numberoftimes -1; countertime ++) { 10 i f ( countertime ==0) { String string = " time "+ " "; 104 for ( int counterpath = 0; counterpath < numberofpaths ; counterpath ++) { string = string + " path no." + String. valueof ( counterpath )+ " "; 106 System. out. println ( string ); 108 String nextstring = String. valueof ( times [ countertime ]) + " "; 110 for ( int counterpath = 0; counterpath < numberofpaths ; counterpath ++) { 11 nextstring = nextstring + String. valueof ( formattersci6. format ( previousvalue. get ( counterpath )))+ " "; 114 System. out. println ( nextstring ); 116 else { double dt = mytimes. gettimestep ( countertime );//times[ countertime+1] times[countertime];// 118 RandomVariableInterface dw = brownianw. getbrownianincrement ( countertime, 0);// 6

7 getincrementatindex(countertime); RandomVariableInterface nextvalue = previousvalue. mult (1+ mu*dt). add (dw. mult ( previousvalue. mult ( sigma ))); 10 String nextstring = String. valueof ( mytimes. gettime ( countertime ))+ " ";//times[countertime])+ " "; 1 for ( int counterpath = 0; counterpath < numberofpaths ; counterpath ++) { 14 nextstring = nextstring + String. valueof ( formattersci6. format ( nextvalue. get ( counterpath )) )+ " "; 16 System. out. println ( nextstring ); Exercise All we need to check is that the coefficient functions µ(x) = cos(x) and σ(x) cos (x) satisfy the standard hypothesis for the existence and uniqueness of strong solutions. For y < x we have x µ(x) µ(y) = µ (z)dz y x x = sin(z)dz sin(z) dz x y y y and analogously for the diffusion part σ, since σ (z) = cos(z) sin(z) = sin(z) 1. This shows that µ and σ are globally Lipschitz continuous. Linear growth is immediate since µ(x) 1 and σ(x) 1 for all x R. We can now easily deduce the Euler-Maruyama discretization of the SDE as X k+1 = X k X 0 = 0. + cos(x k ) t k cos (X k ) B k The strong convergence of the Euler-Maruyama discretization is a direct consequence of the regularity conditions on the drift and diffusion part which are ensuring existence and uniqueness for the solution of the SDE. 7