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Black-Scholes price is integral

over (0,1)

â€¢ We assumed

â€“ current stock price=exercise price=10

â€“ Annual vol (sigma)=0.2.

â€“ Interest rate =5%.

â€“ T=1/4 years.

â€¢ Monte-Carlo integral obtained by

â€“ u=rand(1,500000); mean(fn(u));

Accuracy of Crude Monte-Carlo

option price:

n

1 1

â€¢ Recall that Var ( âˆ‘ f (U )) = Var ( f (U ))

i 1

n n

i =1

â€¢ The standard error or standard deviation

of the estimated Monte-Carlo integral is

estimated by

â€¢ se=sqrt(var( fn(u))/length(u));

â€¢ True val=.4615, standard error approx

.002

The true value by Black-Scholes

formula ( Matlab)

â€¢ [CALL,PUT]=BLSPRICE(S0,K,r,T,sigma,0)

last argument = dividend rate assumed 0

in this problem. CALL= price of call option,

PUT=price of put option both assuming

Black-Scholes model.

â€¢ [CALL,PUT]=BLSPRICE(10,10,.05,.25,.2,0)

Example of crude Monte Carlo

â€¢ u=[.1 .25 .4 .6 .7 .8] mean(fn(u)) gives 0.3146

â€“ This is total area under bars (approximating integral)

Antithetic Random Numbers

â€¢ If we use a given uniform random

number u, also use 1-u. Achieves a

balance between large values of f(u) and

small values. Example u=[.1 .25 .4 .9 .75

.6] gives mean(fn(u)) 0.4222

Why are antithetic random numbers

better? Consider independent uniform

Consider just two uniform random numbers

U 1 , U 2 both uniform on [0,1] and independen t.

1

Estimator is ( f (U 1 ) + f (U 2 )).

2

1

1 1

âˆ« f (u ) du .

Expected value is E[ ( f (U 1 ) + f (U 2 ))] = ( Ef (U 1 ) + Ef (U 2 )) =

2 2 0

Estimator has Variance

1 1

var[ ( f (U 1 ) + f (U 2 ))] = (var( f (U 1 )) + var( f (U 2 )) + 2 cov( f (U 1 ), f (U 2 )))

2 4

1

= (var( f (U 1 )) since var( f (U 1 )) = var( f (U 2 ))

2

and by independen ce, cov( f (U 1 ), f (U 2 )) = 0

Expected value when they are

antithetic

â€¢ The expected value is the same when

antithetic:

U 1 , U 2 both uniform on [0,1] and U 1 = 1 âˆ’ U 2

1

Estimator is ( f (U 1 ) + f (1 âˆ’ U 1 )).

2

Expected value is

1 1

E[ ( f (U 1 ) + f (1 âˆ’ U 1 ))] = ( Ef (U 1 ) + Ef (1 âˆ’ U 1 ))

2 2

1

âˆ« f (u ) du .

=

0

Variance of Antithetic estimator

(n=2)

Estimator has Variance

1 1

var[ ( f (U1 ) + f (1 âˆ’ U1 ))] = (var( f (U1 )) + var( f (1 âˆ’ U1 )) + 2 cov( f (U1 ), f (1 âˆ’ U1 )))

2 4

1

= (var( f (U1 ) + cov( f (U1 ), f (1 âˆ’ U1 ))) since var( f (U1 )) = var( f (U 2 ))

2

This is better than crude if cov( f (U1 ), f (1 âˆ’ U1 )) < 0.

In other words when f (U1 ) is large, f (1 âˆ’ U1 ) tends to be smaller.

TRUE if f(u) is monotone function of u.

The antithetic estimator has smaller variance provided f is monotone

Summary of antithetic

â€¢ Instead of using n uniform[0,1], use n/2 ( u, say)

together with n/2 values of 1-u. This is better than n

crude IF the function f(u) is monotonic (increasing or

decreasing).

â€“ How much better?? For crude, u=rand(1,500000);

mean(fn(u)) (0.4620)

â€“ var(fn(u))/500000 = 8.7e-007

â€¢ ANTITHETIC********************

â€“ u=rand(1,500000); e=(fn(u)+fn(1-u))/2; mean(e) =

0.4630

â€“ var(e)/length(e) (2.23e-7â€”equivalent to about 2 million

crude)

Efficiency Gain

â€¢ The same number of calls to the function fn

using antithetic gives ratio of variances

2.1723/1.1244 relative to crude. Nearly twice

as efficient (half the sample size necessary)

â€¢ In this example, if we insist on a certain

precision (variance of estimator), antithetic

method will get it with half as many calls to the

function. (script8)

Antithetic: efficiency higher if

function more linear

1

Find âˆ« (1 âˆ’ x ) dx.

2 3/ 4

0

The true value is approx 0.71888.

â€¢ Using crude

â€“ u=rand(1,100000); f= (1-u.^2).^(3/4);

â€“ Mean(f)

â€“ Var(f)/100000

%mean(f) Var(f)/100000= 0.7193 7.1783e-007

â€¢ Using Antithetic

â€“ u=rand(1,50000); f= .5*((1-u.^2).^(3/4)+ (1-(1-u).^2).^(3/4));

â€“ mean(f) %mean(f) Var(f)/50000 = 0.7184 1.3704e-007

â€“ Var(f)/50000

â€¢ Efficiency gain=7.18/1.37= 5.24

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