Christophe Chorro - Variance reduction techniques: Around B&Schristophe.chorro.fr/docs/MC4.pdf ·...

Variance reduction techniques: Around B&S

Christophe Chorro ([email protected])

University Paris 1

July 10 2008

Christophe Chorro ([email protected]) (University Paris 1)Introduction to Monte Carlo Methods (Lecture 3 and 4) July 10 2008 1 / 45

Study Plan

Variance reduction techniquesIntroductionReminder on B&S ModelControl VariateAntithetic VariablesImportance samplingConditioning

Biblio: • Stochastic Calculus and Black Scholes:

http://christophe.chorro.fr/docs/CSangl.pdf

• A.G.Z Kemna and A.C.F. Vorst, A pricing method for options based onaverage asset values, J. Banking Finan., 1990, March, 113–129.


Plan

1 Variance reduction techniquesVariance reduction techniques:IntroductionReminder on B&S modelControl VariateAntithetic VariablesImportance samplingConditioning


Variance reduction techniques: Introduction

Let (Xn)n∈N be i.i.d random variables with values in R such that E[|X1|2] < ∞.

For large n, with a confidence of 95%,

E[X1] ∈[

Snn− 1.96σ√

n,

Snn

+1.96σ√

n

]with σ2 = Var(X1).

The magnitude of the error is given by 1.96σ√n : the size of σ is fundamental forthe speed of convergence.

Idea: Reduce σ

Find Y such that E[X1] = E[Y ] and Var(X1) > Var(Y ).


Plan



Brownian motion

We consider a probability space (Ω,A, P).

DefinitionStandard Brownian motion (B.M) is a stochastic process (Bt)t∈[0,T ] fulfilling :

a) B0 = 0 P-a.s.

b) B is continuous i.e t → Bt(w) is continuous for P almost all w.

c) B has independent increments: For Si t > s, Bt − Bs is independent ofFBs = σ(Bu, u ≤ s).

d) the increments of B are stationary and gaussian: For t ≥ s, Bt − Bs followsa N (0, t − s).


Brownian motion

We consider a subdivision 0 = t0 < ... < tn = T of [0, T ]. We want to simulate

(Bt0 , ...Btn).

Idea:Btk = Btk−1 + Btk − Btk−1︸︷︷︸

N (0,tk−tk−1)⊥Btk−1 ,...,B0

.

PropositionIf (G1, ...Gn) are i.i.d N (0, 1), we define

X0 = 0, Xi =i∑

j=1

√tj − tj−1Gj i > 0.

Then(X0, ..., Xn) =

D(Bt0 , ...Btn).


Brownian motion

0.0 0.4 0.8−

0.8

−0.

20.

40.0 0.4 0.8

−1.

5−

0.5

0.0 0.4 0.8

−0.

60.

00.

4

0.0 0.4 0.8−

0.5

0.5

Figure: 4 paths of the Brownian motion on [0, 1] generated using the precedingmethod with the regular subdivision of step 0.001.


Black Scholes model

We consider the time interval [0, T ] and r the risk free rate (supposed to beconstant) during this period.

Non-risky asset: Its dynamic is given by

S00 = 0, S0t = e

rt .

Risky asset: Under the historical probability P its dynamic is given by thefollowing SDE:

dSt = µStdt + σStdBt (1)

with initial condition S0 = x0 > 0 and where B is a standard BM under P.

Itô formula ⇒ St = x0e(µ−12 σ

2)t+σBt .


Black Scholes model

0.0 0.4 0.81

35

sigma=1, mu=1

0.0 0.4 0.8

0.4

0.7

1.0

sigma=1, mu=−1

0.0 0.4 0.8

1.0

2.5

4.0

sigma=0.5, mu=1

0.0 0.4 0.80.

40.

71.

0

sigma=0.5, mu=−1

Figure: Simulation of a path of the risky asset in the B&S model for differentparameters


Black Scholes model

What is, in this model, the price of a contingent claim with payoff ΦT at T?

PropositionIn the B&S model there exists a unique probablity Q ∼ P such that the priceat t = 0 of a contingent claim with payoff ΦT at T is given by

price = e−rT EQ[ΦT ].

Moreover the dynamic of the risky asset under Q is given by

dSt = rStdt + σStdWt (µ ⇔ r) (2)

where W is a standard BM under Q.


Black Scholes model

Examples: The Black Scholes Formulas

For Call options (ΦT = (ST − K )+) one has

EQ[(ST − K)+] = S0N(d1)− Ke−rTN(d2) (3)

where

d1 =log(S0K ) + (r +

σ2

2 )T

σ√

Tand d2 =

log(S0K ) + (r −σ2

2 )T

σ√

T(4)

and where N is the distribution function of a N (0, 1).

For Put options (ΦT = (K − ST )+) one has

EQ[(K− ST)+] = −S0N(−d1) + Ke−rTN(−d2). (5)


Black Scholes model

This formulas are fundamental because

They are easy to compute in practice

They are a Benchmark to test numerical methods (for example variancereduction techniques)

In the sequel, all the variance reduction methods will be test computing byMonte carlo simulations

E[(K − eσG)+] or E[(eσG − K )+]

where G ↪→ N (0, 1) i.e computing (up to some coefficients) the price of a call

or a put option in a B&S model where S0 = 1 and r = 0.


Black Scholes model: Numerical example

Price of a call E[(eσG − K )+] with σ = 0.5 and K = 1Exact value=0.28353Estimated value (N=100) = 0.249, confidence interval at 95% : [0.164; 0.334]Estimated value (N=1000) = 0.279, confidence interval [0.248; 0.308]Estimated value (N=10000) = 0.276, confidence interval [0.267; 0.285]

Price of a put E[(K − eσG)+] with σ = 0.5 and K = 1.Exact value=0.15038Estimated value (N=100) = 0.154, confidence interval at 95% : [0.112; 0.195]Estimated value (N=1000) = 0.155, confidence interval [0.143; 0.167]Estimated value (N=10000) = 0.149, confidence interval [0.145; 0.152]


Black Scholes model

0.5 1.0 1.5 2.0 2.5

0.0

0.1

0.2

0.3

0.4

0.5

0.6

Call (green) and put (red)

K

Sta

ndar

d de

viat

ion

Figure: Standard deviation of the preceding payoffs (σ = 0.5) for different strikes


Black Scholes model

In practical cases, the moneyness S0K of most liquid options (in our exampleS0 = 1) belongs in general to [0.7, 1.3].

Thus, It’s better to price put than call by Monte Carlo methods.

To recover the price of the call we may use the Call-Put parity:

E[(eσG − K )+] = E[(−eσG + K )+] + eσ22 − K .

Price of a call E[(eσG − K )+] with σ = 0.5 and K = 1 computed byCall-Put parity and MC

Exact value=0.28353Estimated value (N=100) = 0.281, confidence interval at 95% : [0.240; 0.320]Estimated value (N=1000) = 0.285, confidence interval [0.272; 0.297]Estimated value (N=10000) = 0.287, confidence interval [0.283; 0.291]


Plan



Controle Variate

We want to compute by Monte Carlo method E[f (X )].

Idea:E[f (X )] = E[f (X )− h(X )] + E[h(X )]

where h is chosen to ensure that

E[h(X )] may be computed explicitly

Var(f (X )− h(X )) � Var(f (X )) (intuitively we look for f (X )− h(X ) small).

Example: Call-Put parity


Controle Variate: Basket options (Homework 1)

Aim: Price by MC method a Basket put in the B&S framework

Let G1 and G2 be two independent N (0, 1) and (σ11, σ12, σ21, σ22) ∈ R4. Wedefine

S1 = eσ11G1+σ12G2

S2 = eσ21G1+σ22G2 .

For a1, a2 > 0 with a1 + a2 = 1, we are interested in the numericalcomputation (no closed form formula) of

E

K − (a1S1 + a2S2)︸︷︷︸

X

+

.



First Idea: Classical MC methods

With a1 = a2 = 0.5, K = 1, σ11 = σ22 = 0.1 and σ12 = σ21 = 0.15,

Box Muller Method to generate gaussian random variables

Estimated value (N=10000) =6, 334.10−2 , confidence interval at 95% :

[6, 159.10−2; 6, 508.10−2]

Inversion Method to generate gaussian random variables:

Estimated value (N=10000) = 6, 336.10−2, confidence interval at 95% :

[6, 161.10−2; 6, 511.10−2]



Second Idea: Control variate when (σ11, σ12, σ21, σ22) are small.

We haveX ∼ eY

where

Y = a1Log(S1) + a2Log(S2) ↪→ N(0, σ2

)with

σ2 = (a1σ11 + a2σ21)2 + (a1σ12 + a2σ22)2

And we remark that

E[(K − X )+

]= E

[(K − X )+ −

(K − eY

)+

]+ C

where

C = E[(

K − eY)+

]= e

σ22 N(

log(K )σ

− σ) + KN( log(K )σ

).



We use a classical Monte Carlo Method (N = 10000) to approximate

E[(K − X )+ −

(K − eY

)+

]and deduce E

[(K − X )+

].

With a1 = a2 = 0.5, K = 1, σ11 = σ22 = 0.1 and σ12 = σ21 = 0.15,

Estimated value (N=10000) =6, 312.10−2 , confidence interval at 95% :

[6, 311.10−2; 6, 313.10−2]


Controle Variate: Asian Options with Kemna and VorstMethodAim: Price by MC method an asian call in the B&S model

Thus we have to approximate (no closed form formula)

e−rT E

[(1T

∫ T0

Ssds − K

)+

]where ∀t ∈ [0, T ]

St = x0e(r−12 σ

2)t+σWt .

First Idea:1T

∫ T0

Ssds ≈1p

p∑k=0

S kTp

+ MC

With σ = 0.2, x0 = K = 100, r = 0.1, T = 1 and p = 50 one obtainsEstimated value (N=10000) = 6.93, confidence interval at 95% :[6.769; 7.101]


Controle Variate: Asian Options with Kemna and VorstMethod

Second Idea: When σ and r are small Kemna and Vorst tell us that

1T

∫ T0

Ssds ≈ exp

(1T

∫ T0

log(Ss)ds

)= AT .

We are going to use the identity

e−rT E[(

1T

∫ T0 Ssds − K

)+

]= e−rT E

[(1T

∫ T0

Ssds − K

)+

− (AT − K )+

]︸︷︷︸

(1)

+ e−rT E [(AT − K )+]︸︷︷︸(2)


Controle Variate: Asian Options with Kemna and VorstMethod

For (1) We use the first idea (Approximation by Riemann sums)

For (2) We just have to remark that

AT = exp

(1T

∫ T0

log(Ss)ds

)↪→ exp

(N ((r̃ − σ̃

2

2)T , σ̃2T )

)

where r̃ = r2 −σ2

12 and σ̃ =σ√3.

Thus we may use Black Scholes formula to compute explicitly (2).

With σ = 0.2, x0 = K = 100, r = 0.1, T = 1 and p = 50 one obtains

Estimated value (N=10000) = 6.981, confidence interval at 95% :[6.969; 6.993].


Plan



Antithetic Variables

If (Xi)i∈N∗ are independent with the same distribution than X we mayapproximate E[g(X )] by

E[g(X )] ≈ 12n

2n∑k=1

g(Xk ) =1

2n

n∑k=1

g(X2k−1) + g(X2k ).

Suppose that there exists a transformation T : R → R such that

T (X ) has the same distribution than X .

In this case we may use

E[g(X )] =E[g(X ) + g(T (X ))]

2≈ 1

2n

n∑k=1

g(Xk ) + g(T (Xk )).

Question: Is Var [g(X1) + g(X2)] > Var [g(X1) + g(T (X1))]?



We haveVar [g(X1) + g(X2)] = 2Var [g(X1)]

and

Var [g(X1) + g(T (X1))] = 2Var [g(X1)] + 2Cov [g(X1), g(T (X1))].

So,

Cov [g(X1), g(T (X1))] < 0 ⇒ Var [g(X1) + g(X2)] > Var [g(X1) + g(T (X1))] .

Under some simple conditions on g and T ,

Cov [g(X1), g(T (X1))] < 0 holds.



LemmaIf X is a random variable and f , h : R → R two non decreasing (or nonincreasing) mappings,

E[h(X )f (X )] ≥ E[h(X )]E[f (X )].

CorollaryIf g is monotonic and T non increasing,

Cov [g(X1), g(T (X1))] ≤ 0.



Proof of the lemma: Let Y be a random variable independent from X withthe same distribution.

FromE[(h(X )− h(Y ))(f (X )− f (Y ))] ≥ 0

we deduce

E[h(X )f (X )] + E[h(Y )f (Y )] ≥ E[h(X )f (Y )] + E[h(Y )f (X )].

Using independence and equi-distribution

E[h(X )f (X )] ≥ E[h(X )]E[f (X )]

thus

Cov [f (X ), h(X )] ≥ 0.


Antithetic Variables for call options in B&S

We want to apply the preceding method to

c = E[(eσG − K )+].

Here we have to remark that

g(x) = (eσx − K )+ is non decreasing (σ > 0),

T (x) = −x is non increasing

G and −G have the same distribution,

thus, we may use the following approximation

E[g(G)] ≈ 12n

n∑k=1

g(Gk ) + g(T (Gk )).


Antithetic Variables for call options in B&S

Numerical results for σ = 0.5 and K = 1

Without antithetic variablesExact value=0.28353Estimated value (n=100) = 0.238, confidence interval at 95% : [0.151; 0.324]Estimated value (n=10000) = 0.285, confidence interval [0.275; 0.295]

With antithetic variablesExact value=0.28353Estimated value (n=100) = 0.283, confidence interval at 95% : [0.227; 0.338]Estimated value (n=10000) = 0.283, confidence interval [0.276; 0.289]


Plan



Importance sampling

Let X having the distribution fX (x)dx , we want to approximate

E[g(X )].

We consider another random variable Y with distribution fY (y)dy . One has

E[g(X )] =∫

Rg(x)fX (x)dx =

∫R

g(x)fX (x)fY (x)fY (x)

dx = E[g(Y )

fX (Y )fY (Y )

].

If we are able to simulate de distribution of Y we may use MC method toapproximate E

[g(Y ) fX (Y )fY (Y )

].

Question: If Z = g(Y ) fX (Y )fY (Y ) , is Var(Z ) < Var(g(X )?


Importance sampling

But

Var(Z ) =∫

Rg(x)2

f 2X (x)fY (x)

dx − E[g(X )]2.

When g ≥ 0, fY (x) = g(x)fX (x)E[g(X)] ⇒ Var(Z ) = 0

Problem: E[g(X )] is unknown...

Idea: Take fY (x) ≈ cste | fX (x)g(x) |.


Importance sampling: example 1

We want to compute by MC methods∫ 10

cos(πx2

)dx = E[cos(πU2

)] whereU ↪→ U([0, 1]).

g(x) = cos(πx2 ) is odd, g(0) = 1, g(1) = 0 thus we take

fY (x) = cste (1− x2) 1[0,1](x).

fY density ⇒ cste = 32

If Y has the density fY ,∫ 10

cos(πx2

)dx = E[Z ] where Z =2cos(πY2 )3(1− Y 2)

.

In this case

Var(Z ) ≈Var(cos(πU2 ))

100...


Importance sampling: example 1

Remark: In the preceding example we are able to simulate Y with density

fY (x) =32

(1− x2) 1[0,1](x)

using the next result we have already seen (rejection method...)

PropositionLet fY be a density on R and

DfY = {(x , u) ∈ R× R+ | 0 ≤ u ≤ fY (x)}.

Let (Y , U) be a random variable on R× R+, then,

(Y , U) ↪→ U(DfY ) ⇔ Y has density fY and ∀x ∈ R+, U | x ↪→ U([0, fY (x)]).


Importance sampling: Call option in B&S

We want to use the preceding method to compute

c = E[(eσG − 1)+]

where G ↪→ N (0, 1).

Idea: Since σ is small, eσx − 1 ≈ σx thus

c =∫

R+(eσx − 1) 1√

2πe−

x22 dx =

∫R+

(eσx − 1)cste σx

cste σx√2π

e−x22︸︷︷︸

≈fY (x)

dx

y = x2,

c =∫

R+

(eσ√

y − 1)√2πy

e−y2

dx2

= E

[(eσ

√Y − 1)√2πY

]where Y ↪→ E(0.5).


Importance sampling: Call option in B&S

Numerical results for σ = 0.5 :

Without importance samplingExact value=0.28353Estimated value (N=100) = 0.249, confidence interval at 95% : [0.164; 0.334]Estimated value (N=10000) = 0.276, confidence interval [0.267; 0.285]

With importance samplingExact value=0.28353Estimated value (N=100) = 0.287, confidence interval at 95% : [0.275; 0.300]Estimated value (N=10000) = 0.284, confidence interval [0.283; 0.285]


Plan



Conditioning

We want to compute by M.C methods

E[g(X , Y )].

If h(X ) = E[g(X , Y ) | X ] we have from classical properties that

E[g(X , Y )] = E[h(X )]

From Jensen inequality for conditional expectation:

h2(X ) ≤ E[g2(X , Y ) | X ], so

Var [g(X , Y )] ≥ Var [h(X )].

Pb: MC for E[h(X )] but we have to know explicitly h.....


ConditioningSuppose that the pair (X , Y ) has the distribution fX ,Y (x , y)dxdy . We know inthis case that:

The distribution of X is given by fX (x)dx where fX (x) =∫

R fX ,Y (x , y)dy .

The distribution of Y is given by fY (y)dy where fY (y) =∫

R fX ,Y (x , y)dx .

The conditional distribution of Y given X = x is given by fY |X (x , y)dywhere

fY |X (x , y) = 1fX (x)>0fX ,Y (x , y)

fX (x).

Thus

h(X ) = E[g(X , Y ) | X ]

where

h(x) =∫

Rg(x , y)fY |X (x , y)dy .


Conditioning: Best of call option

We want to approximate

p = E[(Max(eσ1G1 , eσ2G2)− K )+]

where G1 and G2 are two i.i.d N (0, 1).

First idea: Classical MC...

Second idea: Conditioning by G1 + MC...

We have p = E[h(eσ1G1)] where

h(x) = E[(Max(x , eσ2G2)− K )+]

where h may be explicitly known.



In fact h(x) = E[(Max(x , eσ2G2)− K

)+] but

(Max(x , eσ2G2)− K

)+

= (eσ2G2 − K )+ 1{x≤K}

+(x − K + (eσ2G2 − x)+

)1{x>K}.

Thus

h(x) =(

eσ222 N(− log(K )σ2 + σ2)− KN(−

log(K )σ2

)

)1{x≤K}

+(

x − K + eσ222 N(− log(x)σ2 + σ2)− xN(−

log(x)σ2

)

)1{x>K}.

h is explicit.



With σ1 = 0.2, σ2 = 0.5, K = 1,

Without conditioning (Exact value 0.338)

Estimated value (N=100) = 0.318, confidence interval at 95% : [0.219; 0.417]

Estimated value (N=10000) = 0.327, confidence interval at 95% :[0.318; 0.336]

With conditioning (Exact value 0.338)

Estimated value (N=100) = 0.344, confidence interval at 95% : [0.323; 0.365]

Estimated value (N=10000) = 0.339, confidence interval at 95% :[0.338; 0.341]


Variance reduction techniquesVariance reduction techniques:IntroductionReminder on B&S modelControl VariateAntithetic VariablesImportance samplingConditioning

Christophe Chorro - Variance reduction techniques: Around B&Schristophe.chorro.fr/docs/MC4.pdf ·...

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