Consistency of areas 1 and 2

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Shiryaev’s problem Our model Preliminaries The value function Numerical results

Detecting the maximum of a mean-reverting scalar process

GE.Espinosa Joint work with N.Touzi

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Problem Results

Limits of this model

LetX a Brownian motion. We want to track the maximum ofX over a fixed period ([0,1]).

References :

I Quickest Detection Problems in the Technical Analysis of the Financial Data, A. Shiryaev

I Optimal Stopping of the Maximum Process: the Maximality Principle, G. Peskir

I Predicting the Time of the Ultimate Maximum for Brownian Motion with Drift, J. du Toit and G. Peskir

Only for Brownian motions: importance of the explicit law of (X,X^∗)

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Problem Results

minimizeE`(X₁^∗−Xθ) for a certain loss function `

→ horizon objective References :

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Problem Results

→ horizon objective

Financial Data, A. Shiryaev

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Problem Results

→ horizon objective References :

I Predicting the Time of the Ultimate Maximum for Brownian

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Problem Results

I Maximum process: given z ≥x >0, Zt = sup

s≤t

Xs∨z

V(x) = inf

0≤τ≤1E[`(Z₁−X_τ)|X₀ =Z₀ =x] =V(0)

I The value function would be: v(t,x,z) = inf

t≤τ≤1E[`(Z1−Xτ)|X_t =x,Zt=z] 0≤x ≤z

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Problem Results

s≤t

Xs∨z (Xt,Zt) is a Markov process

I Optimization problem: V(x) = inf

0≤τ≤1E[`(Z₁−X_τ)|X₀ =Z₀ =x] =V(0)

I The value function would be: v(t,x,z) = inf

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s≤t

Xs∨z

(Xt,Zt) is a Markov process

I Optimization problem:

V(x) = inf

0≤τ≤1E[`(Z₁−X_τ)|X₀ =Z₀ =x] =V(0)

I The value function would be:

v(t,x,z) = inf

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Problem Results

Theorem

Let`(x) =x^p,p >0,p 6= 1. Then there exists z_p>0 such as the stopping time:

τ^p= inf{t ≤1, Z_t−X_t ≥z_p√ 1−t}

is optimal.

In the case p = 2,z2 is the unique solution of the equation:

4φ(z)−2zϕ(z)−3 = 0 whereϕ(z) = 1

√ 2πe⁻^z

2

2 and φ(z) = Z z

−∞

ϕ(u)du. The value function is given byV = 2φ(z2)−1

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Theorem

Let`(x) =x^p,p >0,p 6= 1. Then there exists z_p>0 such as the stopping time:

τ^p= inf{t ≤1, Z_t−X_t ≥z_p√ 1−t}

is optimal. In the casep = 2,z₂ is the unique solution of the equation:

4φ(z)−2zϕ(z)−3 = 0 whereϕ(z) = 1

√ 2πe⁻^z

2

2 and φ(z) = Z z

−∞

ϕ(u)du.

The value function is given byV = 2φ(z2)−1

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Problem Results

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Idea of the proof

I First step: use properties of the Brownian motion and its maximum to reduce the problem from 3 dimensions to 1.

I Second step: derive the HJB equation associated to the new problem and solve explicitly this ODE. Then use verification arguments.

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Problem Results

First step

IfE|`(Z₁−Xt)|<∞, then for any stopping time τ in [0,1]:

E(`(Z1−Xτ)|F_τ) =E(`(Zτ∨ max

τ≤r≤1Xr −Xτ)|F_τ)

=E(`(Z_τ∨ { max

0≤r≤1−τB_r +X_τ} −X_τ)|τ,X_τ,Z_τ)

=E(`(max(Zτ−Xτ, ξ))|τ,Xτ,Zτ) whereξ has the following distribution density:

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So that:

V = inf

τ E

"

`(Z_τ −X_τ)(2φ(Z_τ−X_τ

√1−τ )−1) + 2 Z ∞

Zτ√−Xτ 1−τ

`(y√

1−τ)ϕ(y)dy

#

Now asx =z andX is a Brownian motion,Z−X has the same law as|X|. So that (`(x) =x^p):

V = inf

τ E

(1−τ)^p²Hp( |X_τ|

√1−τ)

withHp(y) =y^p+ 2p Z ∞

y

u^p−1(1−φ(u))du

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Problem Results

Now we introduce the ”new” times ∈[0,∞) by:

s =−1

2ln(1−t) LetBs =e^sX_1−e^−2s = ^√^X_1−t^t , we have:

dBs =Bsds+

√ 2dWs

whereWs = 1

√

Z 1−e^−2s √dBu is a Brownian motion.

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We have reduced the problem to the following one:

W(b) = inf

σ≥0Eze^−pσH_p(|B_σ|) whereB is given by B₀=b and the previous dynamics.

And we will haveV =W(0).

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Problem Results

Second step:

We try to solve this problem by considering the following free boundary problem: findzp andW such as:







W⁰⁰(b) +bW⁰(b)−pW(b) = 0 forb ∈(−z_p,z_p) W(±z_p) =Hp(zp)

W⁰(±z_p) =±H⁰(z_p)

And we can solve it explicitly: forb ∈(−z_p,zp), W(b) =Hp(zp)e

z2 p−b2

2 M(^p+1₂ ,¹₂,^b₂²) M(^p+1₂ ,¹₂,^z₂^p²) whereM(a,b,x) = 1 + ^a_bx+_b(b+1)^a(a+1)^x_2!² +...

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Second step:

We try to solve this problem by considering the following free boundary problem: findzp andW such as:







W⁰⁰(b) +bW⁰(b)−pW(b) = 0 forb ∈(−z_p,z_p) W(±z_p) =Hp(zp)

W⁰(±z_p) =±H⁰(z_p)

And we can solve it explicitly: forb ∈(−z_p,zp), W(b) =Hp(zp)e

z2 p−b2

2 M(^p+1₂ ,¹₂,^b₂²) M(^p+1₂ ,¹₂,^z₂^p²) whereM(a,b,x) = 1 +_b^ax+_b(b+1)^a(a+1)^x_2!² +...

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Problem Results

Andz_p is the unique positive solution of:

H_p⁰(z)

Hp(z) +z = (p+ 1)zM(^p+3₂ ,³₂,^z₂²) M(^p+1₂ ,³₂,^z₂²) Outside (−z_p,z_p), we will have W(b) =H_p(|b|).

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Problem Results

I True only for a Brownian motion (possibly with drift as investigated by Peskir)

I How to choose the period (which T)? What is its financial meaning?

I A trader does not want to miss several crossings

I A trader does not want to be in a position for a time of infinite expectation

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Several crossings:

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Problem Results

Infinite expected time of holding:

I sell at timeτ^∗, then wait (at least) until the process crosses 0 to buy back. For a Brownian motion, the average waiting time is ExT0 = +∞, which is a problem for a trader.

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Problem Results

1. `(Z1−Xτ) : horizon objective →interesting but not a natural criteria for trading

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Problem Results

1. `(Z1−Xτ) : horizon objective →interesting but not a natural criteria for trading

2. `(ZT0−Xτ) : gain objective: one defines the trading range (in

%) → natural framework→ never studied before

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Problem Results

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Problem Results

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The process Loss function

Dynamic programming framework

Consider a process starting atx >0 given by the following dynamics:

dX_t =µ(X_t)dt+σ(X_t)dW_t whereµ(x)≤0 for x≥0

X_t = max

0≤s≤tX_s The time of hitting 0 is:

T0 = inf{t≥0,Xt = 0}

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We define the associated current maximum process by:

X_t^∗ = max

0≤s≤tX_s

The time of hitting 0 is:

T0 = inf{t≥0,Xt = 0}

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We define the associated current maximum process by:

X_t^∗ = max

0≤s≤tX_s The time of hitting 0 is:

T0 = inf{t≥0,Xt = 0}

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Consider a loss function`:R+ →R,C², increasing and convex.

We try to solve the following optimization problem V₀(x) = inf

θ≤T0

E`(X_T^∗₀−X_θ)

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Consider a loss function`:R+ →R,C², increasing and convex.

We try to solve the following optimization problem V₀(x) = inf

θ≤T0

E`(X_T^∗₀−X_θ)

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Letz ≥x (the inherited maximum), we introduce the following process

Zt =z∨ max

0≤s≤tXs

The value function of our problem is given by: V(x,z) = inf

θ≤T0

Ex,z`(Z_T₀−X_θ) Notice that it is a stationary problem

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Zt =z∨ max

0≤s≤tXs

The value function of our problem is given by:

V(x,z) = inf

θ≤T0

Ex,z`(Z_T₀−X_θ)

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Zt =z∨ max

0≤s≤tXs

The value function of our problem is given by:

V(x,z) = inf

θ≤T0

Ex,z`(Z_T₀−X_θ) Notice that it is a stationary problem

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Scale function Obstacle function Finiteness ofV A degenerate case

We introduce the functionα(x) =−^2µ(x)_σ(x)₂(≥0) and the scale function of the processX is defined by:

S(x) = Z _x

0

e

Ru

0α(v)dvdu

Assume thatX0=x>0 and Z0 =z ≥x, then the law ofZ_T₀ is given by:

Px,z[Z_T₀≤y] =

(0 ify <z 1−^S(x)_S(y) else In particular,Px,z[Z_T₀ =z] = 1−^S(x)_S_(z)

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We introduce the functionα(x) =−^2µ(x)_σ(x)₂(≥0) and the scale function of the processX is defined by:

S(x) = Z _x

0

e

Ru

0α(v)dvdu We can compute explicitly the law ofZ_T₀: Proposition

Assume thatX0=x>0 and Z0 =z ≥x, then the law ofZ_T₀ is given by:

P [Z ≤y] =

(0 ify <z

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Define the obstacle function

g(x,z) =Ex,z`(Z_T₀−x)

g(x,z) =`(z−x) +S(x) Z +∞

z

`⁰(u−x) S(u) du We defineL(X) as the set of loss functions `such as ^`_S⁰ is integrable in a neighborhood of +∞

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g(x,z) =Ex,z`(Z_T₀−x) From the law ofZT0, we get:

Proposition

g(x,z) =`(z−x) +S(x) Z +∞

z

`⁰(u−x) S(u) du

We defineL(X) as the set of loss functions `such as ^`_S⁰ is integrable in a neighborhood of +∞

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g(x,z) =Ex,z`(Z_T₀−x) From the law ofZT0, we get:

Proposition

g(x,z) =`(z−x) +S(x) Z +∞

z

`⁰(u−x) S(u) du We defineL(X) as the set of loss functions `such as ^`_S⁰ is integrable in a neighborhood of +∞

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We have the following result:

Proposition

Assume that for allx≥0, sup

u≥x

`(u)

`(u−x <∞.

Then there is equivalence between the following assertions:

I V is finite everywhere

I g is finite everywhere

I `∈L(X)

Typically, the additional assumption removes`of the form

`(x) =e^x² (without it we still have (iii)⇒(ii)⇒(i)).

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Proposition

Assume that for allx≥0, sup

u≥x

`(u)

`(u−x <∞.

Then there is equivalence between the following assertions:

I V is finite everywhere

I g is finite everywhere

I `∈L(X)

Typically, the additional assumption removes`of the form

`(x) =e^x² (without it we still have (iii)⇒(ii)⇒(i)).

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If`(x) =x then the problem can be solved directly.

V(x,z) = inf

θ Ex,z(Z_T₀−X_θ) =Ex,zZ_T₀−sup

θ ExX_θ

So we now consider the new value function: W(x) = sup

θ≤T0

ExX_θ And we have the following result:

Proposition

If`(x) =x then V =g

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V(x,z) = inf

θ ExX_θ So we now consider the new value function:

W(x) = sup

θ≤T0

ExX_θ

If`(x) =x then V =g

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V(x,z) = inf

θ ExX_θ So we now consider the new value function:

W(x) = sup

θ≤T0

ExX_θ And we have the following result:

Proposition

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HJB equation

Reduction to a free boundary problem Study ofLg

Equation of the obstacle Other areas

From now on, we assume that`(x) = ^x₂²

The HJB equation of the problem is:







min{LV,g −V}= 0 V(0,z) =`(z) V_z(z,z) = 0

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HJB equation

From now on, we assume that`(x) = ^x₂² We writeL= _∂x^∂²2 −α(x)_∂x^∂







min{LV,g −V}= 0 V(0,z) =`(z) V_z(z,z) = 0

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From now on, we assume that`(x) = ^x₂² We writeL= _∂x^∂²2 −α(x)_∂x^∂







min{LV,g −V}= 0 V(0,z) =`(z) V_z(z,z) = 0

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HJB equation

We guess that the solution should be of the following form: find γ:R+ →R+, andv such as:

γ(x)≥x And:









Lv = 0 if 0<z < γ(x)

v =g andLg(x,z)≥0 ifz ≥γ(x) v_z(z,z) = 0

v(0,z) =`(z)

v(x, γ(x)) =g(x, γ(x))

We will use a verification argument in the end

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We guess that the solution should be of the following form: find γ:R+ →R+, andv such as:

γ(x)≥x And:











Lv = 0 if 0<z < γ(x)

v =g andLg(x,z)≥0 ifz ≥γ(x) v_z(z,z) = 0

v(0,z) =`(z)

v(x, γ(x)) =g(x, γ(x)) v_x(x, γ(x)) =g_x(x, γ(x)) We will use a verification argument in the end

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HJB equation

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HJB equation

In the area whereV =g, the HJB equation implies that Lg ≥0.

Therefore we need to study the set Γ⁺={(x,z), Lg(x,z)≥0}.

we have:

I Γ⁺={(x,z), Lg(x,z)≥0}is above the graph of Γ

I Γ⁻={(x,z), Lg(x,z)≤0}is below the graph of Γ

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HJB equation

In the area whereV =g, the HJB equation implies that Lg ≥0.

Therefore we need to study the set Γ⁺={(x,z), Lg(x,z)≥0}.

We call Γ(x) = min{z ≥x, Lg(x,z)≥0}, which is a C⁰ and piecewiseC¹ function, decreasing and then increasing. Moreover, we have:

I Γ⁺={(x,z), Lg(x,z)≥0} is above the graph of Γ

I Γ⁻={(x,z), Lg(x,z)≤0} is below the graph of Γ

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HJB equation

We call:

I Γ₁ the decreasing part and Γ₂ the increasing part of Γ.

I z₀ = Γ(0)

I z1 = sup{z >0, Lg(z,z)<0}(possibly infinite)

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HJB equation

Recall thatL= _∂x^∂²2 −α(x)_∂x^∂ → only partial derivatives wrtx.

FromLv = 0 together withv(x, γ(x)) =g(x, γ(x)),

v_x(x, γ(x)) =g_x(x, γ(x)) and v_z(x,x) = 0 we get the following ODE:

γ⁰(x) = Lg(x, γ(x)) 1−_S(γ(x))^S(x) And we want (x, γ(x))∈Γ⁺.

Problem: we do not have an a priori initial condition (not a Cauchy problem). →We must choose the right one that will allow us to derive a verification result.

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HJB equation

γ⁰(x) = Lg(x, γ(x)) 1−_S(γ(x))^S(x)

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HJB equation

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HJB equation

We show that it is possible to chooseγ to be the sup of the solutions that cross Γ2 and that we then have:

x→zlim1

γ(x)−x= 0 Moreover, thisγ may cross Γ₁ but not Γ₂.

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HJB equation

In general,γ defined before might cross Γ1. Therefore, the boundary will be made of 2 distinct parts (as we need (x, γ(x))∈Γ⁺).

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HJB equation

As before, but replacing the Neumann condition by the Dirichlet condition on{x = 0} we get the following equation:

z²

2 −g(x,z) +gx(x,z)S(x) S⁰(x) = 0

There existsz^∗ <z₀ such as: for anyz ∈[z^∗,z₀] the solution of f(x,z) = 0 admits a unique solution x(z) that satisfies

- Γ⁻¹₁ (z)≤x(z)≤Γ⁻¹₂ (z)(≤z) - if z <z₀,x(z)>0

- if z =z^∗, then (x(z^∗),z^∗)∈Γ2

- x(z) is C¹

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As before, but replacing the Neumann condition by the Dirichlet condition on{x = 0} we get the following equation:

z²

2 −g(x,z) +gx(x,z)S(x) S⁰(x) = 0 We have the following result:

There existsz^∗ <z₀ such as: for any z ∈[z^∗,z₀] the solution of f(x,z) = 0 admits a unique solutionx(z) that satisfies

- Γ⁻¹₁ (z)≤x(z)≤Γ⁻¹₂ (z)(≤z) - if z <z₀,x(z)>0

- if z =z^∗, then (x(z^∗),z^∗)∈Γ2

- x(z) is C¹

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HJB equation

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HJB equation

Consistency of areas 1 and 2

The previous results guarantee that we are in one of the following situations:

I the graph of γ defined in area 2 does not cross Γ

I the graphs of the two previous γ will intersect on a (unique) point (¯x,¯z)

We now call γ the following function:

γ=







γ₁ on [0,¯x] γ2 on [¯x,z1]

id on [z₁,+∞) ifz₁ <∞

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HJB equation

Consistency of areas 1 and 2

I the graph of γ defined in area 2 does not cross Γ then area 1 is degenerate, we write (¯x,¯z) = (0,z₀)

We now call γ the following function:

γ=







γ₁ on [0,¯x] γ2 on [¯x,z1]

id on [z₁,+∞) ifz₁ <∞

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HJB equation

Consistency of areas 1 and 2

γ=





γ₁ on [0,¯x] γ2 on [¯x,z1]

id on [z₁,+∞) ifz₁ <∞

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HJB equation

Consistency of areas 1 and 2

We now callγ the following function:

γ₁ on [0,¯x]

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Now thatγ, ¯x and ¯z are defined, we define v on areas 1 and 2 by:

ifx ≤¯x,z ∈[¯z, γ(x)]

v(x,z) =v₁(x,z) = z²

2 +g_x(γ₁⁻¹(z),z) S(x) S⁰(γ₁⁻¹(z)) ifx ≥¯x,z ∈[¯z, γ(x)]

v(x,z) =v2(x,z) =g(γ₂⁻¹(z),z)

+gx(γ₂⁻¹(z),z)S(x)−S(γ₂⁻¹(z)) S⁰(γ₂⁻¹(z))

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HJB equation

Area 3

Here we want to solveLv = 0 with the Dirichlet condition on {x= 0}, the Neumann condition on {x =z}, and 2 Dirichlet conditions on{(x,¯z),0≤x ≤¯x} and{(x,¯z),¯x ≤x≤z}

We set:

v(x,z) =v₃(x,z) = z²

2 +S(x)[

Z +∞

z

u

S(u)du−K]

+∞ g (¯x,¯z)

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Area 4

Herev =g, and by construction,Lg ≥0

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HJB equation

Verification

Finally, we must show thatv defined previously satisfiesv ≤g and v≥0. Then we are able to claim the following result:

Theorem(Verification theorem)

The functionv defined previously is indeed the value function V of our initial problem andτ^∗= inf{t ≥0,Z_t≥γ(X_t)}is an optimal stopping time

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γ has the following form for an OU process :

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Plot ofγ

Theoretical simulations Market data simulations

We simulate an OU (α(x) =x) and use the following strategy:

when we ”detect” the maximum, we sell 1 stock, when we cross 0, we close the position, we do the same for the minimum

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We simulate an OU (α(x) =x) and use the following strategy:

when we ”detect” the maximum, we sell 1 stock, when we cross 0, we close the position, we do the same for the minimum

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Plot ofγ

We compute the following ”mean-reverting” process:

X =

BNP Socgen

MA(_Socgen^BNP ) −1

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We compute the following ”mean-reverting” process:

X =

BNP Socgen

MA(_Socgen^BNP ) −1

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Plot ofγ

Extension

Inhomogeneous problem:

dX_t =µ(t,X_t)dt+σ(t,X_t)dW_t whereµ(t,x)≤0 forx ≥0,t ≥0

Problem: Hard to find a regular upper bound forV in this case (necessary for a comparison result).

But we can find one using the homogeneous case if the inhomogeneity is only on the drift:

dX_t =µ(t,X_t)dt+σ(X_t)dW_t

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Plot ofγ

Extension

dX_t =µ(t,X_t)dt+σ(X_t)dW_t

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Plot ofγ

Extension

But we can find one using the homogeneous case if the inhomogeneity is only on the drift:

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Short Bibliography

- The Pricing of the American Option, R. Myneni, The Annals of Applied Probability (1992)

- Quickest Detection Problems in the Technical Analysis of the Financial Data, A. Shiryaev, Mathematical Finance - Bachelier Congress (2000), Springer Finance

- Optimal Stopping of the Maximum Process : the Maximality Principle, G. Peskir, Ann. Prob. (1998)

- Predicting the Time of the Ultimate Maximum for Brownian Motion with Drift, J. du Toit and G. Peskir, Proc. Math.

Control Theory Finance (2007), Springer

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Plot ofγ

Special thanks to J.Lebuchoux - Reech Aim