Stochastic Target Problems
Stochastic Target H. Mete SONER, Department of Mathematics ETH Zürich Swiss Finance Institute INFORMS 2015 Koç Üniversitesi, 5 Temmuz, 2015
0
Collaborators
I have benefited the collaboration of many people including : Erdinç Akyıldırım, Albert Altarovici, Yan Dolinsky, Romuald Elie, Selim Gökay, Ludovic Moreau, Johannes Muhle-Karbe, Marcel Nutz, Dylan Possamaï, Max Reppen, Jianfeng Zhang and
Nizar Touzi
Bruno Bouchard
1
Foreword
In the classical Markovian optimal control theory, one tries to maximize (or minimize) an objective functional of the form [∫ E
0
T
] f(Xαt , αt )dt
+
Φ(XαT )
over all admissible controls α. Here Xα is the controlled state process.
2
Foreword
For the stochastic target problems, one would like to satisfy a constraint of the form XαT ∈ T ,
P − a.s.,
where T ⊂ Rd is a given deterministic target. As such the first is an L1 theory while the stochastic target is an L∞ approach. The stochastic target problem is closely related to stochastic viability theory as developed by Aubin, Cardaliaguet, DaPrato, Frankowska,.... 3
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
4
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
5
Problem
Given are ▶
▶ ▶ ▶
d Controlled state process (Xα,x t )t≥0 ∈ R with initial data Xα,x 0 = x and an admissible control process α ∈ A defined on a filtered probability space (Ω, {Ft }) ; A set of probability measures (or models) P ; Desired success probability z ∈ [0, 1] ; Deterministic target set T ⊂ Rd .
Goal is to characterise the reachability set { ( ) V(t) := x : ∃ α ∈ A so that P Xα,x ∈ T ≥ z, t
∀P∈P
}
.
6
Problem
Given are ▶
▶ ▶ ▶
d Controlled state process (Xα,x t )t≥0 ∈ R with initial data Xα,x 0 = x and an admissible control process α ∈ A defined on a filtered probability space (Ω, {Ft }) ; A set of probability measures (or models) P ; Desired success probability z ∈ [0, 1] ; Deterministic target set T ⊂ Rd .
Goal is to characterise the reachability set { ( ) V(t) := x : ∃ α ∈ A so that P Xα,x ∈ T ≥ z, t
∀P∈P
}
.
I concentrate on the case z = 1 and P = {P} at the beginning.
6
Black-Scholes
7
Black-Scholes
Consider a financial market consisting of a liquidly traded asset with a random price process Su and a constant interest rate of r. Assume that S is a geometric Brownian motion, i.e., dSt = St [µdt + σdWt ],
t > 0,
where µ > r, volatility σ > 0 and W is a Brownian motion. Further given is a general European option that will pay g(St ) at time t with a known deterministic function g. ˆt := e−rt St , Given S0 = s, consider the function with S { } ∫ t −rt ˆ v(t, s) := inf y : ∃ θ such that y + θu dSu ≥ e g(St ) a.s. 0
7
Black-Scholes, II
In the notation of the general problem, control is the portfolio process θ and the state process is ( [ ] ) ∫ t θ,(y,s) θ,(y,s) s rt s s ˆ u , St , Xt = (Yt , St ) = e y + θu dS 0
and the target set is the epigraph of g, { } T = (y, s) ∈ R2+ : y ≥ g(s) . Then, the reachability set V(t) is also an epigraph and v(t, s) = inf { y : (y, s) ∈ V(t) } . The function v(t, s) is the smallest all prices y that allows the seller to hedge the claim g(St ) with probability one. 8
Jane and Paul Game
9
Jane and Paul Game
▶
Instead of explaining the continuous time stochastic problem, I describe a simple deterministic game. This has the advantage of avoiding technical constructions.
▶
It has an immediate stochastic interpretation as well.
▶
In continuous time studied by myself and Touzi in (2002).
▶
▶
Different derivation in one co-dimension was given by Buckdahn, Cardaliaguet, Quimcampoix (2002) using viability theory. Discrete game was introduced by Kohn & Serfaty in 2006. They studied the small step size limit together with Barles & Da Lio using viscosity solutions. 9
Description
▶
▶
We are given a region O on the plane and an initial point x0 ∈ O. Jane wants to leave the region as quickly as possible and can move one unit step in any direction : x 1 = x 0 + ν1 , where ν1 ∈ S1 is chosen by Jane ;
▶
Paul can only reverse Jane’s direction and wants to keep her in the region. So he can choose between the two points, x± 1 = x 0 ± ν1 . 10
Choices
▶ ▶
Jane wants to minimize the exit time. Paul can be considered a random walk and Jane would like to exit with probability one.
11
Choices
Jane may choose the direction heading directly towards the boundary.
11
Choices
But if she does that, Paul may reserve her and she will end up further to the boundary.
11
Choices
Optimal direction of Jane is to leave Paul irrelevant. This forces Jane to choose the tangential direction rather than the normal. Hence curvature is essential.
11
Quantile Hedging
12
Quantile Hedging
Again consider the Black-Scholes structure, i.e., dSt = St [µdt + σdWt ] and an option pay-off g(St ). Instead of hedging with probability one, given z ∈ [0, 1], one wants to hedge with at ˆt := e−rt St , least probability z. Using the notation, S0 = s and S { ( ) } ∫ t −rt ˆ v(t, s, z) := inf y : ∃ θ s.t. P y + θu dSu ≥ e g(St ) ≥ z . 0
12
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
13
Dynamic Programming
Here we take one probability measure and z = 1, i.e, { } V(t) := x : ∃ α ∈ A so that Xα,x ∈ T , P − a.s. . t Then, for any t > 0 and any stopping time τ ≤ t, V(t) = {x : ∃ α ∈ A so that Xα,x ∈ V(t − τ ), P − a.s. }. τ When V is an epigraph, v(t, s) = inf{y : (y, s) ∈ V(t)} satisfies, v(t, s) = inf{y : ∃ α ∈ A s.t. Yα,(y,s) ≥ v(t − τ, Sτ ), P − a.s. }. τ In the next slides, I demonstrate this in the simpler discrete example of Jane-Paul game. 14
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t)
⇔
u(x) = t.
u at the point below is 3.
15
Dynamic Programming
There is a simple geometric dynamic programming
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
The set {u(x) = 1} is between the black and the red curves.
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
The set {u(x) = 2} is between the red and the green curves.
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
The set {u(x) = 3} is between the purple and the green curves.
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
Our point moves from {u(x) = 3} into {u(x) = 2} into {u(x) = 1} and then out. This is dynamic programming. {u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
17
DPE
Once a dynamic programming principle is established, one uses it to obtain the dynamic programming equation. In the case of the original target problem, the solution is the reachability set V and the equation is a geometric equation. In the case of the Jean - Paul game, asymptotically this equation is the mean curvature flow. In the case of an epigraph, one obtains a parabolic equation.
18
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
19
Black-Scholes
Set interest rate r = 0.Recall that dSu = Su [µdu + σdWu ], { } ∫ t v(s, t) := inf x | ∃ θ admissible x + θu dSu ≥ g(St ) a.s. . 0
Dynamic programming states that for any stopping time τ , { } ∫ τ v(s, t) := inf x | ∃ θ s.t. x + θu dSu ≥ v(t − τ, Sτ ) a.s. . 0
20
Black-Scholes
Set interest rate r = 0.Recall that dSu = Su [µdu + σdWu ], { } ∫ t v(s, t) := inf x | ∃ θ admissible x + θu dSu ≥ g(St ) a.s. . 0
Dynamic programming states that for any stopping time τ , { } ∫ τ v(s, t) := inf x | ∃ θ s.t. x + θu dSu ≥ v(t − τ, Sτ ) a.s. . 0
Assume a minimiser θ∗ exists. Then, by Ito calculus, ∫ τ v(t, s) + θu∗ dSu ≥ v(t − τ, Sτ ) 0 ∫ τ = v(t, s) + [vs (t − u, Su )dSu − Lv(t − u, Su )du] , 0
Lv(t, s) = vt (t, s) −
s2 σ 2 vss (t, s). 2
20
Black-Scholes - II
∫ v(t, s)+
0
τ
θu∗ dSu
∫ ≥ v(t, s)+
τ
0
[vs (t − u, Su )dSu − Lv(t − u, Su )du] ,
Hence dS terms must be equal and we conclude that θu∗ = vs (t − u, Su ). Moreover du term, Lv, on the right hand side is zero. Hence, v is the unique solution of the famous Black-Scholes equation, rv + vt − rsvs −
s2 σ 2 vss = 0, 2
t, s > 0,
with initial data v(0, s) = g(s). 21
Black-Scholes - III
Since v solves rv + vt − rsvs −
s2 σ 2 vss = 0, 2
t, s > 0,
with initial data v(0, s) = g(s), Feyman - Kac formula implies that v(t, s) = e−rt EQ∗ [g(Sst )] , where Q∗ is the (unique) probability measure under which ˆt = e−rt St is martingale. S
22
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
23
DPE
Recall that u(x) is the minimal time to exit. Then, u(x) = 0 outside the target T and by dynamic programming, u(x) = inf max {u(x + ν), u(x − ν)} + 1, ν∈S1
x∈T.
This is the dynamic programming equation in this context. To obtain something more tractable, we do asymptotics by making the step size to be ϵ rather than one. But then the number of steps required to leave would be large and we define the scaled minimal time function uϵ (x) to be ϵ2 times the number of required steps. 24
Asymptotics
Using the equation for uϵ and Taylor expansion, inf max {uϵ (x + ϵν), uϵ (x − ϵν)} + ϵ2 { } ϵ2 2 ϵ ϵ ϵ ≈ u (x) + inf max ± ϵν · ∇u (x) + D u (x)ν · ν + ϵ2 . 2 ν∈S1
uϵ (x) =
ν∈S1
Then, optimal ν ∗ satisfies ν ∗ · ∇uϵ (x) = 0,
⇒
ν∗ =
(∇uϵ (x))⊥ . |∇uϵ (x)|
Moreover, formally the limit function u should satisfy (using the ϵ2 terms), [ ] 1 D2 u(x)∇u(x) · ∇u(x) ∆u − + 1 = 0, x ∈ T . 2 |∇u(x)|2 This is the mean curvature flow equation.
25
Continuous Time
One can define the game directly in continuous time as well. { } V(t) = x ∈ R2 : ∃{νu }u∈[0,t] ∈ S1 s.t. Xx,ν t ∈T , where Xx,ν t
∫ := x +
t 0
νu (νu · dWu )
and W is a standard two dimensional Brownian motion. Multi dimensional versions with higher co-dimension is also available. 26
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
27
Problem
Recall dSt = St [µdt + σdWt ] and set r = 0. Given desired success probability z ∈ [0, 1] and S0 = s, { ( ) } ∫ t v(t, s, z) := inf y : ∃ θ s.t. P y + θu dSu ≥ g(St ) ≥ z . 0
With fixed z, the above does not satisfy the dynamic programming unless z = 1. However, Bouchard, Elie, Touzi transformed the above problem to a problem with z = 1 but with one more state variable accounting for the success probability. 28
Transformation
∫t
s 0 θu dSu
θ,(y,s)
A process θ is admissible if Yt := y + ( ) θ,(y,s) Zu := P Yt ≥ g(St )| Fu ,
≥ 0. Set
u ∈ [0, t].
Then, since F0 is trivial, Z0 = z and Zt = χ{Yθ,(y,s) ≥g(Ss )} . t
t
Hence, Z is a martingale. By the martingale representation Theorem, there exist a process α so that ∫ t Zt = z + αu dWu =: Zα,z t . 0
Then, the original quantile hedging problem is equivalent to { } ≥ χ , P − a.s. v(t, s, z) = inf y : ∃ (θ, α) s.t. Zα,z . θ,(y,s) s t ≥g(S )} {Y t
t
29
Binomial Example
Consider quantile hedging in a Binomial model with equal up and down probabilities of 1/2. This means Sn+1 = Sn [1 + ξn+1 ] ,
αα00==1/2 0 S0 Z0 = 0.5
ξn+1 = ±1 with equal probability.
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
When z ∈ (0.75, 1] we cannot miss any nodes, z ∈ (0.5, 0.75] we are allowed to miss one of the nodes, z ∈ (0.25, 0.5] we are allowed to miss two nodes, ...
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Sn+1 = Sn [1 + ξn+1 ] and the probability process is given by Zn+1 = Zn + αn ξn+1 .
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Consider the case z = 0.5. So we can choose 2 nodes to miss.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Suppose we decide to miss the middle 2. Then, the probabilities Z evolve as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
If we decide to miss the lower two, then the probabilities Z evolve as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Sn+1 = Sn [1 + ξn+1 ],
Zn+1 = Zn + αn ξn+1 ,
In the first one, α’s are given as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Sn+1 = Sn [1 + ξn+1 ],
Zn+1 = Zn + αn ξn+1 ,
In the second one, α’s are given as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
31
Uncertain Volatility
Consider again the Black-Scholes model, with r = 0 and dSσt = Sσt [µdt + σt dWt ]. Here were assume that σt is not constant and can be any adapted process in a given interval [σ, σ]. Let A be the set of all such processes. For a given Ft measurable, bounded ξ, set { } ∫ t v(ξ) := inf y : ∃ θ s.t. y + θu dSσu ≥ ξ a.s., ∀σ· ∈ A . 0
One could show (rather technical) that { } ∫ t v(ξ) := sup inf y : ∃ θ s.t. y + θu dSσu ≥ ξ a.s. . σ· ∈A
0
32
Black-Scholes - recalled
Recall the Black-Scholes with r = 0, { } ∫ t v(s, t) := inf x | ∃ θ admissible x + θu dSu ≥ ξ a.s. , 0
where ξ is a given bounded Ft measurable random variable. Then, v(t, s) = EQ∗ [ξ] , where Q∗ is the (unique) probability measure under which St is martingale. 33
Uncertain Volatility II
dSσt = Sσt [µdt + σt dWt ], σt ∈ [σ, σ], A be the set of all such processes and ξ is Ft measurable, bounded. Using the result from Black-Scholes, { } ∫ t σ v(ξ) := inf y : ∃ θ s.t. y + θu dSu ≥ ξ a.s., ∀σ· ∈ A 0 { } ∫ t σ = sup inf y : ∃ θ s.t. y + θu dSu ≥ ξ a.s., σ· ∈A
=
0
sup EQ [ ξ ],
Q∈M
¯σ for where Q ∈ M if and only if Q is the “distribution” of S ¯σu = S ¯σu σu dWu . some σ ∈ A and dS 34
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
35
Problem
In the previous problem, if we take the set of probability measures Q to be the whole set this corresponds to complete model independence. However, then for many functions ξ, we might have sup E[ ξ ] = ∥ξ∥∞ , Q∈Q
and the problem trivialises. But a version of this problem is very close to the celebrated Monge-Kantarovich optimal transport problem.
36
Kantarovich dual
The Monge-Kantarovich problem is in discrete time with two time steps and its dual, derived by Kantarovich is given by, {∫ ∫ v(ξ) := sup hdµ + gdν : Rd Rd } h(x) + g(y) ≥ ξ(x, y), ∀(x, y) ∈ R2d , where µ, ν are given probability measures on Rd and ξ is bounded function of R2d .
37
Duality
{∫ v(ξ)
:= sup
∫
Rd
hdµ +
Rd
gdν :
h(x) + g(y) ≥ ξ(x, y), ∀(x, y) ∈ R2d
} ,
The primal is the celebrated optimal transport problem, v(ξ) =
sup
Q∈Q(µ,ν)
EQ [ξ],
where Q ∈ Q(µ, ν) if it is a probability measure whose marginals are µ and ν, respectively. 38
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
39
FORWARD START
We consider a direct generalisation of the Monge-Kantarovich problem motivated by a financial problem, the so-called forward-start options. The model independent version of this problem is {∫ ∫ gdν : ∃γ : Rd → Rd v(ξ) := sup hdµ + Rd
Rd
h(x) + g(y) + γ(x) · (y − x) ≥ ξ(x, y), ∀(x, y) ∈ R2d
} ,
40
Dual elements
The dual is v(ξ)
{∫
:= sup
Rd
∫ hdµ +
Rd
gdν : ∃γ : Rd → Rd
h(x) + g(y) + γ(x) · (y − x) ≥ ξ(x, y), ∀(x, y) ∈ R2d =
sup
Q∈M(µ,ν)
} ,
EQ [ξ],
where Q ∈ M(µ, ν) if it is in Q(µ, ν) (i.e., satisfies the marginals) and also it is a martingale measure, i.e., EQ [γ(x) · (y − x)] = 0, for any bounded measurable γ. 41
Martingale Measure
If we see y as the value of a process at time 2, S2 , and x as S1 , then above simply states that EQ [γ(x) · (y − x)] = 0,
∀γ,
⇔
EQ [S2 |S1 ] = S1 .
Hence, S is a martingale under the measure Q. This new constraint is due to the term γ(x) · (y − x) in the definition of the martingale optimal transport problem.
42
Comparing
In uncertain volatility, v(ξ) = sup EQ [ξ], Q∈M
where M is a subset of martingale measures ; In optimal transport, v(ξ) =
sup
Q∈Q(µ,ν)
EQ [ξ],
where Q(µ, ν) satisfies marginal constraints ; In martingale optimal transport, v(ξ) =
sup
Q∈M(µ,ν)
EQ [ξ],
where M(µ, ν) satisfies marginal and martingale constraints .
43
Skorokhod Space
Let Ω be the Skorokhod space of all Rd+ -valued càdlàg processes on [0, t], i.e., process that are continuous from right hand have left limits with the standard Skorokhod topology. Further let S be the canonical map, i.e. Su (ω) = ωu for all u ∈ [0, t] and Fu be the filtration generated by S. Given a probability measure ν on Rd+ , set {∫ v(ξ)
:= sup
Rd
gdν : ∃(γu )u∈[0,t]
g(St (ω)) +
∫ 0
t
} γu (ω) · dSu (ω) ≥ ξ(ω), ∀ω ∈ Ω
. 44
Duality
Theorem (Dolinsky, S. : cont. (2012) PTRF - càdlàg (2015) SPA) Assume that ξ is bounded and uniformly continuous with respect to the Skorokhod metric. Then, v(ξ) =
sup
Q∈M(µ)
EQ [ξ].
The set M(µ) is the set of all measures Q under which the canonical map S is a martingale and the marginal of Q at time t is ν. Note that is not compact. Many marginal version is also proved. 45
Concluding
▶
▶
▶
The stochastic target type problems can be solved by geometric dynamic principle. The quantile type problems can be transferred into a standard target problems. One could also consider cases in which many possibly orthogonal measures are given representing the model uncertainty.
THANK YOU FOR YOUR ATTENTION. 46
0
Collaborators
I have benefited the collaboration of many people including : Erdinç Akyıldırım, Albert Altarovici, Yan Dolinsky, Romuald Elie, Selim Gökay, Ludovic Moreau, Johannes Muhle-Karbe, Marcel Nutz, Dylan Possamaï, Max Reppen, Jianfeng Zhang and
Nizar Touzi
Bruno Bouchard
1
Foreword
In the classical Markovian optimal control theory, one tries to maximize (or minimize) an objective functional of the form [∫ E
0
T
] f(Xαt , αt )dt
+
Φ(XαT )
over all admissible controls α. Here Xα is the controlled state process.
2
Foreword
For the stochastic target problems, one would like to satisfy a constraint of the form XαT ∈ T ,
P − a.s.,
where T ⊂ Rd is a given deterministic target. As such the first is an L1 theory while the stochastic target is an L∞ approach. The stochastic target problem is closely related to stochastic viability theory as developed by Aubin, Cardaliaguet, DaPrato, Frankowska,.... 3
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
4
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
5
Problem
Given are ▶
▶ ▶ ▶
d Controlled state process (Xα,x t )t≥0 ∈ R with initial data Xα,x 0 = x and an admissible control process α ∈ A defined on a filtered probability space (Ω, {Ft }) ; A set of probability measures (or models) P ; Desired success probability z ∈ [0, 1] ; Deterministic target set T ⊂ Rd .
Goal is to characterise the reachability set { ( ) V(t) := x : ∃ α ∈ A so that P Xα,x ∈ T ≥ z, t
∀P∈P
}
.
6
Problem
Given are ▶
▶ ▶ ▶
d Controlled state process (Xα,x t )t≥0 ∈ R with initial data Xα,x 0 = x and an admissible control process α ∈ A defined on a filtered probability space (Ω, {Ft }) ; A set of probability measures (or models) P ; Desired success probability z ∈ [0, 1] ; Deterministic target set T ⊂ Rd .
Goal is to characterise the reachability set { ( ) V(t) := x : ∃ α ∈ A so that P Xα,x ∈ T ≥ z, t
∀P∈P
}
.
I concentrate on the case z = 1 and P = {P} at the beginning.
6
Black-Scholes
7
Black-Scholes
Consider a financial market consisting of a liquidly traded asset with a random price process Su and a constant interest rate of r. Assume that S is a geometric Brownian motion, i.e., dSt = St [µdt + σdWt ],
t > 0,
where µ > r, volatility σ > 0 and W is a Brownian motion. Further given is a general European option that will pay g(St ) at time t with a known deterministic function g. ˆt := e−rt St , Given S0 = s, consider the function with S { } ∫ t −rt ˆ v(t, s) := inf y : ∃ θ such that y + θu dSu ≥ e g(St ) a.s. 0
7
Black-Scholes, II
In the notation of the general problem, control is the portfolio process θ and the state process is ( [ ] ) ∫ t θ,(y,s) θ,(y,s) s rt s s ˆ u , St , Xt = (Yt , St ) = e y + θu dS 0
and the target set is the epigraph of g, { } T = (y, s) ∈ R2+ : y ≥ g(s) . Then, the reachability set V(t) is also an epigraph and v(t, s) = inf { y : (y, s) ∈ V(t) } . The function v(t, s) is the smallest all prices y that allows the seller to hedge the claim g(St ) with probability one. 8
Jane and Paul Game
9
Jane and Paul Game
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Instead of explaining the continuous time stochastic problem, I describe a simple deterministic game. This has the advantage of avoiding technical constructions.
▶
It has an immediate stochastic interpretation as well.
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In continuous time studied by myself and Touzi in (2002).
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▶
Different derivation in one co-dimension was given by Buckdahn, Cardaliaguet, Quimcampoix (2002) using viability theory. Discrete game was introduced by Kohn & Serfaty in 2006. They studied the small step size limit together with Barles & Da Lio using viscosity solutions. 9
Description
▶
▶
We are given a region O on the plane and an initial point x0 ∈ O. Jane wants to leave the region as quickly as possible and can move one unit step in any direction : x 1 = x 0 + ν1 , where ν1 ∈ S1 is chosen by Jane ;
▶
Paul can only reverse Jane’s direction and wants to keep her in the region. So he can choose between the two points, x± 1 = x 0 ± ν1 . 10
Choices
▶ ▶
Jane wants to minimize the exit time. Paul can be considered a random walk and Jane would like to exit with probability one.
11
Choices
Jane may choose the direction heading directly towards the boundary.
11
Choices
But if she does that, Paul may reserve her and she will end up further to the boundary.
11
Choices
Optimal direction of Jane is to leave Paul irrelevant. This forces Jane to choose the tangential direction rather than the normal. Hence curvature is essential.
11
Quantile Hedging
12
Quantile Hedging
Again consider the Black-Scholes structure, i.e., dSt = St [µdt + σdWt ] and an option pay-off g(St ). Instead of hedging with probability one, given z ∈ [0, 1], one wants to hedge with at ˆt := e−rt St , least probability z. Using the notation, S0 = s and S { ( ) } ∫ t −rt ˆ v(t, s, z) := inf y : ∃ θ s.t. P y + θu dSu ≥ e g(St ) ≥ z . 0
12
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
13
Dynamic Programming
Here we take one probability measure and z = 1, i.e, { } V(t) := x : ∃ α ∈ A so that Xα,x ∈ T , P − a.s. . t Then, for any t > 0 and any stopping time τ ≤ t, V(t) = {x : ∃ α ∈ A so that Xα,x ∈ V(t − τ ), P − a.s. }. τ When V is an epigraph, v(t, s) = inf{y : (y, s) ∈ V(t)} satisfies, v(t, s) = inf{y : ∃ α ∈ A s.t. Yα,(y,s) ≥ v(t − τ, Sτ ), P − a.s. }. τ In the next slides, I demonstrate this in the simpler discrete example of Jane-Paul game. 14
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t) u at the point below is 3.
⇔
u(x) = t.
PJ : Minimal time function
Let u(x) be the number steps to leave the region. Then, x ∈ V(t)
⇔
u(x) = t.
u at the point below is 3.
15
Dynamic Programming
There is a simple geometric dynamic programming
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
The set {u(x) = 1} is between the black and the red curves.
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
The set {u(x) = 2} is between the red and the green curves.
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
The set {u(x) = 3} is between the purple and the green curves.
{u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Dynamic Programming
Our point moves from {u(x) = 3} into {u(x) = 2} into {u(x) = 1} and then out. This is dynamic programming. {u = 1} {u = 2}
{u ≥ 3}
{u = 3}
16
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
17
DPE
Once a dynamic programming principle is established, one uses it to obtain the dynamic programming equation. In the case of the original target problem, the solution is the reachability set V and the equation is a geometric equation. In the case of the Jean - Paul game, asymptotically this equation is the mean curvature flow. In the case of an epigraph, one obtains a parabolic equation.
18
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
19
Black-Scholes
Set interest rate r = 0.Recall that dSu = Su [µdu + σdWu ], { } ∫ t v(s, t) := inf x | ∃ θ admissible x + θu dSu ≥ g(St ) a.s. . 0
Dynamic programming states that for any stopping time τ , { } ∫ τ v(s, t) := inf x | ∃ θ s.t. x + θu dSu ≥ v(t − τ, Sτ ) a.s. . 0
20
Black-Scholes
Set interest rate r = 0.Recall that dSu = Su [µdu + σdWu ], { } ∫ t v(s, t) := inf x | ∃ θ admissible x + θu dSu ≥ g(St ) a.s. . 0
Dynamic programming states that for any stopping time τ , { } ∫ τ v(s, t) := inf x | ∃ θ s.t. x + θu dSu ≥ v(t − τ, Sτ ) a.s. . 0
Assume a minimiser θ∗ exists. Then, by Ito calculus, ∫ τ v(t, s) + θu∗ dSu ≥ v(t − τ, Sτ ) 0 ∫ τ = v(t, s) + [vs (t − u, Su )dSu − Lv(t − u, Su )du] , 0
Lv(t, s) = vt (t, s) −
s2 σ 2 vss (t, s). 2
20
Black-Scholes - II
∫ v(t, s)+
0
τ
θu∗ dSu
∫ ≥ v(t, s)+
τ
0
[vs (t − u, Su )dSu − Lv(t − u, Su )du] ,
Hence dS terms must be equal and we conclude that θu∗ = vs (t − u, Su ). Moreover du term, Lv, on the right hand side is zero. Hence, v is the unique solution of the famous Black-Scholes equation, rv + vt − rsvs −
s2 σ 2 vss = 0, 2
t, s > 0,
with initial data v(0, s) = g(s). 21
Black-Scholes - III
Since v solves rv + vt − rsvs −
s2 σ 2 vss = 0, 2
t, s > 0,
with initial data v(0, s) = g(s), Feyman - Kac formula implies that v(t, s) = e−rt EQ∗ [g(Sst )] , where Q∗ is the (unique) probability measure under which ˆt = e−rt St is martingale. S
22
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
23
DPE
Recall that u(x) is the minimal time to exit. Then, u(x) = 0 outside the target T and by dynamic programming, u(x) = inf max {u(x + ν), u(x − ν)} + 1, ν∈S1
x∈T.
This is the dynamic programming equation in this context. To obtain something more tractable, we do asymptotics by making the step size to be ϵ rather than one. But then the number of steps required to leave would be large and we define the scaled minimal time function uϵ (x) to be ϵ2 times the number of required steps. 24
Asymptotics
Using the equation for uϵ and Taylor expansion, inf max {uϵ (x + ϵν), uϵ (x − ϵν)} + ϵ2 { } ϵ2 2 ϵ ϵ ϵ ≈ u (x) + inf max ± ϵν · ∇u (x) + D u (x)ν · ν + ϵ2 . 2 ν∈S1
uϵ (x) =
ν∈S1
Then, optimal ν ∗ satisfies ν ∗ · ∇uϵ (x) = 0,
⇒
ν∗ =
(∇uϵ (x))⊥ . |∇uϵ (x)|
Moreover, formally the limit function u should satisfy (using the ϵ2 terms), [ ] 1 D2 u(x)∇u(x) · ∇u(x) ∆u − + 1 = 0, x ∈ T . 2 |∇u(x)|2 This is the mean curvature flow equation.
25
Continuous Time
One can define the game directly in continuous time as well. { } V(t) = x ∈ R2 : ∃{νu }u∈[0,t] ∈ S1 s.t. Xx,ν t ∈T , where Xx,ν t
∫ := x +
t 0
νu (νu · dWu )
and W is a standard two dimensional Brownian motion. Multi dimensional versions with higher co-dimension is also available. 26
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
27
Problem
Recall dSt = St [µdt + σdWt ] and set r = 0. Given desired success probability z ∈ [0, 1] and S0 = s, { ( ) } ∫ t v(t, s, z) := inf y : ∃ θ s.t. P y + θu dSu ≥ g(St ) ≥ z . 0
With fixed z, the above does not satisfy the dynamic programming unless z = 1. However, Bouchard, Elie, Touzi transformed the above problem to a problem with z = 1 but with one more state variable accounting for the success probability. 28
Transformation
∫t
s 0 θu dSu
θ,(y,s)
A process θ is admissible if Yt := y + ( ) θ,(y,s) Zu := P Yt ≥ g(St )| Fu ,
≥ 0. Set
u ∈ [0, t].
Then, since F0 is trivial, Z0 = z and Zt = χ{Yθ,(y,s) ≥g(Ss )} . t
t
Hence, Z is a martingale. By the martingale representation Theorem, there exist a process α so that ∫ t Zt = z + αu dWu =: Zα,z t . 0
Then, the original quantile hedging problem is equivalent to { } ≥ χ , P − a.s. v(t, s, z) = inf y : ∃ (θ, α) s.t. Zα,z . θ,(y,s) s t ≥g(S )} {Y t
t
29
Binomial Example
Consider quantile hedging in a Binomial model with equal up and down probabilities of 1/2. This means Sn+1 = Sn [1 + ξn+1 ] ,
αα00==1/2 0 S0 Z0 = 0.5
ξn+1 = ±1 with equal probability.
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
When z ∈ (0.75, 1] we cannot miss any nodes, z ∈ (0.5, 0.75] we are allowed to miss one of the nodes, z ∈ (0.25, 0.5] we are allowed to miss two nodes, ...
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Sn+1 = Sn [1 + ξn+1 ] and the probability process is given by Zn+1 = Zn + αn ξn+1 .
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Consider the case z = 0.5. So we can choose 2 nodes to miss.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Suppose we decide to miss the middle 2. Then, the probabilities Z evolve as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
If we decide to miss the lower two, then the probabilities Z evolve as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Sn+1 = Sn [1 + ξn+1 ],
Zn+1 = Zn + αn ξn+1 ,
In the first one, α’s are given as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Binomial Example
Sn+1 = Sn [1 + ξn+1 ],
Zn+1 = Zn + αn ξn+1 ,
In the second one, α’s are given as below.
αα00==1/2 0 S0 Z0 = 0.5
αα1 1==1/2 0 S1 ZZ1 1==0.5 1
S2 Z 2 = 1
αα 1 1==−1/2 0 S1 ZZ1 1==0.5 0
S2 Z 2 = 0
S2 Z2 = 10
1 S2 Z 2 = 0 30
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
31
Uncertain Volatility
Consider again the Black-Scholes model, with r = 0 and dSσt = Sσt [µdt + σt dWt ]. Here were assume that σt is not constant and can be any adapted process in a given interval [σ, σ]. Let A be the set of all such processes. For a given Ft measurable, bounded ξ, set { } ∫ t v(ξ) := inf y : ∃ θ s.t. y + θu dSσu ≥ ξ a.s., ∀σ· ∈ A . 0
One could show (rather technical) that { } ∫ t v(ξ) := sup inf y : ∃ θ s.t. y + θu dSσu ≥ ξ a.s. . σ· ∈A
0
32
Black-Scholes - recalled
Recall the Black-Scholes with r = 0, { } ∫ t v(s, t) := inf x | ∃ θ admissible x + θu dSu ≥ ξ a.s. , 0
where ξ is a given bounded Ft measurable random variable. Then, v(t, s) = EQ∗ [ξ] , where Q∗ is the (unique) probability measure under which St is martingale. 33
Uncertain Volatility II
dSσt = Sσt [µdt + σt dWt ], σt ∈ [σ, σ], A be the set of all such processes and ξ is Ft measurable, bounded. Using the result from Black-Scholes, { } ∫ t σ v(ξ) := inf y : ∃ θ s.t. y + θu dSu ≥ ξ a.s., ∀σ· ∈ A 0 { } ∫ t σ = sup inf y : ∃ θ s.t. y + θu dSu ≥ ξ a.s., σ· ∈A
=
0
sup EQ [ ξ ],
Q∈M
¯σ for where Q ∈ M if and only if Q is the “distribution” of S ¯σu = S ¯σu σu dWu . some σ ∈ A and dS 34
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
35
Problem
In the previous problem, if we take the set of probability measures Q to be the whole set this corresponds to complete model independence. However, then for many functions ξ, we might have sup E[ ξ ] = ∥ξ∥∞ , Q∈Q
and the problem trivialises. But a version of this problem is very close to the celebrated Monge-Kantarovich optimal transport problem.
36
Kantarovich dual
The Monge-Kantarovich problem is in discrete time with two time steps and its dual, derived by Kantarovich is given by, {∫ ∫ v(ξ) := sup hdµ + gdν : Rd Rd } h(x) + g(y) ≥ ξ(x, y), ∀(x, y) ∈ R2d , where µ, ν are given probability measures on Rd and ξ is bounded function of R2d .
37
Duality
{∫ v(ξ)
:= sup
∫
Rd
hdµ +
Rd
gdν :
h(x) + g(y) ≥ ξ(x, y), ∀(x, y) ∈ R2d
} ,
The primal is the celebrated optimal transport problem, v(ξ) =
sup
Q∈Q(µ,ν)
EQ [ξ],
where Q ∈ Q(µ, ν) if it is a probability measure whose marginals are µ and ν, respectively. 38
Outline
Introduction Examples Dynamic Programming Solutions Black-Scholes Jean - Paul Game Quantile Hedging Martingale Optimal Transport Model Ambiguity Optimal Transport Martingale Optimal Transport
39
FORWARD START
We consider a direct generalisation of the Monge-Kantarovich problem motivated by a financial problem, the so-called forward-start options. The model independent version of this problem is {∫ ∫ gdν : ∃γ : Rd → Rd v(ξ) := sup hdµ + Rd
Rd
h(x) + g(y) + γ(x) · (y − x) ≥ ξ(x, y), ∀(x, y) ∈ R2d
} ,
40
Dual elements
The dual is v(ξ)
{∫
:= sup
Rd
∫ hdµ +
Rd
gdν : ∃γ : Rd → Rd
h(x) + g(y) + γ(x) · (y − x) ≥ ξ(x, y), ∀(x, y) ∈ R2d =
sup
Q∈M(µ,ν)
} ,
EQ [ξ],
where Q ∈ M(µ, ν) if it is in Q(µ, ν) (i.e., satisfies the marginals) and also it is a martingale measure, i.e., EQ [γ(x) · (y − x)] = 0, for any bounded measurable γ. 41
Martingale Measure
If we see y as the value of a process at time 2, S2 , and x as S1 , then above simply states that EQ [γ(x) · (y − x)] = 0,
∀γ,
⇔
EQ [S2 |S1 ] = S1 .
Hence, S is a martingale under the measure Q. This new constraint is due to the term γ(x) · (y − x) in the definition of the martingale optimal transport problem.
42
Comparing
In uncertain volatility, v(ξ) = sup EQ [ξ], Q∈M
where M is a subset of martingale measures ; In optimal transport, v(ξ) =
sup
Q∈Q(µ,ν)
EQ [ξ],
where Q(µ, ν) satisfies marginal constraints ; In martingale optimal transport, v(ξ) =
sup
Q∈M(µ,ν)
EQ [ξ],
where M(µ, ν) satisfies marginal and martingale constraints .
43
Skorokhod Space
Let Ω be the Skorokhod space of all Rd+ -valued càdlàg processes on [0, t], i.e., process that are continuous from right hand have left limits with the standard Skorokhod topology. Further let S be the canonical map, i.e. Su (ω) = ωu for all u ∈ [0, t] and Fu be the filtration generated by S. Given a probability measure ν on Rd+ , set {∫ v(ξ)
:= sup
Rd
gdν : ∃(γu )u∈[0,t]
g(St (ω)) +
∫ 0
t
} γu (ω) · dSu (ω) ≥ ξ(ω), ∀ω ∈ Ω
. 44
Duality
Theorem (Dolinsky, S. : cont. (2012) PTRF - càdlàg (2015) SPA) Assume that ξ is bounded and uniformly continuous with respect to the Skorokhod metric. Then, v(ξ) =
sup
Q∈M(µ)
EQ [ξ].
The set M(µ) is the set of all measures Q under which the canonical map S is a martingale and the marginal of Q at time t is ν. Note that is not compact. Many marginal version is also proved. 45
Concluding
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▶
▶
The stochastic target type problems can be solved by geometric dynamic principle. The quantile type problems can be transferred into a standard target problems. One could also consider cases in which many possibly orthogonal measures are given representing the model uncertainty.
THANK YOU FOR YOUR ATTENTION. 46
