Partial hedging: numerical methods. Cyril Benezet* Joint work with - PowerPoint PPT Presentation

Partial hedging: numerical methods. Cyril Benezet* Joint work with Jean-Fran¸ cois Chassagneux and Christoph Reisinger Universit´ e Paris Diderot LPSM *Research supported by a Joint Research Initiative, AXA Research Fund May 02, 2018 S´ eminaire des Doctorants, CERMICS 1 / 20

Motivation, framework, example 1 Motivations Framework The Black&Scholes case PDE characterisation, comparison theorem 2 PDE characterisation Numerical method 3 Control space truncation Control space discretisation Piecewise constant policy timestepping scheme The numerical scheme - Black & Scholes setting 4 Motivation, framework, example Motivations 2 / 20

What is partial hedging ? Partial hedging aims to determine: prices to sell products with respect to a risk constraint (e.g. Value at Risk, or the probability of a successful hedge: quantile hedging), associated strategies to satisfy this constraint. Why is it useful? Super-replication price can be high for insurance (complex, long-term and with high notional) products: quantile hedging allows a price reduction. Insurance companies need to control their balance sheet with Value at Risk constraints. Motivation, framework, example Motivations 2 / 20

The Markovian model Consider a risky asset with price given, for an initial condition ( t , x ) ∈ [0 , T ] × (0 , ∞ ) d , by: � s � s X t , x diag( X t , x u ) µ ( X t , x diag( X t , x u ) σ ( X t , x = x + u ) d u + u ) d W u , s t t � s � s µ X ( X t , x σ X ( X t , x = x + u ) d u + u ) d W u , s ∈ [ t , T ] . t t where W is a Brownian motion. Given an initial wealth y ≥ 0 and a process ν modeling the amount of wealth invested in the asset, the wealth process is: � s � s Y t , x , y ,ν f ( u , X t , x u , Y t , x , y ,ν ν u σ ( X t , x = y + , ν u ) d u + u ) d W u , s ∈ [ t , T ] . s u t t Assume that the coefficients are Lipschitz continuous: we then have existence and uniqueness for every initial condition and control such that the wealth stays non-negative (such a control is called admissible). Motivation, framework, example Framework 3 / 20

The stochastic control problem The stochastic control problem we are interested in takes the form � � � � ℓ ( Y t , x , y ,ν − g ( X t , x v ( t , x , p ) = inf y ≥ 0 : ∃ ν, E T )) ≥ p , T where ℓ satisfies hypothesis so that v is finite and with polynomial growth, and conv ℓ ( R ) is compact. For example, if ℓ ( x ) = ✶ R + ( x ), we find the quantile hedging problem: � � � � Y t , x , y ,ν ≥ g ( X t , x v ( t , x , p ) = inf y ≥ 0 : ∃ ν : P T ) ≥ p . T In the sequel we will only consider quantile hedging, so conv ℓ ( R ) = [0 , 1]. A few basic properties about v : v ( t , x , · ) is increasing, � � g ( X t , x v ( t , x , p ) = 0 if p ≤ p min ( t , x ) = P T ) = 0 . v ( t , x , 1) = v ( t , x ) is the super-replication price of the derivative. Motivation, framework, example Framework 4 / 20

An example The first work about quantile hedging was done by F¨ ollmer and Leukert [5], in the Black&Scholes model. Suppose the underlying is a 1-dimensional geometric Brownian motion: � s � s X t , x = x + µ X u d u + σ X u d W u , s ∈ [ t , T ] . s t t Suppose a hedging strategy is only possible by buying and selling the underlying in a linear market (with zero interest rate for simplicity). Given such a strategy ν and an initial wealth y ≥ 0, the associated wealth process Y y ,ν is given by (recall: ν is the wealth invested in the asset): � s � s Y t , y ,ν = y + µν u d u + σν u d W u , s ∈ [ t , T ] . s t t They provide, thanks to the Neyman-Pearson lemma from statistics, closed-form expressions for the quantile hedging problem for vanilla options. Motivation, framework, example The Black&Scholes case 5 / 20

An illustration 3.5 Partial hedging of vanilla put in Black-Scholes model 3 2.5 2 V(p) 1.5 1 0.5 0 0 0.2 0.4 0.6 0.8 1 p The parameters used are: µ = 0 . 05 , σ = 0 . 25, and we are plotting the graph of p �→ v (0 , 30 , p ) for a put of maturity T = 1 and strike price K = 30. Motivation, framework, example The Black&Scholes case 6 / 20

General case : reduction to a stochastic target problem If α is a control, let P t , p ,α be the process defined by: � s P t , p ,α = p + α u d W u , s ∈ [ t , T ] . s t The control α is admissible if P t , p ,α ∈ [0 , 1] a.s.. T Then, by the martingale reprensentation theorem, Bouchard, Elie and Touzi [3] prove the following: Lemma (The associated stochastic target problem) For every ( t , x , p ) ∈ [0 , T ] × (0 , ∞ ) × [0 , 1] , we have: � � T ) } ≥ P t , p ,α v ( t , x , p ) = inf y ≥ 0 : ∃ ( ν, α ) , ✶ { Y t , x , y ,ν a.s. . − g ( X t , x T T This lemma allows them to obtain a PDE representation for v , but with a discontinuous operator. PDE characterisation, comparison theorem PDE characterisation 7 / 20

The PDE Following an idea from Bokanowski et al. [2], Bouveret and Chassagneux [4] obtain a new PDE representation for v , together with a comparison theorem which implies uniqueness: Theorem v is the unique positive viscosity solution, on [0 , T ) × (0 , ∞ ) d × (0 , 1) , of: � 1 − ∂ t ϕ − µ X ( x ) ⊤ D x ϕ + f ( t , x , ϕ, ∇ a ϕ ) sup (1) 1 + | a | 2 a ∈ R d � � � − | a | 2 − 1 σ X ( x ) σ X ( x ) ⊤ D 2 2 ∂ 2 pp ϕ − a ⊤ σ X ( x ) ⊤ D 2 2 Tr xx ϕ xp ϕ = 0 , where ∇ a ϕ = D x ϕ ⊤ diag( x ) + ∂ p ϕ a ⊤ σ − 1 ( x ) , satisfying to the following: v ( t , x , 0) = 0 on [0 , T ] × (0 , ∞ ) d , v ( t , x , 1) = v ( t , x ) on [0 , T ] × (0 , ∞ ) d , v ( T , x , p ) = g ( x )1 p � =0 on (0 , ∞ ) d × [0 , 1] , v ( T − , x , p ) = pg ( x ) on (0 , ∞ ) d × [0 , 1] . PDE characterisation, comparison theorem PDE characterisation 8 / 20

Numerical approximation of v : remarks and strategy Our goal is to provide a numerical method to numerically approximate the solution of this PDE. First, in view of the discontinuity at time t = T , it is convenient for us to take the v ( T , x , p ) = pg ( x ) on (0 , ∞ ) d × (0 , 1) as our terminal condition. There are several numerical difficulties we have to deal with. Unboundedness of the control space, 1 The “nonlinearity” 1+ | a | 2 in front of the ∂ t term, which makes it impossible to use usual schemes. The semilinear term f in the PDE. In a first step, we truncate the control space in order to solve the two first issues. Then, it allows us to discretise the control space. Last, we introduce a piecewise constant policy iteration scheme to solve the PDE numerically. Numerical method 9 / 20

First step: control space truncation For each n ≥ 1, let K n := [ − n , n ] d ⊂ R d , and we define v n as the unique viscosity solution of: � � − ∂ t ϕ − µ X ( x ) ⊤ D x ϕ − 1 σ X ( x ) σ X ( x ) ⊤ D 2 2 Tr xx ϕ � � f ( t , x , ϕ, ∇ a ϕ ) − | a | 2 2 ∂ 2 pp ϕ − a ⊤ σ X ( x ) ⊤ D 2 + sup xp ϕ = 0 , a ∈ K n satisfying to the same boundary conditions as v . It is straightforward to see that v n is the solution of (1) where the supremum is to be taken over K n . Then, we prove the following: Theorem The sequence ( v n ) n ≥ 1 converges to v uniformly on compact sets, as n goes to infinity. A key ingredient in the proof Dini’s theorem, as the sequence of operators ( H n ) such that H n ( v n ) = 0 is increasing and simply converging. Numerical method Control space truncation 10 / 20

Second step: control space discretisation For each m ≥ 1, let K n , m be a finite subset of K n satisfying: b ∈ K n , m | a − b | ≤ m − 1 . a ∈ K n min max Let v n , m be the unique viscosity solution of the following PDE: � � − ∂ t ϕ − µ X ( x ) ⊤ D x ϕ − 1 σ X ( x ) σ X ( x ) ⊤ D 2 2 Tr xx ϕ � � f ( t , x , ϕ, ∇ a ϕ ) − | a | 2 2 ∂ 2 pp ϕ − a ⊤ σ X ( x ) ⊤ D 2 + sup = 0 , xp ϕ a ∈ K n , m satisfying the same boundary conditions as v . Then we have: Theorem As m → ∞ , v m , n → v n , uniformly on compact sets. Numerical method Control space discretisation 11 / 20

Piecewise constant policy timestepping scheme We now introduce the piecewise constant policy timestepping scheme. We fix a grid π = { t 0 = 0 < · · · < t j < · · · < t κ = T } ( κ ≥ 1) for the time-discretisation. The backward algorithm for the approximation � v of v ( T , x , p ) = g ( x ) p on (0 , ∞ ) d × [0 , 1], and v n , m is given by � v ( t , x , p ) = min a ∈ K v a ( t , x , p ) on [ t j , t j +1 ) × (0 , ∞ ) × [0 , 1] for j < κ . � Here, for each control a ∈ K and j < κ , v a is the solution on [ t j , t j +1 ) to: � � − ∂ t ϕ − µ X ( x ) ⊤ D x ϕ − 1 σ X ( x ) σ X ( x ) ⊤ D 2 + f ( t , x , ϕ, ∇ a ϕ ) 2 Tr xx ϕ − | a | 2 2 ∂ 2 pp ϕ − a ⊤ σ X ( x ) ⊤ D 2 xp ϕ = 0 , with the boundary conditions: v ( t j +1 , x , p ) on (0 , ∞ ) d × [0 , 1] , ϕ ( t j +1 , x , p ) = � ϕ ( t , x , p ) = v ( t , x )1 p =1 on [ t j , t j +1 ) × (0 , ∞ ) d × { 0 , 1 } . Piecewise constant policy timestepping Numerical method scheme 12 / 20