A3 : Skorokhod Problem

Proof.If f ∈Φ, (X_t^f)_t∈[0,∞) is a positive supermartingale. We deduce that W(x)≤EX∞^f ≤f(x)by virtue of Fatou’s lemma.

From a), the inequality 0 ≤ W(x_n) ≤ f(x_n) allows us to immediately conclude about the second statement.

Theorem 4.4.16. If the HJB equation (4.2.10) has an unique global visco- sity solution in C₁(K), then the Bellman function is concave.

Proof. If we define the Bellman function only on the class of admissible strategies generating portfolio processes evolving inK all the time, i.e

Abx :={π∈ A:V_t^π ∈K∀t >0} 6=∅,¹ and

Wc(x) := sup

π∈Abx

EJ_∞^π(x), x∈intK,

then the Bellman functioncW is a global viscosity solution to the same HJB equation as W. By assumption, the global viscosity solution of this HJB equation is unique, i.e Wc = W. It is well known that the function Wc is concave (the proof of it is straightforward and given in Framstad et al. [40]).

Therefore,W is also concave.

4.4 Appendix 113 The aim of this section is to show that this r.s.d.e has a solution on the set K₀ which is trapped at zero. To do so, we shall prove several intermediate lemmas. The main proof is based on the existence of a solution to a r.s.d.e.

on a bounded domain G when the direction of the reflection is given by a C²-functionγ satisfying the following condition (see [30]) :

C1 :γ ∈C²(R²,R²)and there is b∈(0,1)such that [

0≤t≤b

B(x−tγ(x), tb)⊆G^c, forx∈∂G.

Theorem 4.4.17. The Skorokhod problem (4.4.51) has a solution which is trapped at zero.

Proof.

Denote by D the bisector of the cone K₀ⁿ and we put d > 0 such that D:={−dx+y= 0}. Let us introduce the polygons

K₀ⁿ:=K₀∩ {ǫ⁻¹_n ≤x+dy≤ǫ_n}

where ǫ_n → ∞. Let x ∈ ∂K₀ⁿ be a starting point. The case x = 0 being trivial, we assume thatx6= 0 hencex∈K₀ⁿ if nis large enough.

• Step 1. There exists closed regions K˜₀ⁿ such that K₀ⁿ ⊆ K˜₀ⁿ ⊆ K₀ⁿ⁺¹ verifying the conditionC1for some reflection functionγnsatisfyingγn(x)→ γ(x) for all x ∈ ∂K₀. Indeed, we denote by a_n and b_n the two points of

∂K₀∩{ǫ_n=x+dy}such thaty_an > y_bn. Observe thatb_nis the symmetric of anwith respect to the bisectorD. Similarly,cnand dnare the two symmetric points of∂K₀∩ {ǫ⁻¹_n =x+dy}. Denotee˜_n:=D∩ {(ǫ_n+ǫ_n+1)/2 =x+dy}. We then defineK˜₀ⁿ as the polygon

K˜₀ⁿ:=K₀∩ {(ǫ⁻¹_n +ǫ⁻¹_n+1)/2≤x+dy≤(ǫ_n+ǫ_n+1)/2},

and denote by˜a_n and˜b_n the two points of∂K₀∩ {(ǫ_n+ǫ_n+1)/2 =x+dy} such thaty_˜an> y_˜b

n. Similarly,˜c_nandd˜_nare the two points of∂K₀∩{(ǫ⁻¹_n + ǫ⁻¹_n+1)/2 =x+dy} such thaty_˜cn > y_d˜

n.

Letη₁ be the outward normal to∂K₀∩K₁ andη₂ be the outward normal to

∂K₀∩K₂. We considerg₃ a unit vector such that g₃η₁>0andg₃(1, d)>0.

Similarly, we defineg4 as a unit vector such that g4η2 >0 and g4(1, d)>0, g₅ is a unit vector such thatg₅η₁ >0,g₅η₂ >0.

Let us introduce the smooth function

γ_n(x) := −

g₁χ¹(x) +g₂χ²(x) +g₃(1−χ¹(x))(1−χ⁴(x))χ³(x)

+ g₄(1−χ²(x))(1−χ³(x))χ⁴(x) +g₅(1−χ¹(x))(1−χ²(x))χ⁵(x)

, where, by lemma 4.4.18, the functionsχi∈C^∞(R²,[0,1]),i= 1,· · · ,5and, withγ_n→0 small enough,

χ¹(x) = 1on [d_n, b_n]_γn and χ¹(x) = 0 onR²\[d_n, b_n]_2γn χ²(x) = 1on [cn, an]γn and χ²(x) = 0 onR²\[cn, an]2γn, χ³(x) = 1on(C₃ⁿ)_γn :=

[˜e_n,˜b_n]∪[˜b_n, b_n]

γn

andχ³(x) = 0onR²\(C₃ⁿ)_2γn, χ⁴(x) = 1on(C₄ⁿ)_γn := ([˜e_n,˜a_n]∪[˜a_n, a_n])_γn andχ⁴(x) = 0onR²\(C₄ⁿ)_2γn. χ⁵(x) = 1 on (C₅ⁿ)_γn :=

[c_n,c˜_n]∪[˜c_n,d˜_n]∪[ ˜d_n, d_n]

γn

and χ⁵(x) = 0 on R²\(C₅ⁿ)_2γn.

Let us denote byη(x)the outward normal at each point of ∂K˜₀ⁿ. The map- pingη :∂K˜₀ⁿ7→R² is continuous except at the points˜a_n,˜b_n,˜c_n,d˜_n where it admits left and righ limits we denote by η(x±). Moreover, by construction we haveγn(x)η(x−)>0 and γn(x)η(x+)>0 for allx∈∂K˜₀ⁿ.

Observe that

d(x−tγn(x), ∂K˜₀ⁿ)≥ min

x∈∂K˜₀ⁿ

d(x−tγn(x), ∂K˜₀ⁿ) :=m, ∀x∈∂K˜₀ⁿ. Indeed, by a compactness argument, m= d(x∞−tγn(x∞), ∂K˜₀ⁿ) for some x_∞∈K˜₀ⁿ. Since

d(x_∞−tγ_n(x_∞), ∂K˜₀ⁿ)≥ −tγ_n(x_∞)η(x_∞±) = 2bt,

where 2b := −γ_n(x_∞)η(x_∞±) > 0, we finally deduce that Condition C1 holds.

•Step2. By virtue of Corollary 5.2 [30], there exists a unique strong solution (Vⁿ, kⁿ) starting from x to the reflected s.d.e. (4.4.51) on the domain K˜₀ⁿ. Let

τⁿ := inf{t:V_tⁿ(1, d) =ǫ⁻¹_n }, ρⁿ := inf{t:V_tⁿ(1, d) =ǫn},

andµⁿ:=τⁿ∧ρⁿ.On the intervall[0, µⁿ], the process(Vⁿ, kⁿ)is solution to (4.4.51) on the domainK˜₀ⁿ⁺¹ with respect to γⁿ⁺¹. Indeed, on the intervall [0, µⁿ], the reflection only occurs on the boundary∂K₀on whichγⁿ⁺¹andγⁿ coincides withγ. By the uniqueness property given by Corollary 5.2 [30], we deduce that(Vⁿ, kⁿ) = (Vⁿ⁺¹, kⁿ⁺¹) on [0, µⁿ]. It follows that µⁿ ≤µⁿ⁺¹. The rest of the proof is done as in [61], page229.

For any compact subsetC ofR² and for allǫ >0, let us define C_ǫ:= [

c∈C

B_c(ǫ).

Recall the well known result :

Lemma 4.4.18. For any compact subset C of R² and for all ǫ > 0, there exists χ^ǫ ∈ C^∞(R²,[0,1]) such that χ^ǫ = 1 on C_ǫ and χ^ǫ vanishes outside C_2ǫ.

4.4 Appendix 115 Skorokhod Problem for Pure-Jumps Lévy Processes

Let γ : ∂K₀ 7→ R² be a vector-valued function with g(x) = −g_i on the set (∂K0 ∩∂Ki)\{0} and γ(0) = 0. Recall that a Lévy process has finite activity when ν(R²) < ∞. In this case, it can be represented as the sum of a compound poison process and a scaled Wiener process with drift. So, consider Y a process such thatYt = (Y₀¹, Y₀²,0) + (t, Wt, Nt), t≥ 0, where W is a standard brownian motion and N is a pure jump process of finite activity. This means that

N_t=

N˜t

X

k=1

∆N_Tk,

where ∆N_Tk are i.i.d. random variables and N˜_t = P

k1_Tk≤t is a Poisson process with jump stopping times(T_k)_k.

Let σ = R² 7→ R²×R²×R² be a matrix-valued function assumed to be Lipschitz-continuous.

We consider the Skorokhod problem on K0 formulated as follows : find a pair of adapted làdlàg (resp. càglàd) processes,V, starting fromx∈K₀ and k, real-valued, starting at zero, and increasing such that

dV_t = σ(V_t−)dY_t+γ(V_t)dk_t, dk_t = I_Vt∈K\intK0dk_t,

V_t⁺ ∈ K¯0∀t≥0. (4.4.54)

The aim of this subsection is to show that this r.s.d.e has a solution on the setK¯0.

Theorem 4.4.19. There exists a unique solution to the Skorokhod problem (4.4.54).

Proof.Let(T_k)_k∈N\{0}the jump stopping times of the processY. Assume we have already constructed a solution(V, k) to (4.4.54) on the interval[0, T_k).

Define

V_T^k :=V_T^k₋+σ(V_T^k₋)∆Y_T^k. Let us set

V_T^k₊ =P^−K_K

0 (V_T^k)∈L⁰(R²,FT^k)∈∂K₀

where the projection operator P_K0 is defined in lemma 4.4.20. We define

∆⁺kTk by the equality

∆⁺V_Tk :=γ(V_Tk)∆⁺k_Tk.

Applying theorem 4.4.17 and the strong markov property, there exists a solution ( ˜V ,k)˜ to (4.4.54) from the starting point V˜₀ :=V_Tk+ with respect toN˜t:=N_T^k_+t−N_T^k,W˜t:=W_T^k_+t−W_T^k on the interval[0, T^k+1−T^k]. We then setVt:= ˜V_t−T^k and kt:=kTk++ ˜k_t−T^k on(T^k, T^k+1). The uniqueness follows from the uniqueness on each interval[T_k, T_k+1).

In the following, we assume that K ⊆R² is a constant cone satisfying the hypothesis of the introduction and K₀ ⊆ K is a closed cone with ∂K₀ ⊆ intK and intK0 6=∅.

Lemma 4.4.20. [Projection onto K₀ parallel to −K] Given x ∈ K, there exists a uniquey :=P^−K_K

0 (x)∈K₀ such that kx−yk= min

k∈K{kx−kk:x−k∈K}.

We omit the proof which is standard. It is easily observable that the direction ofx−P^−K_K

0 (x) is given by g₂ if x∈K₂ andg₁ if x∈K₁.