Backward stochastic differential equations with respect to general filtrations and applications to insider finance

Backward stochastic differential equations with respect to general filtrations and applications to insider finance

Bernt Øksendal^1,2 and Tusheng Zhang^3,1 3 September 2009

Abstract

In this paper, we study backward stochastic differential equations with respect to general filtrations. The results are used to find the optimal consumption rate for an insider from a cash flow modeled as a generalized geometric Itˆo-L´evy process.

AMS Subject Classification: Primary 60H15 Secondary 93E20, 35R60.

1 Introduction

The classical backward stochastic differential equation (BSDE) consists in finding a pair (Y_t, Z_t) of F_t -adapted processes such that

dY_t =−f(t, Y_t, Z_t)dt+Z_tdB_t; t∈[0, T]

YT =ξ. (1.1)

whereB_t is a Brownian motion on a filtered probability space (Ω,F,{F_t}_t≥0, P) , ξ is a givenF_T -measurable random variable andf : [0, T]×R×R→R is a given function.

If f(t, y, z) =f(t, y) does not depend on z, then an equivalent way of writing (1.1) is Y_t=E

ξ+ Z T

t

f(s, Y_s)ds|F_t]; t∈[0, T]. (1.2) In this paper we extend (1.2) to a general filtration H_t and consider the problem to find an H_t- adapted process Yt such that

Y_t=E ξ+

Z T t

f(s, Y_s)ds|H_t]; t∈[0, T], (1.3) where ξ now is a given H_T-measurable random variable. Thus we arrive at a BSDE based on a general filtration H_t, not necessarily the filtrationF_tof Brownian motion.

This turns out to be a useful generalization for certain applications, for example in connection with insider trading in finance.

1CMA, University of Oslo, Box 1053 Blindern, NO-0316 Oslo, Norway

2Norwegian School of Economics and Business Administration (NHH), Helleveien 30, NO-5045, Bergen, Norway

3School of Mathematics, University of Manchester, Oxford Road, Manchester M13 PL, U.K.

Here is an outline of the paper. In Section 2 we give a more detailed presentation of our BSDE based on a given filtration. In Section 3 we prove existence and uniqueness of solutions of such equations. In Section 4 we study reflected BSDEs based on a given filtration. We prove existence and uniqueness of solution and we show that it coincides with the solution of an optimal stopping problem (forH-stopping times). In Section 5 we give an application to finance. We show that the optimal consumption problem for an insider can be transformed into a BSDE with respect to the information filtration H_t of the insider. Then we apply results from previous sections to find the optimal consumption rate explicitly.

2 Statement of the problem

Let (Ω,H,H_t, P) be a complete filtrated probability space with a right continuous filtration{H_t, t≥ 0}. Let T >0 and let ξ be anH_T measurable random variable withE[|ξ|]<∞, where E denotes expectation with respect toP. Letf(ω, t, y) : Ω×[0, T]×R^d→R^dbe a givenP ×B(R^d)-measurable function, whereP is the predictableσ-field associated with the filtration{H_t, t≥0}. Consider the following backward stochastic differential equation (BSDE):

BSDE(1): Find an H_t- predictable processY_t such that E

Z T 0

|f(s, Ys)|ds

<∞. (2.1)

and

Y_t=E ξ+

Z T t

f(s, Y_s)ds|H_t]; t∈[0, T]. (2.2) Next, consider the following BSDE:

BSDE(2): Find an H_t- predictable processYt and anH_t-local martingale Mt such that dY_t =−f(t, Y_t)dt+dM_t

Y_T =ξ. (2.3)

An equivalent formulation to (2.3) is that Z T

0

|f(s, Y_s)|ds <∞ a.s. (2.4) and

Yt=ξ+ Z T

t

f(s, Ys)ds−(MT −Mt); t∈[0, T]. (2.5) There is a close relation between BSDE(1) and BSDE(2): First note that if Y_t satisfies BSDE(1), then we can define

Mt=E[ξ+ Z T

0

f(s, Ys)ds|H_t] and we see from (2.2) that

Yt = E[ξ+ Z T

0

f(s, Ys)ds− Z t

0

f(s, Ys)ds|H_t]

= −

Z t 0

f(s, Ys)ds+Mt.

Moreover,YT =ξ. Hence (Yt, Mt) satisfies BSDE(2).

Conversely, if (Yt, Mt) satisfies both (2.5) and the stronger version (2.1) of (2.4), then (1.2) follows by taking conditional expectation of (2.5) with respect to H_t ( stopping if necessary).

Hence Y_t satisfies BSDE(1).

We now proceed to study BSDE(2).

Definition 2.1 We say that a pair (Yt, Mt, t≥0)is a solution to BSDE(2) if (i). Y_t is an H_t-predictable, right continuous R^d-valued process.

(ii). Mt, t≥0 is a right continuous R^d-valuedH_t-local martingale.

(iii). For every t≥0,

Y_t=ξ+ Z T

t

f(s, Y_s)ds−(M_T −M_t) (2.6)

P-almost surely.

3 Backward Stochastic Differential Equations

3.1 Existence and Uniqueness Theorem 3.1 Suppose ξ ∈ L²(Ω) and E[RT

0 |f(t,0)|²dt]<∞. Assume that f is uniformly Lips- chitz with respect toy, i.e., there exists a constant C such that

|f(t, y1)−f(t, y2)| ≤C|y₁−y2| (3.1) Then there exists a unique solution (Y, M) to the BSDE(2) satisfying

E[ sup

0≤t≤T

|Y_t|²]<∞. (3.2)

Proof. LetB denote the Banach space of R^d-valued, H_t-adapted processes X such that

||X||_B := sup

0≤t≤T

(E[X_t²])¹² <∞.

Define recursively a sequence Y_tⁿ, t≥0 of processes inB by Y⁰ = 0 and Y_tⁿ⁺¹=E[ξ+

Z T t

f(s, Y_sⁿ)ds

H_t] (3.3)

It is easy to see thatYⁿ∈B for all n≥1. Moreover, E

|Y_tⁿ⁺¹−Y_tⁿ|²

≤ T E Z T

t

|f(s, Y_sⁿ)−f(s, Y_sⁿ⁻¹)|²ds

≤ CT Z _T

t

E[|Y_sⁿ−Y_sⁿ⁻¹|²]ds (3.4) Setφ_n(t) =E[|Y_tⁿ−Y_tⁿ⁻¹|²]. Then (3.4) becomes

φn+1(t)≤CT Z T

t

φn(s)ds (3.5)

Repeating the above inequality, we get sup

0≤t≤T

φ_n+1(t)≤ sup

0≤s≤T

φ₁(s)(CT)ⁿTⁿ

n! (3.6)

This implies that Yⁿ, n ≥ 1 is a Cauchy sequence in B. Denote the limit of Yⁿ by ˆY. Letting n→ ∞ in (3.3) we obtain

Yˆt=E[ξ+ Z T

t

f(s,Yˆs)ds

H_t] (3.7)

Next we show that ˆY_t, t ≥ 0 admits a right continuous version which will be the solution to BSDE(2). LetMt, t≥0 be the right continuous version of the square integrable martingaleE[ξ+ RT

0 f(s,Yˆs)ds

H_t]. Put

Y_t=M_t− Z _t

0

f(s,Yˆ_s)ds, t≥0 Then Y_t is right continuous and for everyt≥0,

Y_t=E[ξ+ Z T

t

f(s,Yˆ_s)ds

H_t] = ˆY_t P-almost surely. By the Fubini theorem, it follows that

Y_t = M_t−M_T +ξ+ Z T

0

f(s,Yˆ_s)ds− Z t

0

f(s,Yˆ_s)ds

= ξ+ Z T

t

f(s,Yˆs)ds−(MT −Mt)

= ξ+ Z T

t

f(s, Y_s)ds−(M_T −M_t) (3.8) P-almost surely. This shows that (Y, M) is a solution to the BSDE(2). Let us now prove (3.2).

Using Doob’s inequality, we have E[ sup

0≤t≤T

|Y_t|²] ≤ 2E[ sup

0≤t≤T

|M_t|²] + 2T E[

Z T 0

|f(s, Y_s)|²ds]

≤ C2E[|M_T|²] + 4T E[

Z T 0

|f(s,0)|²ds] + 4T Z T

0

E[|Y_s|²]ds

= C₂E[|ξ+ Z T

0

f(s, Y_s)ds|²] + 4T E[

Z T 0

|f(s,0)|²ds] + 4T Z T

0

E[|Y_s|²]ds

≤ C(E[|ξ|²] + sup

0≤t≤T

E[|Y_t|²] +E[

Z T 0

|f(s,0)|²ds]<∞. (3.9) It remains to prove the uniqueness. Let (X, Z) be another solution to equation BSDE(2). Then both Y and X satisfy

Yt=E[ξ+ Z T

t

f(s, Ys)ds

H_t] (3.10)

Xt=E[ξ+ Z T

t

f(s, Xs)ds

H_t] (3.11)

Using the Lipschitz continuity off, as the proof of (3.4), we have E

|Y_t−X_t|²

≤CT Z T

t

E[|Y_s−X_s|²]ds (3.12)

By Gronwall’s inequality, it follows that Y_t = X_t, which in turn gives M_t = Z_t. The proof is complete. 2

Next theorem states a result on existence and uniqueness under some monotone conditions on the coefficients.

Theorem 3.2 Suppose 1. ξ∈L²(Ω)and E[RT

0 |f(t,0)|²dt]<∞.

2. There exists a constant C such that

(y1−y2)(f(t, y1)−f(t, y2))≤C|y₁−y2|² (3.13) 3. f(t, y) is continuous in y and

|f(t, y)| ≤C1(t), (3.14)

withE[RT

0 C₁(s)ds]<∞.

Then there exists a unique solution (Y, M) to the BSDE(2) satisfying E[ sup

0≤t≤T

|Y_t|²]<∞. (3.15)

Proof. Take an even, non-negative functionφ∈C₀^∞(R) with R

Rφ(x)dx= 1. Define f_n(t, y) =

Z

R

f(t, z)φ_n(y−z)dz,

whereφ_n(z) =nφ(nz). Sincef is continuous iny, it is easy to see thatf_n(t, y)→f(t, y) asn→ ∞.

Furthermore, for every n≥1,

|f_n(t, y₁)−f_n(t, y₂)| ≤C_n|y₁−y₂|, (3.16) for some constantCn. Consider the BSDE:

Y_tⁿ=ξ+ Z T

t

fn(s, Y_sⁿ)ds+M_Tⁿ−M_tⁿ; t∈[0, T]. (3.17) Equation (3.17) has a unique solution (Yⁿ, Mⁿ) according to Theorem 2.1. Next we show thatY_tⁿ is a Cauchy sequence. By Itˆo’s formula, we have

|Y_tⁿ−Y_t^m|²+ [Yⁿ−Y^m, Yⁿ−Y^m]T −[Yⁿ−Y^m, Yⁿ−Y^m]t

= 2 Z T

t

(Y_sⁿ−Y_s^m)(fn(s, Y_sⁿ)−fm(s, Y_s^m))ds−2 Z T

t

(Y_s−ⁿ −Y_s−^m)d(M_sⁿ−M_s^m) (3.18)

In view of (3.13), (3.14),

(Y_sⁿ−Y_s^m)(fn(s, Y_sⁿ)−fm(s, Y_s^m))

= Z

R

(Y_sⁿ−Y_s^m)(f(s, Y_sⁿ− 1

nz)−f(s, Y_s^m− 1

mz))φ(z)dz

= Z

R

[(Y_sⁿ− 1

nz)−(Y_s^m− 1

mz)](f(s, Y_sⁿ− 1

nz)−f(s, Y_s^m− 1

mz))φ(z)dz +

Z

R

(1 nz− 1

mz))(f(s, Y_sⁿ− 1

nz)−f(s, Y_s^m− 1

mz))φ(z)dz

≤ C Z

R

((Y_sⁿ− 1

nz)−(Y_s^m− 1

mz))²φ(z)dz+C1(s) Z

R

(1

n|z|+ 1

m|z|)φ(z)dz

≤ C(Y_sⁿ−Y_s^m)²+C Z

R

( 1 n² + 1

m²)z²φ(z)dz+C1(s) Z

R

(1

n|z|+ 1

m|z|)φ(z)dz (3.19) Substitute (3.19) into (3.18), take expectation to obtain

E[|Y_tⁿ−Y_t^m|²] +E{[Yⁿ−Y^m, Yⁿ−Y^m]T −[Yⁿ−Y^m, Yⁿ−Y^m]t}

≤ C Z T

t

E[(Y_sⁿ−Y_s^m)²]ds+CT Z

R

( 1 n² + 1

m²)z²φ(z)dz +CE[

Z T t

C1(s)ds]

Z

R

(1

n|z|+ 1

m|z|)φ(z)dz (3.20)

Applying the Gronwall’s inequality, it follows from (3.20) that E[|Y_tⁿ−Y_t^m|²]≤CT{

Z

R

( 1 n² + 1

m²)z²φ(z)dz+E[

Z T t

C1(s)ds]

Z

R

(1

n|z|+ 1

m|z|)φ(z)dz} (3.21) Hence,

n,m→∞lim sup

0≤t≤T

E[|Y_tⁿ−Y_t^m|²] = 0 (3.22)

By (3.20) and the Burkholder inequality, (3.22) further implies

n,m→∞lim E[ sup

0≤t≤T

|M_tⁿ−M_t^m|²]

≤ lim

n,m→∞E([Mⁿ−M^m]_T)

= lim

n,m→∞E([Yⁿ−Y^m]_T) = 0. (3.23)

Consequently, there exist a square integrable, predictable processYtand a square integrable, right continuous martingale M_tsuch that

n→∞lim sup

0≤t≤T

E[|Y_tⁿ−Y_t|²] = 0 (3.24)

n→∞lim E[ sup

0≤t≤T

|M_tⁿ−M_t|²] = 0 (3.25)

In view of (3.14), use the dominated convergence theorem and let n→ ∞ in (3.17) to get Yt=ξ+

Z T t

f(s, Ys)ds+MT −Mt; t∈[0, T]. (3.26)

Since the right hand side of (3.26) is right continuous, we can takeY to be right continuous. Thus Yt, t≥0 is a solution to BSDE(2).

Now we prove the uniqueness. Suppose that (Y¹, M¹) and (Y², M²) are two solutions to BSDE(2). Similar to the calculations for (3.18), we have

|Y_t¹−Y_t²|²+ [M¹−M², M¹−M²]_T −[M¹−M², M¹−M²]_t

= 2 Z T

t

(Y_s¹−Y_s²)(f(s, Y_s¹)−f(s, Y_s²))ds−2 Z T

t

(Y_s−¹ −Y_s−² )d(M_s¹−M_s²) (3.27) Taking expectation and keeping (3.13) in mind, we get from (3.27) that

E{|Y_t¹−Y_t²|²+ [M¹−M², M¹−M²]T −[M¹−M², M¹−M²]t} ≤CE[

Z T t

(Y_s¹−Y_s²)²ds]

By Gronwall’s inequality, we deduce that Y_t¹ = Y_t², M_t¹ = M_t² for t ≥0, thereby completing the proof.

3.2 Comparison theorem

Let (Y, M) be the solution to the following linear BSDE:

Yt=ξ+ (φT −φt) + Z T

t

βsYsds−(MT −Mt), (3.28) where φt, t ≥0 is a given, right continuous process of bounded variation with φ0 = 0 and βt is a bounded predictable process. We have the following result.

Theorem 3.3 Assume the total variation of φ is integrable. The following representation holds Yt=E[L^T_tξ+

Z T t

L^s_tdφs|H_t], (3.29)

where

L^s_t =exp(

Z s t

β_udu)

In particular, if ξ ≥0, then Yt≥0. Moreover Y0 = 0 implies ξ = 0 and φ= 0.

Proof. Put L_t=exp(Rt

0β_udu). By Itˆo’s formula, we find that Y_tL_t+

Z _t

0

L_sdφ_s=Y₀− Z _t

0

L_sdM_s

is a martingale. Consequently, Y_tL_t+

Z _t

0

L_sdφ_s=E[Y_TL_T + Z _T

0

L^s_tdφ_s|H_t]

=E[ξLT + Z T

0

L^s_tdφs|H_t].

(3.29) follows. 2

Let both (ξ¹, f¹(s, y)) and (ξ², f²(s, y)) satisfy the conditions in Theorem 2.1. Denote by (Y¹, M¹) and (Y², M²) the solutions of the BSDEs associated with (ξ¹, f¹(s, y)) and (ξ², f²(s, y)), respectively.

Theorem 3.4 Suppose f¹(s, Y_s²) ≥ f²(s, Y_s²) almost surely on Ω×[0, T] and ξ¹ ≥ ξ². Then, Y_t¹ ≥ Y_t² P-almost surely for all t ≥ 0. Furthermore, if Y_t¹ = Y_t² P-almost surely on an event A∈ H_t, then ξ¹=ξ² on A and Y_s¹=Y_s² onA for s≥t.

Proof. Define

βs=

( _f1(s,Y_s¹)−f¹(s,Y_s²)

Y_s¹−Y_s² ifY_s¹ 6=Y_s²,

0 otherwise. (3.30)

Then β_s is bounded. Moreover, we have Y_t¹−Y_t² =ξ¹−ξ²+

Z _T

t

(f¹(s, Y_s²)−f²(s, Y_s²))ds+ Z _T

t

β_s(Y_s¹−Y_s²)ds−[(M_T¹−M_T²)−(M_t¹−M_t²)]

(3.31) Using Theorem 2.2, we have

Y_t¹−Y_t²=E[L^T_t(ξ¹−ξ²) + Z T

t

L^s_t(f¹(s, Y_s²)−f²(s, Y_s²))ds|H_t] (3.32) (3.32) implies the desired results.2

As a corollary to Theorem 2.4, we have the following

Theorem 3.5 If f(t,0)≥0 dP ×dt, then the solution Yt(ξ) gives rise a price system, that is, 1. At any time t, the price Yt(ξ) for a positive contingent claim ξ is positive.

2. At any time t, the price Yt(ξ) is an increasing function with respect to ξ.

3. No-arbitrage holds, i.e., if the prices Y_t¹ and Y_t² coincide on an event A ∈ F_t, then on A, ξ¹=ξ², a.s.

4 Reflected Backward Stochastic Differential Equations

Consider the reflected backward stochastic differential equation:

dY_t=−f(t, Y_t)dt+dM_t−dK_t (4.1) Definition 4.1 Let Lt;t≥ 0 be a given H_t-adapted process. We say that (Yt, Mt, Kt, t ≥0) is a solution to RBSDE(3.1) with lower barrier Lt, t≥0 if

(i). Y_t is an H_t-predictable, right continuous real-valued process (ii). Y_t≥L_t P-a.s. for every t≥0.

(iii). Mt, t≥0 is a right continuous real-valued H_t-local martingale.

(iv). K_t, t≥0 is an increasing, continuous H_t-adapted process withK₀ = 0.

(v). For every t≥0, Y_t=ξ+

Z T t

f(s, Y_s)ds−(M_T −M_t) +K_T −K_t P−almostly surely. (4.2) (vi). RT

0 (Yt−Lt)dKt= 0.

In the following we let T_t,T^H denote the set ofH-stopping times τ such that t≤τ ≤T a.s.

Theorem 4.2 Let f(t, y) and ξ be as in Theorem 2.1. Assume ξ ≥LT and one of the following conditions hold:

(i). L_t is a right continuous, increasing, square integrable predictable process withE[L²_T]<∞.

(ii).L_t is absolutely continuous and E[RT

0 (L⁰_t)²dt]<∞.

Then :

a) The RBSDE(4.1) admits a unique solution.

b) The solution process Y_t can be given the optimal stopping representation Yt=esssup_τ∈T^H

t,TE[

Z τ t

f(s, Ys)ds+Lτχτ <T +ξχτ=T|H_t];t∈[0, T] (4.3) c) The solution process Kt is given by

KT−t−KT =maxs≤t(ξ+ Z T

T−s

f(u, Yu)du−(MT −MT−s)−LT−s)⁻;t∈[0, T] (4.4) where x⁻=max(−x,0).

Proof.

a). We first prove the uniqueness. Suppose that (Y_t¹, M_t¹, K_t¹) and (Y_t², M_t², K_t²) are two solutions to the RBSDE(2). By Itˆo’s formula, we have

|Y_t¹−Y_t²|²+ [Y¹−Y², Y¹−Y²]T −[Y¹−Y², Y¹−Y²]t

= 2 Z T

t

(Y_s¹−Y_s²)(f(s, Y_s¹)−f(s, Y_s²))ds−2 Z T

t

(Y_s−¹ −Y_s−² )d(M_s¹−M_s²) +2

Z T t

(Y_s¹−Y_s²)d(K_s¹−K_s²) (4.5)

Take expectation in the above equation, use (ii), (vi) in the definition 3.1 to obtain E[|Y_t¹−Y_t²|²] +E{[Y¹−Y², Y¹−Y²]T −[Y¹−Y², Y¹−Y²]t}

≤ C Z T

t

E[(Y_s¹−Y_s²)²]ds−2E[

Z T t

(Y_s²−L_s)dK_s¹]

−2E[

Z _T

t

(Y_s¹−L_s)dK_s²]

≤ C Z T

t

E[(Y_s¹−Y_s²)²]ds (4.6)

(4.6) and Gronwall’s inequality implies thatE[|Y_t¹−Y_t²|²] = 0 fort≥0, proving the uniqueness.

To prove the existence, we will use the penalization method. For n≥1, consider the penalized backward stochastic differential equation:

Y_tⁿ=ξ+ Z T

t

f(s, Y_sⁿ)ds−(M_Tⁿ−M_tⁿ) +n Z T

t

(Y_sⁿ−Ls)⁻ds (4.7) Equation (4.7) admits a unique solution according to Theorem 2.1. By the comparison Theorem 2.4, we know that the sequenceYⁿ, n≥1 is increasing, i.e., Y_tⁿ ≤Y_tⁿ⁺¹ P-a.s. Set Y_t := limn→∞Y_tⁿ. Similar to the proof of Theorem 4.2 of [E], we next give an a priori estimate for the L² bound of Yⁿ. PutK_tⁿ=nRt

0(Y_sⁿ−Ls)⁻ds. By Itˆo’s formula, we have

|Y_tⁿ|²+ [Mⁿ, Mⁿ]T −[Mⁿ, Mⁿ]t

= ξ²+ 2 Z T

t

Y_sⁿ(f(s, Y_sⁿ)ds−2 Z T

t

Y_s−ⁿdM_sⁿ

+2n Z T

t

Y_sⁿ(Y_sⁿ−Ls)⁻ds (4.8)

Asf has a linear growth in the variabley, it follows that Z T

t

|Y_sⁿ(f(s, Y_sⁿ)|ds≤CT(1 + Z T

t

(Y_sⁿ)²ds) (4.9)

For anyδ >0,

2nE Z T

t

Y_sⁿ(Y_sⁿ−Ls)⁻ds

= 2nE Z T

t

(Y_sⁿ−L_s)(Y_sⁿ−L_s)⁻ds

+ 2nE Z T

t

L_s(Y_sⁿ−L_s)⁻ds

≤ 1 δE

sup

0≤s≤T

(L_s)²

+δE

(K_Tⁿ−K_tⁿ)²

(4.10) On the other hand, in view of (4.7), we see that

E

(K_Tⁿ−K_tⁿ)²

≤ CE[|ξ|²] +CE[|Y_tⁿ|²] +C(1 + Z T

t

E[(Y_sⁿ)²]ds) +CE

(M_Tⁿ−M_tⁿ)²

≤ CE[|ξ|²] +CE[|Y_tⁿ|²] +C(1 + Z _T

t

E[(Y_sⁿ)²]ds) +CE

[Mⁿ, Mⁿ]T −[Mⁿ, Mⁿ]t

(4.11)

Take expectation in (4.8) and substitute (4.9)–(4.11) into (4.8) to get E[|Y_tⁿ|²] +E

[Mⁿ, Mⁿ]_T −[Mⁿ, Mⁿ]_t

≤ CδE[|ξ|²] +CδE

sup

0≤s≤T

(Ls)²

+Cδ(1 + Z T

t

E[(Y_sⁿ)²]ds) +Cδ

E[|Y_tⁿ|²] +E

[Mⁿ, Mⁿ]_T −[Mⁿ, Mⁿ]_t

(4.12) Selectδ so thatCδ <1 and Apply Gronwall’s inequality to deduce that

sup

n

sup

0≤t≤T

(E[|Y_tⁿ|²] +E([Mⁿ, Mⁿ]T))≤CTE[|ξ|²] +CTE

sup

0≤s≤T

(Ls)²

(4.13) This implies sup_nE[(M_Tⁿ)²]< ∞. Thus, there exists a subsequence n_k such that M_T^nk converges weakly to some random variableM_T inL²(Ω) ask→ ∞. LetM_t, t≥0 denote the martingale with terminal value MT. Then it is easy to see that M_t^nk converges weakly to Mt in L²(Ω) for every t≤T. Replacingnby n_k in (4.7) we get

K_T^nk −K_t^nk =Y_t^nk−ξ− Z T

t

f(s, Y_s^nk)ds+ (M_T^nk −M_t^nk) (4.14) Since each term on the right hand side converges, we deduce that there exists an increasing process K_t, t ≥ 0 such that K_t^nk converges weakly to K_t. Moreover, (Y, M, K) satisfies the following backward equation:

Yt=ξ+ Z T

t

f(s, Ys)ds−(MT −Mt) +KT −Kt (4.15) By Lemma 2.2 in [P], it follows from the equation (4.15) thatY_t, K_t are right continuous with left limits. Furthermore, using Fatou Lemma it follows that

E[

Z T 0

(Y_t−L_t)⁻dt]

≤ lim inf

n→∞ E[

Z T 0

(Y_tⁿ−Lt)⁻dt]

≤ lim inf

n→∞

1

nE[(K_Tⁿ−K_tⁿ)]≤C lim

n→∞

1

n = 0 (4.16)

As both Y and L are right continuous, (4.16) implies thatYt≥Lt P-a.s. for evertt≥0. To show that (Y, M, K) is a solution to the RBSDE(3.1), it remains to prove

Z _T

0

(Y_t−L_t)dK_t= 0 (4.17)

To this end, we need to strengthen the convergence ofKⁿ toK. Define φ(u, x) =n[(x−Lu)⁻]²

Then φ(u, x) is convex in x for every u ≥0. By smooth approximation, we may assume φ⁰⁰(u, x) exists and φ⁰⁰(u, x)≥0, whereφ⁰ stands for the derivative ofφw.r.t. x. By Itˆo’s formula, we have

φ(t, Y_tⁿ) = ∂tφ(t, Y_tⁿ) +φ⁰(t, Y_tⁿ)dY_tⁿ +1

2φ⁰⁰(t, Y_tⁿ)d[Yⁿ, Yⁿ]^c_t +d

X

0<s≤t

{φ(s, Y_sⁿ)−φ(s, Y_s−ⁿ)−φ⁰(s, Y_s−ⁿ)∆Y_sⁿ}

(4.18) Hence,

φ(t, Y_tⁿ) + Z T

t

[n(Y_uⁿ−L_u)⁻]²du+ Z T

t

1

2φ⁰⁰(u, Y_uⁿ)d[Yⁿ, Yⁿ]^c_u

+ X

0<s≤t

{φ(s, Y_sⁿ)−φ(s, Y_s−ⁿ)−φ⁰(s, Y_s−ⁿ)∆Y_sⁿ}

= −2n

Z T t

|_{Lu>Yn

u}(Lu−Y_uⁿ)dLu−2n Z T

t

(Y_uⁿ−Lu)⁻f(u, Y_uⁿ)du

−2n Z _T

t

(Y_uⁿ−L_u)⁻dM_uⁿ (4.19)

Since φ(u, x) is convex inx, we have Z T

t

1

2φ⁰⁰(u, Y_uⁿ)d[Yⁿ, Yⁿ]^c_u≥0, X

0<s≤t

{φ(s, Y_sⁿ)−φ(s, Y_s−ⁿ)−φ⁰(s, Y_s−ⁿ)∆Y_sⁿ} ≥0 (4.20) By virtue of the linear growth off, it is easy to see that

−2n Z T

t

(Y_uⁿ−L_u)⁻f(u, Y_uⁿ)du≤ 1 3

Z T t

[n(Y_uⁿ−L_u)⁻]²du+C_T +C_T Z T

t

(Y_uⁿ)²du (4.21) If condition (a) holds,−2nRT

t χ_{Lu>Yⁿ

u}(L_u−Y_uⁿ)dL_u≤0. In this case, it follows from (4.19)–(4.21) that

2 3E

Z T t

[n(Y_uⁿ−Lu)⁻]²du

≤C+E Z T

t

(Y_uⁿ)²du

(4.22) On the other hand, if condition (b) is true, then

−2n Z T

t

|_{Lu>Yn

u}(L_u−Y_uⁿ)dL_u ≤ 1 3

Z T t

[n(Y_uⁿ−L_u)⁻]²du+C Z T

t

(L⁰_u)²du In this case, we deduce from (4.19)–(4.21) that

1 3E

Z T t

[n(Y_uⁿ−Lu)⁻]²du

≤C+CE Z T

t

(Y_uⁿ)²du

+CE Z T

t

(L⁰_u)²du

(4.23) In view of (4.13), we obtain both from (4.22) and (4.23) that

sup

n

E Z T

t

[n(Y_uⁿ−Lu)⁻]²du

<∞. (4.24)

Choosing a further subsequence if necessary, (4.24) implies that nk(Y_u^nk−Lu)⁻ converges weakly to some functiongu inL²(Ω×[0, T], P×dt) and Kt defined above is given by Kt=R_t

0 gudu. Now we are in a position to prove (4.17). Write

Z T 0

(Yu−Lu)dKu− Z T

0

(Y_u^nk−Lu)dK_u^nk

= Z T

0

(Y_u−L_u)[n_k(Y_u^nk −L_u)⁻−g_u]du +

Z T 0

(Yu−Y_u^nk)[n_k(Y_u^nk−Lu)⁻]du (4.25) Because of the weak convergence, we have

k→∞lim Z T

0

(Yu−Lu)[nk(Y_u^nk−Lu)⁻−gu]du= 0 (4.26) By the monotone convergence theorem and (4.24), it follows that

k→∞lim | Z _T

0

(Yu−Y_u^nk)[n_k(Y_u^nk −Lu)⁻]du|

≤ lim

k→∞

Z T 0

(Y_u−Y_u^nk)²du

¹₂ Z T 0

[n_k(Y_u^nk−L_u)⁻]²du ¹₂

= 0 (4.27)

Combining (4.26) and (4.27) we obtain Z T

0

(Yu−Lu)dKu = lim

k→∞

Z T 0

(Y_uⁿ−Lu)dK_u^nk ≤0 AsYu≥Lu, (4.17) follows. The proof of a) is complete.

b) Next we prove that the unique solution process Yt of (4.3) can be given the representation (4.4). We do this by adapting the argument used in [EKPPQ] to our setting: First note that if τ ∈ T_t,T^H, then by (4.2) we have

Y_τ =ξ+ Z T

τ

f(s, Y_s)ds−(M_T −M_τ) +K_T −K_τ (4.28) Subtracting (4.28) from (4.2) and taking conditional expectation with respect to H_t we get

Yt = E[

Z τ t

f(s, Ys)ds+Yτ +Kτ −Kt|H_t]

≤ E[

Z τ t

f(s, Y_s)ds+L_τχ_{τ <T} +ξχ_τ=T|H_t].

Since τ ∈ T_t,T^H was arbitrary, this proves that Yt≤esssup_τ∈T^H

t,TE[

Z T t

f(s, Ys)ds+Lτχτ <T +ξχτ=T|H_t];t∈[0, T] (4.29)

On the other hand, if we define ˆ

τ_t=inf{s∈[t, T];Y_s=L_s} then ˆτ ∈ T_t,T^H and

E[

Z τˆt

t

f(s, Y_s)ds+L_τˆtχ_ˆτt<T +ξχ_τˆt=T|H_t]

=E[

Z τˆt

t

f(s, Ys)ds+Y_τˆt+K_τˆt−Kt|H_t] =Yt

Here we have used that

K_τˆt−K_t= 0,

which is a consequence of the requirement (vi) of Definition 4.1, i.e. of the equation Z T

0

(Y_t−L_t)dK_t= 0.

This completes the proof of b).

To prove c) we use the following result:

Skorohod Lemma. Let x(t) be a real c`adl`ag function on [0,∞) such that x(0) ≥ 0. Then there exists a unique pair (y(t), k(t)) of c`adl`ag functions on [0,∞) such that

(i) y(t) =x(t) +k(t) (ii) y(t)≥0

(iii) k(t) is c`adl`ag and nondecreasing, k(0) = 0 (iv) The functionk(t) is given by

k(t) =sups≤tx⁻(s) (4.30)

wherex⁻(s) =max(−x(s),0).

We say that (y, k) isthe solution of the Skorohod problem.

Comparing with Definition 4.1 we see that if we put y(t) =Y_T−t−L_T−t=ξ+

Z T T−t

f(s, Y_s)ds−(M_T −M_T−t)−L_T−t+K_T −K_T−t, (4.31) x(t) =ξ+

Z T T−t

f(s, Ys)ds−(M_T −M_T−t)−L_T−t, (4.32)

k(t) =K_T−t−K_T, (4.33)

then (y, k) solves the Skorohod problem described in Definition 4.1. By (4.30) we conclude thatK_t is given by

K_T−t−K_T

= maxs≤t(ξ+ Z T

T−s

f(u, Yu)du−(M_T −M_T−s)−L_T−s)⁻;t∈[0, T] (4.34) Since the unique solution K_t of the RBSDE (4.1) is in particular a solution of the corresponding Skorohod problem and this solution is unique and given by (4.34), we can conclude that (4.34) definesKtas anH-adapted process. This completes the proof of c) and hence the proof of Theorem 4.2.

5 Application to finance

Suppose we have a cash flowX_t=X^(λ)(t) given by dXt = Xt−

(µt−λt)dt+σtd⁻Bt

+ Z

R0

θ(t, z) ˜N(d⁻t, dz)

;X0>0 (5.1)

whereµt, σtandθ(t, z) are givenH_t-predictable processes,θ >−1, andd⁻Bt, ˜N(d⁻t, dz) indicates that we use a forward integral interpretation. See e.g. [DMØP] or the monograph [DØP] for a motivation for the use of the forward integral in this context of insider trading. Here c(t) :=λ_tX_t is the consumption rate, λt being our relative consumption rate. We assume that we are given a family A_H of admissible controls λ_t ≥ 0 included in the set of H_t-predictable processes, where H_t⊇ F_t is a given filtration, such that the solutionX_t of (5.1) exists and is given by

Xt = xexp Z t

0

{µ_s−λs−1 2σ_s² +

Z

R0

[log(1 +θ(s, z))−θ(s, z)]ν(dz)}ds+ Z t

0

σ_sd⁻B_s

+ Z t

0

Z

R0

log(1 +θ(s, z)) ˜N(d⁻s, dz)

(5.2) LetU₁, U₂ be given utility functions. Consider the problem to find Φ andλ^∗ ∈ A_H such that

Φ = sup

λ∈A_H

J(λ) =J(λ^∗), (5.3)

where

J(λ) =E[

Z T 0

e^−ρsU1(λsXs)ds+e^−ρTU2(XT)];

whereT >0, ρ >0 are given constants.

To study this problem we use a perturbation argument:

Suppose λis optimal. Chooseβ ∈ A_H,δ >0, and consider g(y) :=J(λ+yβ) for y∈(−δ, δ) Since λis optimal we haveg⁰(0) = 0. Hence

0 = d

dyE Z _T

0

e^−ρsU₁ (λ_s+yβ_s)X_s^(λ+yβ) ds +e^−ρTU₂(X_T^(λ+yβ))

y=0

= E Z T

0

U₁⁰ (λs+yβs)X_s^(λ+yβ) e^−ρs {β_sX_s^(λ+yβ)+ (λ_s+yβ_s) d

dyX_s^(λ+yβ)}ds +e^−ρTU₂⁰(X_T^(λ+yβ)) d

dyX_T^(λ+yβ)

y=0 (5.4)

Now, by (5.2),

d

dyX_t^(λ+yβ)=X_t^(λ+yβ)

− Z _t

0

β_rdr

(5.5) Hence, (5.4) gives

E Z T

0

e^−ρsU₁⁰ λsX_s^(λ)

{β_sX_s^(λ)−λsX_s^(λ) Z s

0

βrdr }ds

−e^−ρTU₂⁰(X_T^(λ))X_T^(λ) Z T

0

β_rdr] = 0 (5.6)

By the Fubini theorem,

Z T 0

h_s Z s

0

β_rdrds= Z T

0

( Z T

s

h_rdr)β_sds

Hence (5.6) can be written as E

Z T 0

{e^−ρsU₁⁰ λsX_s^(λ) X_s^(λ)−

Z T s

U₁⁰(λrX_r^(λ))λrX_r^(λ)e^−ρrdr

−e^−ρTU₂⁰(X_T^(λ))X_T^(λ)}β_sds] = 0 (5.7) Now apply this to

βs :=α(ω)χ_[t,t+h](s) (α H_t−measurable) for a fixedt∈[0, T). Then (5.7) becomes

E Z t+h

t

{e^−ρsU₁⁰ λ_sX_s^(λ) X_s^(λ)−

Z T s

U₁⁰(λ_rX_r^(λ))λ_rX_r^(λ)e^−ρrdr

−e^−ρTU₂⁰(X_T^(λ))X_T^(λ)}αds] = 0 (5.8) Differentiating w.r.t. h ath= 0 and using that (4.12) holds for all H_t -measurableα, we get

E

{e^−ρtU₁⁰ λ_tX_t^(λ) X_t^(λ)−

Z _T

t

U₁⁰(λ_rX_r^(λ))λ_rX_r^(λ)e^−ρrdr

−e^−ρTU₂⁰(X_T^(λ))X_T^(λ)}|H_t] = 0 (5.9) Define

Yt:=e^−ρtU₁⁰ λtX_t^(λ)

X_t^(λ) (5.10)

ξ:=e^−ρTU₂⁰(X_T^(λ))X_T^(λ) (5.11)

f(t, y, ω) =λty. (5.12)

Then (5.9) can be written

Y_t=E[ξ+ Z T

t

f(s, Y_s, ω)ds|H_t]; t∈[0, T]. (5.13) This is an equation of the type considered in Section 2. Hence we can apply the results of that section to study (5.13).

By Theorem 2.2 the solution of (5.13) is Y_t=E

ξexp(

Z T t

λ_sds)|H_t

= E

e^−ρTU₂⁰(X_T^(λ))X_T^(λ)exp(

Z T t

λsds)|H_t ,

which gives

exp(−ρt+ Z t

0

λsds)U₁⁰(λtX_t^(λ))X_t^(λ)

= E

exp(−ρT + Z T

0

λsds)U₂⁰(X_T^(λ))X_T^(λ)|H_t

;t∈[0, T].

Note that

exp(

Z _t

0

λ_sds)X_t^(λ)=X_t⁽⁰⁾,

where X_t⁽⁰⁾ is the solution of (5.1) when there is no consumption (λ= 0). Therefore, if we write Z_t=X_t⁽⁰⁾ we have the following:

Theorem 5.1 The relative consumption rate λ is optimal for problem (4.3) if and only if the following holds:

exp(−ρt)U₁⁰(λtX_t^(λ))Zt=E

exp(−ρT)U₂⁰(X_T^(λ))ZT|H_t

;t∈[0, T]. (5.14) Equation (5.14) gives a relation between the optimal consumption rate

c_t=λ_tX_t^(λ)

and the corresponding optimal terminal wealth X_T^(λ). In some cases this can be used to find both.

To see this, note that by (5.14) we get

U₁⁰(ct) =exp(ρ(t−T))E

U₂⁰(X_T^(λ))ZT

Zt

|H_t or

c_t=I₁(exp(ρ(t−T))E

U₂⁰(X_T^(λ))Z_T Z_t|H_t

), (5.15)

whereI1= (U₁⁰)⁻¹, the inverse ofU₁⁰. Substituting (5.15) into the equation (5.1) we get dX_t^(λ)=X_t−^(λ)

µ_tdt+σ_td⁻B_t+ Z

R0

θ(t, z) ˜N(d⁻t, dz)

−c_tdt. (5.16)

The solution of this equation is

X_t^(λ) =X₀G_t− Z _t

0

Gt

G_sc_sds, (5.17)