First, we consider the following BSDE:
dYt= h(t, Yt, Z1t, · · · , Zdt)dt +
d
P
i=1
ZitdWti, 0 ≤ t ≤ T YT = ξ,
(80)
where ξ ∈ L2F
T(Ω, Rm) and h ∈ L2F(0, T ; W1,∞(Rm × (Rm)d; Rm)) i.e., h : [0, T ] × Rm × (Rm)d×Ω → Rm, such that (t, ω) 7→ h(t, y, z1, · · · , zd; ω) is {Ft}t≥0-progressively measur-able ∀(y, z1, · · · , zd) ∈ Rm× (Rm)d with h(t, 0, 0, · · · , 0; ω) ∈ L2F(0, T ; Rm) and ∃L > 0,
|h(y, z1, · · · , zd) − h(y, z1, · · · , zd)| ≤ L |y − y| +
d
X
i=1
|zi− zi|
! ,
∀y, z1, · · · , zd, y, z1, · · · , zd∈ Rm, a.e. t ∈ [0, T ], a.s. (81) Denote
N [0, T ] , L2F(Ω; C(0, T ; Rm)) × [L2F(0, T ; Rm)]d (82) and
k(Y, Z1, · · · , Zd)kN [0,T ] , E
"
sup
0≤t≤T
|Yt|2+
d
X
i=1
Z T 0
|Zt|2dt
#!12
(83) Then, N [0, T ] is a Banach space under norm (83).
In this chapter, we present another method. It will give a sufficient condition for the unique solvability of (3). We will obtain a Riccati type equation and a BSDE associated with (3). Let us now carry heuristic derivation.
Suppose (X, Y, Z1, · · · , Zd) ∈ M[0, T ] is an adapted solution of (3). We assume that X and Y are related by
Yt= P (t)Xt+ pt, ∀t ∈ [0, T ], a.s. (84) where P : [0, T ] → Rm×nis a deterministic matrix-valued function and p : [0, T ]×Ω → Rm is an {Ft}t≥0-adapted process. We are going to derive the equations for P and p. First of all, from (8) and the terminal condition in (3), we have
g = P (T )XT + pT. (85)
Let us impose
P (T ) = O, pT = g. (86)
Since g ∈ L2FT(Ω; Rm) and p is required to be {Ft}t≥0-adapted, we should assume that p satisfies a BSDE:
dpt = αtdt +
d
P
i=1
qitdWti, 0 ≤ t ≤ T pT = g
(87)
with α, q1, · · · , qd ∈ L2F(0, T ; Rm) being undetermined. Next, by Itbo Formula, we have dYt = ˙P (t)Xtdt + P (t)dXt+ dpt
=[ ˙P (t)Xt+ P (t)(AXt+ BYt) + αt]dt +
d
X
i=1
[P (t)(Ai1Xt+ B1iYt+ C1iZit) + qit]dWti
={[ ˙P (t) + P (t)A + P (t)BP (t)]Xt+ P (t)Bpt+ αt}dt +
d
X
i=1
{[P (t)Ai1+ P (t)B1iP (t)]Xt+ P (t)C1iZit+ P (t)B1ipt+ qit}dWti (88) Now compare (88) with the second equation in (3) (note (84)), we obtain that (drift coefficient)
[ ˙P (t) + P (t)A + P (t)BP (t)]Xt+ P (t)Bpt+ αt = [ bA + bBP (t)]Xt+ bBpt, (89) and (diffusion coefficient)
[P (t)Ai1+ P (t)Bi1P (t)]Xt+ P (t)C1iZit+ P (t)B1ipt+ qit= Zit, i = 1, · · · , d (90) By assuming I − P (t)C1i to be invertible ∀t ∈ [0, T ], i = 1, · · · , d, we have from (90) that Zit= [I − P (t)C1i]−1{[(P (t)Ai1+ P (t)B1iP (t)]Xt+ P (t)B1ipt+ qit}, i = 1, · · · , d (91) Then, (89) can be written as
0 = [ ˙P (t) + P (t)A + P (t)BP (t) − bA − bBP (t)]Xt+ [P (t)B − bB]pt+ αt. (92) Now, we introduce the following differential equation for Mm×n(R)-valued function P :
P (t) + P (t)A + P (t)BP (t) − b˙ A − bBP (t) = O, 0 ≤ t ≤ T
We refer to (93) as a Riccati type equation. Suppose (93) admits a solution P over [0, T ] such that
[I − P (t)C1i]−1 is bounded , i = 1, · · · , d, ∀t ∈ [0, T ]. (94) Then, (92) gives
αt = [ bB − P (t)B]pt
Combining this with (87), we see that one should introduce the following BSDE:
dpt = [ bB − P (t)B]ptdt +
d
P
i=1
qitdWti, 0 ≤ t ≤ T pT = g.
(95)
When (93) admits solution P such that (94) holds, BSDE (95) admits a unique solution (p, q1, · · · , qd) ∈ N [0, T ]. In the form provided here, a proof can be found in several sources, for instance Ma & Yong (2000), pp. 15-16. From (84) and (91), the forward equation (X):
dXt= {[A + BP (t)]Xt+ Bpt}dt +
d
P
i=1
{Ai1+ B1iP (t) + C1i[I − P (t)C1i]−1[P (t)Ai1+ P (t)B1iP (t)]}Xt +B1ipt+ C1i[I − P (t)C1i]−1[P (t)B1ipt+ qit]
dWti, 0 ≤ t ≤ T X0 = 0
Then we can define the following:
A(t) = A + BP (t),e
Aei1(t) = Ai1+ B1iP (t) + C1i[I − P (t)C1i]−1[P (t)Ai1+ P (t)B1iP (t)], i = 1, · · · , d bet = Bpt,
eσti = B1ipt+ C1i[I − P (t)C1i]−1[P (t)B1ipt+ qit], i = 1, · · · , d.
(96)
It is clear that eA and eAi1 are time-dependent matrix-valued function and eb and σei are {Ft}t≥0-adapted processes. Further, under (94), by the Existence and Uniqueness The-orem for Stochastic Differential Equations, the following SDE admits a unique (strong) solution:
dXt = [ eA(t)Xt+ ebt]dt +
d
P
i=1
[ eAi1(t)Xt+eσti]dWti, 0 ≤ t ≤ T, X0 = 0.
(97)
The following Theorem gives a representation of the adapted solution of FBSDE (3).
Theorem 4. Let (93) admits a solution P such that (94) holds. Then FBSDE (3) admits a unique solution (X, Y, Z1, · · · , Zd) ∈ M[0, T ] which is determined by (97), (84) and (91).
Proof. First of all, a direct computation (from above) shows that the process (X, Y, Z1, · · · , Zd) determined by (97), (84) and (91) is an adapted solution of (3). We now prove the unique-ness. Let (X, Y, Z1, · · · , Zd) ∈ M[0, T ] be any adapted solution of (3). Set
Yt = P (t)Xt+ pt,
Zit= [I − P (t)C1i]−1{[P (t)Ai1+ P (t)B1iP (t)]Xt+ P (t)B1ipt+ qit}, i = 1, · · · , d
(98)
where P and (p, q1, · · · , qd) are (adapted) solutions of (93) and (95), respectively. Denote Y = Y − Y and bb Zi = Zi− Zi. By the Itbo Formula,
dYt= ˙P (t)Xtdt + P (t)dXt+ dpt
= [ bA + bBP (t) − P (t)A − P (t)BP (t)]Xtdt + P (t)
"
(AXt+ BYt)dt +
d
X
i=1
(Ai1Xt+ B1iYt+ C1iZit)dWti
#
+ [ bB − P (t)B]ptdt +
d
X
i=1
qitdWti
= [ bAXt+ P (t)B(Yt− Yt) + bBYt]dt +
d
X
i=1
{P (t)B1i(Yt− Yt) + P (t)B1i[P (t)Xt+ pt] + P (t)(Ai1Xt+ C1iZit) + qit}dWti
= [ bAXt+ P (t)B(Yt− Yt) + bBYt]dt +
d
X
i=1
{P (t)B1i(Yt− Yt) + P (t)C1iZit+ [I − P (t)C1i]Zit}dWti
= [ bAXt+ P (t)B(Yt− Yt) + bBYt]dt +
d
X
i=1
{P (t)B1i(Yt− Yt) + Zit+ P (t)C1i(Zit− Zit)}dWti
Then a direct computation shows that (compare to (3))
d bYt= [ bB − P (t)B] bYtdt +
d
P
i=1
{[I − P (t)C1i]bZit− P (t)B1iYbt}dWti, YbT = 0.
(99)
By (94), We may set
to get the following equivalent BSDE (of (99)):
d bYt= [ bB − P (t)B] bYtdt +
d
P
i=1
ZeitdWti, YbT = 0.
(101)
It is clear that such a BSDE admits a unique adapted solution ( bY, eZ1, · · · , eZd) ≡ (0, 0, · · · , 0) a.s. (see Ma & Yong (2000), pp. 15-16). Consequently, bZi ≡ 0 a.s., i = 1, · · · , d (since Zbit= [I − P (t)C1i]−1[eZit+ P (t)B1iYbt], i = 1, · · · , d). Hence, by (98), we obtain
Yt= P (t)Xt+ pt,
Zit = [I − P (t)C1i]−1{[P (t)Ai1+ P (t)B1iP (t)]Xt+ P (t)B1ipt+ qit}, i = 1, · · · , d.
(102) This means that any adapted solution (X, Y, Z1, · · · , Zd) of (3) must satisfy (102). Then, similar to the heuristic derivation above, we have that X has to be the solution of (97).
Hence, we obtain the uniqueness.
The following result tells us something more.
Proposition 2. Let (93) admits a solution P such that (94) holds for t ∈ [T0, T ] (with some T0 ≥ 0). Then, ∀ eT ∈ [0, T − T0], linear FBSDE (3) is uniquely solvable on [0, eT ].
Proof. Let
P (t) = P (t + T − ee T ), 0 ≤ t ≤ eT . (103) Then eP satisfies (93) with [0, T ] replaced by [0, eT ] and
[I − eP (t)C1i]−1 is bounded for 0 ≤ t ≤ eT , i = 1, · · · , d (104)
Thus, Theorem 4.1 applies.
The above proposition tells that if (93) admits a solution P satisfying (94), FBSDE (3) is uniquely solvable over any [0, eT ] (with eT ≤ T ). Then in this case, by Theorem 1, the solvability (3) of FBSDE over [0, eT ] admits a solution ∀g ∈ L2F
Te(Ω; Rm), of which a necessary condition is
det
[ O I ]eAt
O
I
> 0, 0 ≤ t ≤ T. (105)
Therefore, by Theorem 3, compare (105) and (29), we see that the solvability of Riccati type equation (93) is only sufficient condition for the solvability of (3).
We have seen that (105) is necessary condition for (93) having a solution P satisfying (94). The following result gives the inverse of this.
Theorem 5. Let (105) hold. Then (93) admits a unique solution P which has the fol-lowing representation: Moreover, it holds
I−P (t)C1i = Consequently, if in addition to (105), (30) holds, then (94) holds and the linear FBSDE (3) is uniquely solvable with the representation given by (97), (84) and (91).
Proof. We can easily check that (106) is a solution of (93). You can find in Ma & Yong (2000), pp. 49-50.
Uniqueness is obvious since (93) is a terminal value problem with the right hand side of the equation being locally Lipschitz.
I − P (t)C1i = I +
Finally, an easy calculation shows (by (105), (30))
det(I − P (t)C1i) =
det
[ O I ]eA(T −t)C1i
det
[ O I ]eA(T −t)
O
I
> 0, ∀t ∈ [0, T ], i = 1, · · · , d,
and I − P (t)C1i is a continuous function on [0, T ], ∀i = 1, · · · , d, hence (94) holds. Then we complete proof.
5. Conclusion
This study proposed some extension of Ma & Yong (2000). We give the sufficient and necessary conditions of the linear FBSDE (3), and modify their work to prove the closeness of R(K) (with bA = O). Then we find the connection between a Riccati type equation and the linear FBSDE.
There are at least three more possible extensions of our method for future research.
First, we can add nonzero Zis term in drift coefficient and derive the sufficient and nec-essary conditions. Second, one can prove or give a counterexample of the closeness in general case (not just special case in this study). Third, someone can consider some spe-cial nonlinear cases (like quadratic form) and discuss the solvability of the FBSDE. This work may be can apply in the mathematical finance like mean-variance portfolio selection and consider in optimal control and LQ problem (like Zhou and Li (2000)) in the future.
Bibliography
Bjork, T. (2009). Arbitrage theory in continuous time (3rd ed.). New York: Oxford.
Karatzas, I., & Shreve, S., E. (1998). Brownian motion and stochastic calculus (2nd ed.).
New York: Springer.
Karoui, N. E., Peng, S., & Quenez, M. C. (1997). Backward stochastic differential equa-tions in finance. Mathematical Finance, Vol. 7., No. 1, 1-71.
Lax, P. D., (2002). Functional Analysis. Wiley.
Ma, J., & Yong, J. (2000). Forward-backward stochastic equations and their applications.
Springer.
Oksendal, B. (2003). Stochastic differential equations (6th ed.) Springer.
Pham, H. (2000). Continuous-time stochastic control and optimization with Financial Application. Springer.
Zhou, X. Y., & Li, D. (2000). Continuous-time mean-variance portfolio selection: a stochastic LQ framework, Applied Mathematics Optimization, 19-33.
Appendix
Lemma 6. Consider the following SDE:
dXt= (AtXt+ BtYt)dt +
d
P
i=1
(Ai1,tXt+ B1,ti Yt+ C1,ti Zit)dWti, dYt= bBtYtdt +
d
P
i=1
ZitdWti, X0 = 0, Y0 = 0,
0 ≤ t ≤ T,
where
A, Ai1 : [0, T ] × Ω → Mn(R) B, B1i, C1i : [0, T ] × Ω → Mn×m(R) B : [0, T ] × Ω → Mb m(R)
all the processes are {Ft}t≥0-progressively measurable, ∃M ≥ 0 with kAtk, kAi1,tk, kBtk, kB1,ti k, kC1,ti k ≤ M , i = 1 · · · , d, a.e. t ∈ [0, T ], a.s. And ∃bb > 0 such that k bBtk > bb a.e. t ∈ [0, T ], a.s.
Then there exists c > 0 (independent of (Z1, · · · , Zd)), ∀(Z1, · · · , Zd) ∈ H such that E[|XT|2] ≤ cE[|YT|2].
Proof. By the Itbo Formula,
d|Yt|2 = 2Yt· dYt+ dhYit
= 2Yt· bBtYt+
d
X
i=1
|Zit|2
! dt + 2
d
X
i=1
Yt· ZitdWti, hence,
E[|Yt|2] = E
"
Z t 0
2Ys· bBsYs+
d
X
i=1
|Zis|2ds
#
≥ 2bbE
Z t 0
|Ys|2ds
+
d
X
i=1
E
Z t 0
|Zis|2ds
. (108)
Similarly, by the Itbo Formula,
d|Xt|2 = 2Xt· dXt+ dhXit
=
"
2Xt· (AtXt+ BtYt) +
d
X
i=1
|Ai1,tXt+ B1,ti Yt+ C1,ti Zit|2
# dt
+ 2
d
XX · (Ai X + Bi Y + Ci Zi)dWi,
taking expectation we get and use the inequality αa2+ 1
By Gronwall Inequality and (108), E[|Xt|2] ≤ eM (2+αM +3M d)t
Take α sufficient small, we obtain E[|Xt|2] ≤ eM (2+αM +3M d)t 1
, we complete the argument.
The following case is that we want to show in the Chapter 3.
Lemma 7. Consider the following SDE:
the matrices are defined by (4). Then there exists C > 0 (independent of (Z1, · · · , Zd)),
∀(Z1, · · · , Zd) ∈ H such that
E[|XT|2] ≤ CE[|YT|2]. (109) Proof. We use the General Itbo Formula in e− bBtYt,
d(e− bBtYt) = − bBe− bBtYtdt + e− bBtdYt = e− bBt
d
X
i=1
ZitdWti.
Now, we consider ete− bBtYt and use the Itbo Formula again, d(ete− bBtYt) = ete− bBtYtdt + etd(e− bBtYt)
= ete− bBtYtdt + ete− bBt
d
X
i=1
ZitdWti
Let bYt= ete− bBtYt, and bZit= ete− bBtZit, we derive the SDE:
dXt= (AXt+ Be−teBtbYbt)dt +
d
P
i=1
(Ai1Xt+ B1ie−teBtb Ybt+ C1ie−teBtb Zbit)dWti, d bYt= bYtdt +
d
P
i=1
ZbitdWti, X0 = 0, bY0 = 0,
0 ≤ t ≤ T,
by Lemma 6, ∃c > 0 such that
E[|XT|2] ≤ cE[| bYT|2] ≤ ce2Te2k bBkTE[|YT|2] Take C = ce2Te2k bBkT, we prove (109).