We were about to prove existence of solutions to the SDE.
Proof. Existence. Just as in the case with ODE, we use Picard iteration to get a solution.
That is, we put
(0)
Xt = x0 for all t, ω,
and then set
(n+1)
Xt
Z
= x0 +
t
µ(s, Xs(n) )
Z
Lemma 0.1. If
σ(s, Xs(n) ) dBs .
0
0
(n)
(Xt )
t
ds +
2
is L -bounded on [0, T ], measurable, and adapted, then
(n)
(n)
σ(t, Xt ) ∈ H2 [0, T ] and µ(t, Xt ) ∈ L2 (Leb × P).
(n+1)
Moreover, (Xt
) is L2 -bounded on [0, T ], continuous, measurable, and adapted.
(n)
(n)
(n)
Proof. Since (Xt ) is measurable and adapted, so are µ(s, Xs ) and σ(s, Xs ) (these are
(n)
(n)
just measurable functions of s, Xs ). Put B = supt∈[0,T ] E(Xt )2 . Then using |σ(t, x)|2 ≤
K(1 + x2 ), we obtain
Z
t
(n)
|σ(t, Xt )|2 dt ≤ T K(1 + B).
E
0
(n)
This means σ(t, Xt ) ∈ H2 [0, T ]. Isometry implies
2
Z t
(n)
E
σ(s, Xs ) dBs ≤ T K(1 + B).
0
Since also |µ(t, x)|2 ≤ K(1 + x2 ), Cauchy-Schwarz gives for t ∈ [0, T ],
2
Z t
Z t
Z t
(n) 2
(n)
|µ(s, Xs )| ds
1[0,t] ds
µ(s, Xs ) ds ≤ E
E
0
0
0
Z T
≤ T KE
(1 + (Xs(n) )2 ) ds
0
2
≤ T K(1 + B).
(n)
(n+1)
This means µ(t, Xt ) ∈ L2 (Leb × P). Since both integrals in the definition of Xt
are
(n+1)
2
L -bounded, so is (Xt
). Since it is a sum of integrals, we have shown before that this
(n+1)
implies Xt
is adapted and continuous (so measurable).
(n+1)
For the rest of the proof, we have to show that (Xt
) a.s. converges in C[0, T ] and
that the limit satisfies the SDE. This is broken into steps.
Step 1. Preliminary bounds on the difference of X (n) ’s.
Lemma 0.2. There exists C such that for all n and t,
Z t
(n+1)
(n) 2
E sup |Xs
− Xs | ≤ C
E(Xs(n) − Xs(n−1) )2 ds.
0≤s≤t
0
1
(n)
Proof. Set Xt
(n+1)
− Xt
= Dt + Mt , where
Z t
(µ(s, Xs(n) ) − µ(s, Xs(n−1) )) ds
Dt =
0
and
t
Z
(σ(s, Xs(n) ) − σ(s, Xs(n−1) )) dBs .
Mt =
0
So the left side of the lemma is bounded by
2E sup Ms2 + 2E sup Ds2 .
0≤s≤t
0≤s≤t
The second term is bounded using Cauchy-Schwarz as
Z s
2
Ds ≤ s
(µ(u, Xu(n) ) − µ(u, Xu(n−1) ))2 du,
0
so by the Lipschitz condition,
2E sup
0≤s≤t
t
Z
Ds2
(µ(s, Xs(n) ) − µ(s, Xs(n−1) ))2 ds
≤ 2T E
0
t
Z
E(Xs(n) − Xs(n−1) )2 ds.
≤ 2T K
0
For the other terms, since Mt is a martingale, Doob along with isometry and the Lipschitz
condition gives
Z t
2
2
2E sup Ms ≤ 8EMt = 8E (σ(s, Xs(n) ) − σ(s, Xs(n−1) ))2 ds
0≤s≤t
0
Z t
E(Xs(n) − Xs(n−1) )2 ds.
≤ 8K
0
Summing these two bounds gives the result with C = 2T K + 8K.
Step 2. Convergence. Putting
gn (t) = E sup (Xs(n+1) − Xs(n) )2 ,
0≤s≤t
the last step implies
Z
gn (t) ≤ C
t
(Xs(n)
−
Xs(n−1) )2
Z
ds ≤ C
0
t
gn−1 (s) ds.
0
We now want to iterate as we did in the proof of uniqueness. First note that
0 ≤ g0 (t) ≤ 2x20 + 2E sup (Xs(1) )2 .
0≤s≤t
2
By following a similar proof to that in the previous step (and you will on HW), we can show
that the second term is bounded uniformly for t ∈ [0, T ]. Thus, there exists M such that
0 ≤ g0 (t) ≤ M for all t ∈ [0, T ].
Returning to the main inequality, we obtain
Z t
M ds = CtM,
g1 (t) ≤ C
0
and iterating this, we obtain
0 ≤ gn (t) ≤ M C n tn /n!.
By Markov,
P
sup
0≤s≤t
|Xs(n+1)
−
Xs(n) |
−n
≥2
≤ 2n gn (t) ≤ M
C n T 2n
.
n!
This is summable in n, so just as in our proof of the continuity of the Ito integral, Borel(n)
Cantelli tells us that a.s. the continuous functions Xt converge in C[0, T ] to a continuous
(n)
function Xt . Because each Xt is adapted and measurable, and this uniform convergence
implies a.s. convergence for each fixed t, also the limit Xt is adapted and measurable.
Step 3. L2 convergence. By definition of gn ,
1/2
(n+1)
(n)
(n+1)
(n)
kXt
− Xt k2 = E(Xt
− Xt )2
≤ gn (T )1/2 .
Since (gn (T )1/2 ) is summable in n, the sequence is Cauchy, and by uniqueness of the limit, we
(n)
see that for fixed t, Xt → Xt in L2 , and (Xt ) is also L2 -bounded. For the L2 boundedness,
we even get
∞
X
1/2
kXt k2 ≤ |x0 | +
gn (T )1/2 for t ∈ [0, T ].
n=0
(n)
Now that Xt → Xt in L2 for fixed t, we must show the same statement for the coefficients
of the SDE. First, from isometry,
Z t
Z t
σ(s, Xs(n) ) dBs −
= σ(s, Xs(n) ) − σ(s, Xs ) 2
σ(s,
X
)
dB
,
s
s
L (Leb×P)
0
0
2
and the right side squared is
Z T
Z
(n)
2
E(σ(s, Xs ) − σ(s, Xs )) ds ≤ K
0
T
E(Xs(n) − Xs )2 ds
0
The integrand is (using L2 -convergence)
(n)
Xs − Xs 2 ≤
2
∞
X
!2
kXs(k) − Xs(k+1) k2
k=n
≤
∞
X
k=n
3
!2
gk (t)1/2
.
(n)
This converges to 0 uniformly in s, so kσ(s, Xs ) − σ(s, Xs )kL2 (Leb×P) → 0, and we find
Z
t
Z
σ(s, Xs(n) )
dBs →
t
σ(s, Xs ) dBs in L2 .
(1)
0
0
A similar argument works for the µ terms. Indeed,
2
Z t
Z t
2
Z t
(n)
µ(s, Xs(n) ) ds −
µ(s, Xs ) ds = E
(µ(s, Xs ) − µ(s, Xs )) ds
0
0
0
2
Z t
E(µ(s, Xs(n) ) − µ(s, Xs ))2 ds
≤T
0
Z t
E(Xs(n) − Xs )2 ds,
≤ KT
0
and as we saw above, this converges to 0. Therefore
Z t
Z t
(n)
µ(s, Xs ) ds →
µ(s, Xs ) ds in L2 .
0
(2)
0
Step 4. Solution to the SDE. Last, we need to check that (Xt ) solves the SDE. We have
already chosen (Xt ) so that
(n)
→ Xt uniformly on [0, T ].
a.s., Xt
By the L2 convergence of the last step, a.s., one can find a subsequence (nk ) such that as
k → ∞,
Z t
Z t
(nk )
µ(s, Xs ) ds →
µ(s, Xs ) ds a.s. for all t ∈ [0, T ] ∩ Q
0
and
Z
0
t
σ(s, Xs(nk ) )
Z
0
t
σ(s, Xs ) ds a.s. for all t ∈ [0, T ] ∩ Q.
ds →
0
But since a.s. for all k and t ∈ [0, T ] ∩ Q,
Z t
Z t
(nk +1)
(nk )
Xt
=
µ(s, Xs ) ds +
σ(s, Xs(nk ) ) dBs ,
0
0
one has a.s. for all t ∈ [0, T ] ∩ Q,
Z t
Z t
Xt =
µ(s, Xs ) ds +
σ(s, Xs ) dBs .
0
0
Both sides are continuous, so this holds a.s. for all t ∈ [0, T ].
4
Some concluding remarks in the chapter: SDEs can be formulated in general dimensions.
Steele gives the following setup:
dXt = µ(t, Xt ) dt + σ(t, Xt ) dBt , X0 = x0 ∈ Rd .
Here, µ is a vector-valued function


µ1 (t, Xt )
,
···
µ(t, Xt ) = 
µd (t, Xt )
dBt is a vector valued differential
(1) 
dBt
dBt =  · · ·  ,
(d)
dBt

and σ is a matrix-valued function

σ11 (t, Xt ) · · ·
···
σ(t, Xt ) = 
σd1 (t, Xt ) · · ·
σ1d (t, Xt )

.
σdd (t, Xt )
There are similar existence and uniqueness statements. For instance, the process
(1)
(2)
Xt = (Xt , Xt ) = (cos(Bt ), sin(Bt ))
is called BM on the circle, and satisfies the SDE
!
!
(1)
1 Xt(1)
dXt
dt +
=
(2)
2 Xt(2)
dXt
(2)
−Xt
(1)
Xt
0
0
!
(1)
dBt
(2)
dBt
!
.
The fact that the diffusion matrix is not of full rank represents that BM on the circle lives
essentially on a one-dimensional set, the circle.
5