Uniqueness of solutions of stochastic differential equations

Introduction

In this paper we consider the stochastic differential equation

It follows from a theorem of Veretennikov that (1) has a unique strong solution, i.e. there is a unique process $x(t)$, adapted to the filtration of the Brownian motion, satisfying (1). Veretennikov in fact proved this for a more general equation. Here we consider a different question, posed by N. V. Krylov : we choose a Brownian path $W$ and ask whether (1) has a unique solution for that particular path. The main result of this paper is the following affirmative answer:

This theorem can also be regarded as a uniqueness theorem for a random ODE: writing $x(t)=W(t)+u(t)$, the theorem states that for almost all choices of $W$, the differential equation $\frac{du}{dt}=f(t,W(t)+u(t))$ with $u(0)=0$ has a unique solution.

In Section 4, we give an application of this theorem to convergence of numerical approximations to (1). Idea of proof of theorem. The theorem is trivial when $f$ is Lipschitz in $x$, and the idea of the proof is essentially to find some substitute for a Lipschitz condition. The proof splits into two parts, the first (section 2) being the derivation of an estimate which acts as a substitute for the Lipschitz condition, and the second (section 3) being the application of this estimate to prove the theorem. We start with a reduction to a slightly simpler problem. A reduction. It will be convenient to suppose $|f(t,x)|\leq 1$ everywhere, which we can by scaling. Then it will suffice to prove uniqueness of a solution on , as we can then repeat to get uniqueness on and so on.

is a Brownian motion, i.e. $W$ has law $P_{W}$.

For a particular choice of $x$, and with $W$ defined by (2), $x$ will be the unique solution of (1) provided the only solution of

in $X$ is $u=0$. So, to prove the theorem it suffices to show that, for $\mu$-a.a. $x$, (3) has no non-trivial solution, since for such $x$, with $W$ defined by (2) no other $x$ can satisfy (2).

But $\mu$ is absolutely continuous w.r.t. $P_{W}$, so it suffices to show that, for $P_{W}$-a.a. $x$, (3) has no non-trivial solution. In other words, it suffices to show that, if $W$ is a Brownian motion then with probability 1 there is no non-trivial solution $u\in X$ of

We prove this in section 3. Remark. Our proof does not make use of the existence of a strong solution. It is tempting to try to prove the theorem by measure-theoretic arguments based on the strong solution and Girsanov’s theorem. Define $T:X\rightarrow X$ by

The strong solution gives a measurable map $S:E\rightarrow F$ where $E$ and $F$ are Borel subsets of $X$ with $P_{W}(E)=P_{W}(F)=1$, such that $T\circ S$ is the identity on $E$, and $F$ is the range of $S$. It follows that $T$ is (1-1) on $F$ and for any $W\in E$ there is a unique solution of (1) in $F$. But we need a solution which is unique in $X$ and to achieve this we need to show that $T(X\backslash F)$ is a $P_{W}$-null set, and this seems to be a significant obstacle.

Our proof is quite complicated and it seems reasonable to hope that it can be simplified. In particularly one might expect a simpler proof of Proposition 2.2. This seems to be nontrivial even for $p=2$. The bound for $p=2$ follows from the first part of Lemma 2.5 (with $t_{0}=0$ and $r=0$) and I do not know an essentially simpler proof.

In one dimension, in the case when $f(t,x)$ depends only on $x$, a different and shorter proof of Theorem 1.1 can be given, using local time, but it is not clear how to extend it to $d>1$.

The basic estimate

This section is devoted to the proof of the following:

where $C$ is an absolute constant, $|x|$ denotes the usual Euclidean norm and $W(t)$ is a standard $d$-dimensional Brownian motion with $W(0)=0$,

This will be deduced from the following one-dimensional version:

where $C$ is an absolute constant, and here $W(t)$ is one-dimensional Brownian motion with $W(0)=0$.

We start by observing that the LHS can be written as

and using the joint distribution of $W(t_{1}),\cdots,W(t_{p})$ this can be expressed as

where $E(t,z)=(2\pi t)^{-1/2}e^{-z^{2}/2t}$ and here $t_{0}=0$, $z_{0}=0$.

and we shall show that $J_{p}(0,0)\leq C^{p}/\Gamma(\frac{p}{2}+1)$; Proposition 2.2 will then follow since $p!\leq 2^{p}((p/2)!)^{2}$.

In order to estimate $J_{k}$ we use integration by parts to shift the derivatives to the exponential terms. We introduce some notation to handle the resulting terms - we define $B(t,z)=E^{\prime}(t,z)$ and $D(t,z)=E^{\prime\prime}(t,z)$ (where again primes denote differentiation w.r.t. the second variable).

If $S=S_{1}\cdots S_{k}$ is a word in the alphabet $\{E,B,D\}$ then we define

In fact, only certain words in $\{E,B,D\}$ will be required: we say a word is allowed if, when all $B$’s are removed from the word, a word of the form $(ED)^{r}=EDED\cdots ED$, $r\geq 0$, is left. The allowed words of length $k$ correspond to the subsets of $\{1,2,\cdots,k\}$ having an even number of members (namely the set of positions occupied by $E$ and $D$ in the word). Hence the number of allowed words of length $k$ is the number of such subsets of $\{1,2,\cdots,k\}$, namely $2^{k-1}$.

where each $S^{(j)}$ is an allowed word of length $k$ (in fact each allowed word of length $k$ appears exactly once in this sum, but we do not need this fact). The proof will then be completed by obtaining a bound for $I_{S}$.

We prove (5) by induction on $k$. So, assuming (5) for $J_{k}$, we have

We now proceed to the estimation of $I_{S}(t_{0},z_{0})$, when $S$ is an allowed string. We start with some preliminary lemmas.

Now if $|l-m|=k\geq 2$ then for $z\in[l,l+1)$ and $y\in[m,m+1)$ we have $|z-y|\geq k-1$ and then it follows easily that

and hence $I_{lm}\leq C_{2}e^{-l^{2}/8}e^{-(k-2)^{2}/4}$ from which we deduce

Now suppose $|l-m|\leq 1$. We use $\hat{\phi}_{l}(s,u)$ for the Fourier transform in the second variable, and similarly $\hat{h}_{m}$. We note that $\int\hat{\phi}_{l}(s,u)^{2}du=\int\phi_{l}(s,z)^{2}dz\leq C_{4}e^{-|l|^{2}/6}$ for $0\leq s\leq 1$ and similarly $\int\hat{h}_{m}(t,u)^{2}du\leq 1$. We have

Applying $ab\leq\frac{1}{2}(a^{2}c+b^{2}c^{-1}$ with $a=\hat{\phi}_{l}(s,u)$, $b=\hat{h}_{m}(t,-u)$ and $c=e^{l^{2}/12}$, we deduce that

In the first integral we integrate first w.r.t. $t$ and obtain the bound const.$e^{-l^{2}/12}$ for the integral. We get a similar bound for the second integral (integrating w.r.t. $s$ first), and hence

Summing over $l$ and $m$ such that $|l-m|\leq 1$, we obtain

These follow easily from Lemma (2.3), the second using the easily verified fact that $|B(s,z)|\leq Cs^{-1/2}(e^{-z^{2}/3s})$. ∎

Again, we let $C_{1},\cdots$ be absolute constants. By using the change of variables $t^{\prime}=(t-t_{0})/(1-t_{0})$, $s^{\prime}=(s-t_{0})/(1-t_{0})$, $y^{\prime}=y(1-t_{0})^{-1/2}$, it suffices to prove these estimates when $t_{0}=0$. To do this, we start by scaling the first part of Corollary 2.4, and get

for $k=0,1,2\cdots$ and then by summing over $k$, we get

and combining these bounds gives the first result. Similarly, by scaling the second part of Corollary 2.4, we get

for $k=0,1,2\cdots$ and then by summing over $k$, we get

We can now complete the proof of Proposition 2.2 by obtaining the required bound for $I_{S}(t_{0},z_{0})$. Again we use $C_{1},C_{2},\cdots$ for absolute constants. We shall show that, for a suitable choice of $M$, we have for any allowed string $S$ of length $k$

We shall prove (7) by induction on $k$, provided $M$ is chosen large enough. The case $k=0$ is immediate, so assume $k>0$ and that (7) holds for all allowed strings of length less than $k$. Then there are three cases: (1) $S=BS^{\prime}$ where $S^{\prime}$ has length $k-1$; (2) $S=EDS^{\prime}$ where $S^{\prime}$ has length $k-2$; (3) $S=EB^{m}DS^{\prime}$ where $m\geq 1$ and $S^{\prime}$ has length $k-m-2$. In each case $S^{\prime}$ is an allowed string. We consider the three cases separately. Case 1. In this case we have

where we have used the inductive hypothesis to bound $I_{S^{\prime}}$, and then the bound (6). (7) then follows if $M$ is large enough. Case 2. Now we have

We set $h(t,z)=g(t,z)I_{S^{\prime}}(t,z)(1-t)^{1-\frac{k}{2}}$ so that $\|h\|_{\infty}\leq M^{k-2}/\Gamma(k/2)$ by the inductive hypothesis, and then from the first part of Lemma 2.5 we deduce that

and (7) follows if $M$ is large enough. Case 3. In this case have

Now let $h(t,z)=g(t,z)I_{S^{\prime}}(t,z)(1-t_{m+2})^{(2+m-k)/2}$, so that by the inductive hypothesis on $S^{\prime}$ we have $\|h\|_{\infty}\leq M^{k-m-2}/\Gamma(\frac{k-m}{2})$. Then, writing

from which again (7) follows, provided $M$ is large enough. Putting (7) with $t_{0}=0$, $z_{0}=0$ and $k=p$ in (5) completes the proof of Proposition 2.2. ∎

and then the required result follows by averaging over $W_{2},\cdots,W_{d}$.

What we in fact need is a scaled version of Proposition 2.1 for subintervals of . For $s\geq 0$ we denote by ${\mathcal{F}}_{s}$ the $\sigma$-field generated by $\{W(\tau):0<\tau<s\}$. Then we can state the required result:

where $l=b-a$ and $C$ is the constant in Proposition 2.1.

First assume $s=a=0$, $b=1$. Let $\alpha=(2C^{2}|x|^{2})^{-1}$. Then

Now, if $q=\frac{p}{p-1}$ then $\int E(t,z)^{q}dz=O(t^{-(q-1)d/2})$ and $p>1+\frac{d}{2}$ implies $(q-1)d/2<1$, so the result follows from Hölder’s inequality. ∎

Proof of Theorem

We now apply Corollary 2.6 and Lemma 2.7 to the proof of the theorem. First we give a brief sketch of the proof. Outline of proof. The proof is motivated by the elementary case when $f$ is Lipschitz in the second variable. In this case, if $I=[a,b]$ is a subinterval of and $u$ is a solution of (4) satisfying

and $\beta=|u(a)|$, then we deduce from (9) that $|u(t)|\leq\alpha^{\prime}=\beta+L|I|\alpha$ for $t\in I$, where $L$ is the Lipschitz constant, i.e. (9) holds with $\alpha$ replaced by $\alpha^{\prime}$. If $L|I|<1$ it follows that (9) holds with $\alpha=(1-L|I|)^{-1}\beta$, and of course if $\beta=0$ this gives $u=0$ on $I$.

We try to copy this argument using Corollary 2.6 as a substitute for a Lipschitz condition. There are two difficulties: first, Corollary 2.6 is a statement about probabilities and we need an ‘almost sure’ version, and in doing so we lose something; second, in Corollary 2.6, $x$ is a constant, whereas we are dealing with a function $u$ depending on $t$. The way round the second problem is to approximate $u$ by a sequence of step functions $u_{l}$ and then use

where $u_{n}$ is constant on the interval $I$, and then to apply the ‘almost sure’ form of the proposition to each interval of constancy of the terms on the right. Again, we lose something in doing this, but, as it turns out, we still have good enough estimates to prove the theorem. In fact, we need two versions of the ‘almost sure’ (nearly) Lipschitz condition, the first to estimate $\int\{f(W(t)+u_{n}(t))-f(W(t))\}dt$ and the second to estimate $\int\{f(W(t)+u_{l+1}(t))-f(W(t)+u_{l}(t))\}dt$. We also need a third estimate, for sums of integrals of the second type.

The two versions of the ‘almost sure’ nearly-Lipschitz condition are conditions (11) and (12) below, and the third estimate is (20). In Lemmas 3.1, 3.2, 3.5 and 3.6 it is shown that these conditions indeed hold almost surely. Lemmas 3.3 and 3.4 establish a technical condition (15) needed to justify the passage to the limit as $l\rightarrow\infty$ (which is not trivial when $f$ is not continuous). With these preliminaries the above programme is carried out in Lemma 3.7. The analogue of (9) above is (25). We no longer immediately get $\alpha=0$ when $\beta=0$, but we get a good enough bound to prove the uniqueness of the solution to (1), for any $W$ satisfying (11,12,15,20).

for all dyadic $x,y\in Q$ and all choices of integers $n,k$ with $n>0$ and $0\leq k\leq 2^{n}-1$.

and by summing over all possible choices of $n,k,m,x,y$ we find that the probability that

for some choice of $I_{nk}$ and dyadic neighbours $x,y\in Q$ is not more than $\sum_{n=1}^{\infty}\sum_{m=0}^{\infty}2^{n}3^{d}2^{d(m+3)}C_{1}e^{-C_{2}\lambda^{2}(1+m+n)}$ which approaches 0 as $\lambda\rightarrow\infty$.

It follows that, given $\epsilon>0$, we can find $\lambda(\epsilon)$ such that, with probability $>1-\epsilon$, we have

for all choices of $n,k$ and dyadic neighbours in $Q$.

(note that the sums are actually finite, since $x,y$ are dyadic, so that $x=x_{r}$ and $y=y_{r}$ for large $r$). Then applying the above bounds for the case of dyadic neighbours to each term, we get the desired result. ∎

Next we prove a similar estimate for $\sigma_{nk}$, which is analogous to the Law of the Iterated Logarithm for Brownian motion.

The next two lemmas are used to justify the passage to the limit $l\rightarrow\infty$ in (10).

Let $\Phi$ denote the set of $Q$-valued functions $u$ on satisfying $|u(s)-u(t)|\leq|s-t|$, $s,t\in$, and let $\Phi_{n}$ denote the set of $Q$-valued functions on which are constant on each $I_{nk}$ and satisfy $|u(k2^{-n})-u(l2^{-n})|\leq|k-l|2^{-n}$. Then let $\Phi^{*}=\Phi\cup\cup_{n}\Phi_{n}$.

for all pairs of dyadic points $x,y$ in $Q$ and all choices of $n,k$. Then we choose $m$ such that $4K\sum_{n=m}^{\infty}n^{1/2}2^{-n/2}<\epsilon$. Let $\Omega$ be a finite set of dyadic points of $Q$ such that every $x\in Q$ is within distance $2^{-m}$ of some point of $\Omega$.

for each $k,x$. Then the probability that

Now let $u\in\Phi^{*}$. For each $n\geq m$ choose $u_{n}\in\Phi_{n}$ taking a constant dyadic value within $2^{-n}$ of $u(k2^{-n})$ on $I_{nk}$ for $k=0,1,\cdots,2^{n}-1$. Now if $A_{r}$ and $B_{r}$ hold then $\int_{0}^{1}\phi_{r}(t,W(t)+u_{m}(t))dt\leq\epsilon/2$ and

Note that Lemma 3.4 implies that $\rho_{nk}(x,y)$ and $\rho_{nk}(x)$ are continuous, so that the estimates of Lemmas 3.1 and 3.2 will hold for all $x,y\in Q$.

We also need a stronger bound for sums of $\rho_{nk}$ terms than that given by the bounds for individual terms in Lemma 3.1, and the next two lemmas provide this. They are motivated by the idea that any solution of (4) should satisfy the approximate equation $u((k+1)2^{-n})\approx u(k2^{-n})+\sigma_{nk}(u(k2^{-n}))$ which suggests that on a short time interval a solution can be approximated by an ‘Euler scheme’ $x_{k+1}=x_{k}+\sigma_{nk}(x_{k})$.

where $\gamma_{q}=y_{q+1}-y_{q}-\sigma_{n,k+q}(y_{q})$.

Let $\delta_{n}=2^{-2^{n/2}}$. By Lemma 3.1, with probability 1 there exists $C>0$ such that, for any $n,k\geq 0$ and any $x,y\in Q$, we have

for some $n,r,k$ as in the statement and some $x_{0}\in\Omega_{n}$, is bounded above by $C_{1}\sum_{n=0}^{\infty}\lambda^{-p}2^{n(3+d)}2^{-pn/8}$ which approaches 0 as $\lambda\rightarrow\infty$. Hence with probability 1 there exists $C>0$ such that

for all $n,k,r$ as above and $x_{0}\in\Omega_{n}$.

We now suppose, as we may with probability 1, that (21) and (22) hold (with the same $C$). We fix $n,k,r,y_{0}\cdots y_{r},\gamma_{0}\cdots\gamma_{r}$ as in the statement of the lemma. Take the smallest $s$ such that $y_{0}\in Q_{s}$, noting that then $2^{-s-1}\leq|y_{0}|\leq d^{1/2}2^{-s}$. Then we find $x_{0}\in\Omega_{ns}$ with $|x_{0}-y_{0}|<2^{-s-n}\leq 2^{1-n}|y_{0}|$ and define $x_{1}\cdots x_{r}$ by the recurrence relation $x_{q+1}=x_{q}+\sigma_{n,k+q}(x_{q})$. Then by (22)

Using (21) we have $|x_{q+1}|=|x_{q}+\sigma_{n,k+q}(x_{q})|\leq(1+C2^{-n/4})|x_{q}|+\delta_{n}$ so $|x_{q}|\leq C_{1}(|x_{0}|+r\delta_{n})$ and

Now let $u_{q}=x_{q}-y_{q}$. Then $|u_{q+1}-u_{q}|\leq|\rho_{n,k+q}(x_{q},y_{q})|+|\gamma_{q}|$ so

and since $|u_{0}|\leq 2^{1-n}|y_{0}|$ we deduce that $|u_{q}|\leq C_{3}(2^{-n}|y_{0}|+r\delta_{n}+\sum_{q=0}^{r-1}|\gamma_{q}|)$ and so

and we have the same bound for $|\rho_{n,k+q}(x_{q-1},y_{q-1})|$. Now

and then using (23), (24) and the fact that $|x_{0}-y_{0}|\leq 2^{1-n}|y_{0}|$ we deduce that

We now proceed to complete the proof of the theorem. From now on we take $g=f$ in the definition of $\sigma_{nk}$ and $\rho_{nk}$. We consider a Brownian path $W$ satisfying the conclusions of Lemmas 3.1, 3.2, 3.6 and 3.4 for some $C>0$. We shall show that for such a Brownian path the only solution $u$ of (4) in $\Phi$ is $u=0$. This will follow from the following:

Suppose $W$ satisfies the conclusions of Lemmas 3.1, 3.2, 3.6 and 3.4 for some $C>0$. Then there are positive constants $K$ and $m_{0}$ such that, for all integers $m>m_{0}$, if $u$ is a solution of (4) in $\Phi$ and for some $j\in\{0,1,\cdots,2^{m}-1$ and some $\beta$ with $2^{-2^{3m/4}}\leq\beta\leq 2^{-2^{2m/3}}$ we have $|u(j2^{-m})|\leq\beta$, then

We use $C_{1},C_{2},\cdots$ for positive constants which depend only on the constant $C$ and the dimension $d$. Fix $m$, $j$ and $\beta$ as in the statement, and suppose $|u(j2^{-m})|\leq\beta$. Let $N$ be the integer part of $4\log_{2}(1/\beta)$. Suppose $u\in\Phi$ satisfies (4), and let $u_{n}$ be the step function which takes the constant value $u(k2^{-n})$ on the interval $I_{nk}$, for $k=0,1,\cdots,2^{n}-1$.

Let $\alpha$ be the smallest nonnegative number such that

for $n>m$,and since $\psi_{m}=\beta$ it follows that

for all $n$ with $m\leq n\leq N$, where we have used the fact that $m^{1/2}2^{m/2}$ is bounded by const.$N$.

Now fix $n\geq m$. Then for $k=j2^{n-m},\cdots,(j+1)2^{n-m}-1$ we have, using (15)

where $\Omega_{l}=\sum_{r=j2^{l-m}}^{(j+1)2^{l-m}-1}|\rho_{l+1,2r+1}(u(2^{-l-1}(2r+1)),u(2^{-l}r))|$.

We now proceed to estimate the two sums on the right of (29), starting with the easier $\sigma_{nk}$ term. Using Lemma 3.2 and the fact that $N<2^{m}$, we have $|\sigma_{nk}(x)|\leq C_{2}n^{1/2}2^{-n/2}(2^{-N}+|x|)$ and so

Next we bound $\sum\Omega_{l}$, which we do in two stages. We first obtain a relatively crude bound by applying (11) to each term, and then obtain an improved by applying the crude bound together with Lemma (3.6). To start with the crude bound, from (11) we have $|\rho_{nk}(x,y)|\leq C_{3}2^{-n/2}N^{1/2}(2^{-N}+|x-y|)$ and using this together with (25) gives

For $l>N$ we use $|u(t)-u(t^{\prime})|\leq|t-t^{\prime}|$ and (11) to obtain

The second stage is to improve the estimate (34) by applying Lemma 3.6 to obtain a better estimate for $\Omega_{n}$ for larger $n$; we use (34) to bound the $\gamma$ term in Lemma 3.6.

Let $N^{1/6}\leq n\leq N$. We define $\gamma_{nk}=u((k+1)2^{-n})-u(k2^{-n})-\sigma_{nk}(u(k2^{-n}))$, noting that (28) implies that

so that $\Omega_{n}\leq\Lambda_{n+1}$. Let $r=\lfloor 2^{n/4}\rfloor$. In order to apply Lemma 3.6 to estimate $\Lambda_{n}$, we will split the sum into $r$-sized pieces. First we find $i\in\{0,1,\cdots,r-1\}$ such that, writing $s=\lfloor r^{-1}(2^{n-m}-i)\rfloor$, we have $\sum_{t=0}^{s}|u(j2^{-m}+(i+tr)2^{-n})|\leq r^{-1}\psi_{n}$. Now we fix for the moment $t\in\{0,1,\cdots,s\}$ and apply Lemma 3.6 with $y_{q}=u((k+q)2^{-n})$ where $k=j2^{n-m}+i+tr$. We obtain

From the last two inequalities, using (27), (35) and $|u(j2^{-m})|\leq\beta$, we find that

Since $n\geq N^{1/6}$ the first term dominates so $\Lambda_{n}\leq C_{11}2^{-m}(\beta+\alpha 2^{-m}N)$, and the same bound holds for $\Omega_{n}\leq\Lambda_{n+1}$. We deduce that

Using the original bound (31) for $l<N^{1/6}$ we have

Combining these two estimates with (33) we get our improved bound.

To conclude the proof we use this bound along with (30) in (29) and obtain

for all $n$ with $m\leq n\leq N$. Comparing this with (25) we see by the minimality of $\alpha$ that

Then if $m$ is large enough to ensure $C_{15}(N2^{-m}+N^{-1/4}+2^{-m/2}N^{1/2})<1/2$ it follows that $\alpha\leq 2C_{15}\beta$. Then applying (25) with $n=m$ gives $|u((j+1)2^{-m})|\leq\beta+2C_{15}\beta(m^{1/2}2^{m/2}+N)2^{-m}\leq\beta(1+C_{16}N2^{-m})$ from which the required result follows. ∎

To complete the proof of Theorem 1.1, using the notation of Lemma 3.7 let $m>m_{0}$ and $\beta_{0}=2^{-2^{3m/4}}$, and define $\beta_{j}$ for $j=1,2,\cdots,2^{m}$ by the recurrence relation $\beta_{j+1}=\beta_{j}(1+K2^{-m}\log(1/\beta_{j}))$. Writing $\gamma_{j}=\log(1/\beta_{j})$ we then have

so the sequence $(\gamma_{j})$ is decreasing and

for all $j=1,2,\cdots,2^{m}$, provided $m$ is large enough. Then for each $j$, $\beta_{j}$ is in the range specified in Lemma 3.7, and it follows from that lemma by induction on $j$ that $|u(j2^{-m})|\leq\beta_{j}$ for each $j$. Hence $|u(j2^{-m})|\leq 2^{-2^{2m/3}}$ for each $j$. This holds for all large enough $m$, and hence $u$ vanishes at all dyadic points in , and, as $u$ is continuous, $u=0$ on . This completes the proof of the theorem.

An Application

We give an application of Theorem 1.1 to convergence of Euler approximations to (1) with variable step size.

In this section we assume $f$ is continuous and consider (1) on a bounded interval $[0,T]$. Given a partition ${\mathcal{P}}=\{0=t_{0}<t_{1}<\cdots<t_{N}=T\}$ of $[0,T]$ we consider the Euler approximation to (1) given by:

for $n=0,\cdots,N-1$, with $x_{0}=0$. For such a partition ${\mathcal{P}}$ we let $\delta({\mathcal{P}})=\max_{n=1}^{N}(t_{n}-t_{n-1})$. Then we have the following:

For almost every Brownian path $W$, for any sequence

of partitions with $\delta({\mathcal{P}}_{k})\rightarrow 0$, we have

as $k\rightarrow\infty$, where $x(t)$ is the unique solution of (1) and $\{x_{n}^{(k)}\}$ is the Euler approximation using the partition ${\mathcal{P}}_{k}$.

Suppose $W$ is a path for which the conclusion of Theorem 1.1 holds, and suppose there is a sequence of partitions with $\delta({\mathcal{P}}_{k})\rightarrow 0$ such that $\max_{n=1}^{N_{k}}|x_{n}^{(k)}-x(t_{n}^{(k)})|\geq\delta>0$. Then if we let $u_{n}^{(k)}=x_{n}^{(k)}-W(t_{n}^{(k)})$ we have $|u_{n+1}^{(k)}-u_{n}^{(k)}|\leq\|f\|_{\infty}(t_{n+1}^{(k)}-t_{n}^{(k)})$ so by Ascoli-Arzela, after passing to a subsequence we have a continuous $u$ on $[0,T]$ such that $\max_{n=1}^{N_{k}}|u_{n}^{(k)}-u(t_{n}^{(k)})|\rightarrow 0$. Then writing $y(t)=u(t)+W(t)$ we see that $y\neq x$ and, using the continuity of $f$, that $y$ satisfies (1), contradicting the conclusion of the theorem. Corollary 4.1 is proved. ∎

The point of Corollary 4.1 is that the partitions can be chosen arbitrarily, no ‘non-anticipating’ condition is required. For general SDE’s with non-additive noise and sufficiently smooth coefficients Euler approximations will converge to the solution provided the partition points $t_{n}$ are stopping times, but this condition is rather restrictive for numerical practice, and an example is given in section 4.1 of of a natural variable step-size Euler scheme for a simple SDE which converges to the wrong limit. also contains related results and discussion.

Acknowledgement. The author is grateful to Istvan Gyöngy for drawing his attention to Krylov’s question and for valuable discussions.

References