I am currently going through the proof of the existence of a solution of the SDE

\begin{align} dX_t = bdt + \sigma dB_t \end{align}

where $B_t$ is a Brownian motion wrt a filtration $\{\mathcal{F}_t\}_{t\in[0,T]}$. Of course $b$ and $\sigma$ satisfies the linear growth bound and Lipschitz conditions, and they are both $\mathcal{F}_t$-adapted. We consider the Picard iterations \begin{align} \begin{cases} X_t^0 &= X_0 \\ X_t^{n+1} &= X_0 + \int_0^tb(s,X_s^n)ds + \int_0^t \sigma(s,X_s^n)dB_s \end{cases} \end{align} and the growth conditions to show that $X_t^n$ converges almost surely uniformly on $[0,T]$ to some process, say $Y_t$. In order for the limit process $Y_t$ to be a solution we need that the equation

\begin{align} Y_t = X_0 + \int_0^tb(s,Y_s)ds + \int_0^t \sigma(s,Y_s)dB_s \end{align}

is satisfied almost surely. We already have that

\begin{align} Y_t = X_0 + \lim\limits_{n\rightarrow\infty}\int_0^tb(s,X_s^n)ds + \lim\limits_{n\rightarrow\infty}\int_0^t \sigma(s,X_s^n)dB_s \end{align}

so it remains to show that

\begin{align} \lim\limits_{n\rightarrow\infty}\int_0^tb(s,X_s^n) - b(s,Y_s) ds &= 0 \\ \lim\limits_{n\rightarrow\infty}\int_0^t \sigma(s,X_s^n) - \sigma(s,Y_s)dB_s &= 0 \end{align}

almost surely. The first of these results follow for instance by the Lebesgue dominated convergence theorem, but I struggle with how to show the other one. Is there some convergence theorem for the Ito integral I can use?


1) $\sigma(s,x)$ is continuous (in the second argument) by the Lipschitz condition.

2) By the continuity we have the almost sure limit $\lim\limits_{n\rightarrow\infty} \sigma(s,X_s^n) = \sigma(s,X_s)$.

3) The process $\int_0^t \sigma(s,X_s^n) - \sigma(s,X_s)dB_s$ is a martingale wrt $\mathcal{F}_t$ for each $n$.

Some of observations 1) - 3) might be useful to answer my question.


1 Answer 1


First of all, if $(Y_t)_{t \geq 0}$ is a solution to the SDE it should satisfy

$$Y_t = \color{red}{Y_0} + \int_0^t b(s,\color{red}{Y_s}) \, ds + \int_0^t \sigma(s,\color{red}{Y_s}) \, dB_s$$

and not

$$Y_t = X_0 + \int_0^t b(s,X_s) \, ds + \int_0^t \sigma(s,X_s) \, dB_s.$$

Note that it suffices to show that $$\lim_{k \to \infty} \int_0^t \sigma(s,X_s^{n(k)}) \, dB_s = \int_0^t \sigma(s,Y_s) \, dB_s \qquad \text{a.s.}$$ for some subsequence $(X^{n(k)})_{k \in \mathbb{N}}$ of $(X^n)_{n \in \mathbb{N}}$. Therefore it suffices to show that

$$\left\| \int_0^t \sigma(s,X_s^n) \, dB_s - \int_0^t \sigma(s,Y_s) \, dB_s \right\|_{L^2} \to 0 \qquad \text{as} \, \, n \to \infty.$$

By Itô's isometry,

$$\left\| \int_0^t \sigma(s,X_s^n) \, dB_s - \int_0^t \sigma(s,Y_s) \, dB_s \right\|_{L^2}^2 = \mathbb{E} \left( \int_0^t |\sigma(s,X_s^n)-\sigma(s,Y_s)|^2 \, ds \right).$$

Using that $\sigma$ is Lipschitz-continuous (with respect to the $x$-variable), we get

$$\left\| \int_0^t \sigma(s,X_s^n) \, dB_s - \int_0^t \sigma(s,Y_s) \, dB_s \right\|_{L^2}^2 \leq C \mathbb{E} \left( \int_0^t |X_s^n-Y_s|^2 \, ds \right).$$

Now the claim follows from the fact that $X^n$ converges locally uniformly to $Y$.

  • $\begingroup$ @northwiz Well, we know that $$\lim_{n \to \infty} X_t^n = Y_t$$ as well as $$\lim_{n \to \infty} \int_0^t b(s,X_s^n) \, ds = \int_0^t b(s,Y_s) \, ds$$ almost surely and obviously this implies that this convergence also holds for any subsequence. Now if we can choose $(n(k))_{k \in \mathbb{N}}$ as in my answer, then it follows that $$\begin{align*} Y_t -Y_0&= \lim_{k \to \infty} X_s^{n(k)}-Y_0 \\ &= \lim_{k \to \infty} \int_0^t \sigma(s,X_s^{n(k)}) \, dB_s + \int_0^t b(s,X_s^{n(k)}) \, ds \\ &= \int_0^t \sigma(s,Y_s) \, dB_s +\int_0^t b(s,Y_s) \, ds. \end{align*}$$ $\endgroup$
    – saz
    Feb 13, 2015 at 14:50
  • $\begingroup$ @Northwiz Please do not delete your comments ... $\endgroup$
    – saz
    Feb 13, 2015 at 14:56
  • $\begingroup$ Thank you so much for the help. I deleted my first comment since I surpassed the 5 minutes editing limit, and had lots more to add. Didn't see your reply. Sorry. Thanks again for your help, much appreciated! $\endgroup$
    – northwiz
    Feb 13, 2015 at 14:56
  • $\begingroup$ @northwiz I see. You are welcome. $\endgroup$
    – saz
    Feb 13, 2015 at 14:58

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