# Weak Uniqueness of solution of SDE

The following is an excercise where we try to prove Lemma 5.3.1. Oksendal "Stochastic Differential Equations"

Suppose that $$b(t,x)$$ and $$\sigma(t,x)$$ satisfy the hypotheses of the existence/(strong) uniqueness theorem. Suppose that $$X_t,\hat{X}_t$$ are adapted processes, both of which are continuous in mean square. Define

\begin{aligned} Y_t&=\int_0^tb(s,X_s)ds+\int_0^t\sigma(s,X_s)dB_s\\ \hat{Y}_t&=\int_0^tb\left(s,\hat{X}_s\right)ds+\int_0^t\sigma\left(s,\hat{X}_s\right)d\hat{B}_s \end{aligned}

where $$B_t$$ and $$\hat{B}_t$$ are each Brownian motions.

• a) Prove that the processes $$Y_t$$ and $$\hat{Y}_t$$ have the same distribution.

• b) Prove that $$Y_t$$ is continuous in mean square.

• c) Now consider the Ito diffusions

\begin{aligned} dX_t&=b(t,X_t)dt+\sigma(t,X_t)dB_t\\ d\hat{X}_t&=b\left(t,\hat{X}_t\right)dt+\sigma\left(s,\hat{X}_t\right)d\hat{B}_t \end{aligned}

with initial conditions $$X_0$$ and $$\hat{X}_0$$ where $$X_0$$ and $$\hat{X}_0$$ have the same distribution. Let $$X_t^k$$ and $$\hat{X}_t^k$$ be the sequence of approximations to $$X_t$$ and $$\hat{X}_t$$ that come from Picard iteration. Use a) and b) to prove that the processes $$X_t^k$$ and $$\hat{X}_t^k$$ have the same distribution for all $$k$$ and that they are continuous in mean square for all $$k$$. This is easy given a) and b).

• d) Conclude that the proceeses $$X_t$$ and $$\hat{X}_t$$ have the same distribution, i.e., we have weak uniqueness. Can we conclude that the solutions $$X_t$$ and $$\hat{X}_t$$ are continuous in mean square?

Here is the sketch of proof

How do I solve parts a) and b)?

• From what's written, I don't see how $a)$ could possibly be true. We have no assumptions on $X_t$ and $\hat X_t$ other than adaptedness, so $X_t = 1$ and $\hat X_t = 0$ satisfy the hypotheses. There is no way that $\int_0^t b(s,1)ds + \int_0^t \sigma(s,1) dB_s \stackrel{d}{=} \int_0^t b(s,0) ds + \int_0^t \sigma(s,0) d \hat B_s$ for all $b,\sigma$ satisfying the hypotheses of those theorems. Mar 22 at 18:01

We will add a few more details, let me know if you need more.

Let $$X_t$$ and $$Y_t$$ be the strong solutions constructed from $$\tilde{B}_t$$ and $$\hat{B}_t$$, respectively, as above.

Here they simply mean that they obtain strong solutions for the following SDEs \begin{aligned} X_t&=\int_0^tb(s,X_s)ds+\int_0^t\sigma(s,X_s)d\tilde{B}_s\\ Y_t&=\int_0^tb\left(s,Y_s\right)ds+\int_0^t\sigma\left(s,Y_s\right)d\hat{B}_s, \end{aligned} which is possible from Theorem 5.2.1 since the coefficients are assumed to be Lipschitz.

Then the same uniqueness argument as above applies to show that $$X_t = \tilde{X}_t$$ and $$Y_t = \hat{X}_t$$ for all t, a.s.

By studying the mean-square as in the proof of Theorem 5.2.1

$$E|X_{t}-\tilde{X}_{t}|^{2}|,$$ they can again show that this is zero since they correspond to the same Brownian motion $$\tilde{B}_s$$ i.e. they both solve in law \begin{aligned} X_t&=\int_0^tb(s,X_s)ds+\int_0^t\sigma(s,X_s)d\tilde{B}_s\\ \tilde{X}_t&=\int_0^tb(s,\tilde{X}_s)ds+\int_0^t\sigma(s,\tilde{X}_s)d\tilde{B}_s. \end{aligned} The same is done for $$Y,\hat{X}$$.

Therefore it suffices to show that $$X_t$$ and $$Y_t$$ must be identical in law. We show this by proving by induction that if $$X^{(k)}_t , Y^{(k)}_t$$ are the processes in the Picard iteration...

The equality in distribution indeed follows by induction. We start from the base case $$X^{(0)}_t=X_{0} \sim Y_{0} =Y^{(0)}_t$$. Then we define

\begin{aligned} X_t^{(k+1)}&:=\int_0^tb(s,X_s^{(k)})ds+\int_0^t\sigma(s,X_s^{(k)})d\tilde{B}_s\\ Y_t^{(k+1)}&:=\int_0^tb\left(s,Y_s^{(k)}\right)ds+\int_0^t\sigma\left(s,Y_s^{(k)}\right)d\hat{B}_s, \end{aligned} but since we have $$(X_s^{(k)})\stackrel{d}{=}(Y_s^{(k)})$$ and $$b,\sigma$$ are measurable functions, we obtain $$(X_s^{(k+1)})\stackrel{d}{=}(Y_s^{(k+1)})$$.

Finally we discuss proving continuity in mean square using ideas from proof of 5.2.1

$$E[(Y_{s}-Y_{t})^{2}]=E\left[\left(\int_{s}^{t}bdr+\int_{s}^{t}\sigma dB_r\right)^{2}\right].$$

By Itô-isometry and zero quadratic variation for $$dr,dB_{r}$$, we are left with

$$=E\left[\left(\int_{s}^{t}bdr\right)^{2}\right]+\int_{s}^{t}E\left[(\sigma)^{2}\right] dr$$

using Cauchy-Schwartz

$$\leq (t-s)\int_{s}^{t}E[b^{2}]dr+\int_{s}^{t}E\left[(\sigma)^{2}\right] dr$$

using the linear growth and the inequality $$(x+y)^{2}\leq 2x^{2}+2y^{2}$$ we bound by

$$\leq c_{1}(t-s)\int_{s}^{t}E[1+Y_{r}^{2}] dr dr+c_{2}\int_{s}^{t}E\left[1+Y_{r}^{2}\right] dr$$

$$\leq b_{1}(t-s)+b_{2}\int_{s}^{t}E[Y_{r}^{2}] dr.$$

Finally, we use the bound 5.2.14 for L2-moments

$$E[Y_{r}^{2}]\leq \sum_{k=0}^{\infty}\frac{(AT)^{k}}{k!}+E[Y_{0}^{2}].$$

• Could you please explain to me how do we prove part a), that is Prove that the processes $Y_t$ and $\hat{Y}_t$ have the same distribution. and part b), that is Prove that $Y_t$ is continuous in mean square.? Thanks Mar 24 at 22:34
• @IrvingLee The part (a) is the very last part with Picard iteration. Mar 24 at 22:58
• The part (b) follows from Itô isometry as done in proof of Theorem 5.2.1 i.e. we get $$E[(Y_t−E[Y_t])^2]=E(\int^t b−E[b]ds)^2+\int^tE\sigma^2ds,$$ which is a continuous function of t. . Mar 24 at 23:07
• When I say "prove that $Y_t$ is continuous in mean square", I mean to prove that $Y_t$ stafisfies the following property: A stochastic process $Y_t$ is continuous in mean square if for all $t$ we have $\mathbb{E}\left[Y_t^2\right]<\infty$ and $$\lim_{s\rightarrow t}\mathbb{E}\left[\left(Y_s-Y_t\right)^2\right]=0.$$ How do I do that with $Y_t=\int_0^tb(s,X_s)ds+\int_0^t\sigma(s,X_s)dB_s$ ? Mar 24 at 23:46
• @IrvingLee see updated answer. Mar 25 at 0:42