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)?
We will add a few more details, let me know if you need more.
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.
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 $Y,\hat{X}$.
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
$$E[Y_{r}^{2}]\leq \sum_{k=0}^{\infty}\frac{(AT)^{k}}{k!}+E[Y_{0}^{2}].$$