Stuck on proof of Uniqueness of SDE solution (Oksendal Thm. 5.2.1)

Question

First, here is the theorem in question.

Theorem (Oksendal Thm. 5.2.1)

Let $T>0$ and $$ \begin{array}{l} b :[0,T]\times\Bbb R^n \to {\mathbb{R}^n};\\ \sigma :[0,T]\times\Bbb R^n\to {\mathbb{R}^{n \times m}}; \end{array} $$ be measurable functions for which there exist constants $C$ and $D$ such that $$ \begin{array}{l} |b(t,x)|+|\sigma (t,x)|\le C(1+|x|);\\ |b(t,x)-b(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le D|x-y|; \end{array} $$ Let $Z$ be a random variable that is independent of the $\sigma$-algebra generated by $B_s$, $s ≥ 0$, and with finite second moment: $$ E[|Z|^2]<\infty $$ Then the stochastic differential equation/initial value problem $$ \begin{array}{l} {\rm{d}}{X_t} =b(t,X_t)\mathrm dt+\sigma(t,X_t)\mathrm dB_t,\quad \text{for } t \in [0,T];\\ X_0 = Z; \end{array} $$ has a Pr-almost surely unique $t$-continuous solution $(t,ω)\mapsto X_t(ω)$ such that $X$ is adapted to the filtration $\mathcal F_t^Z$ generated by $Z$ and $B_s$, $s\leq t$, and $$ E\left[\int_0^T|X_t|^2\,\mathrm dt\right]<\infty. $$

I am stuck at a few places in the uniqueness part of the proof below.

For the first step, I expanded the squared term to find $$ E[|X_t-\hat X_t|^2]< 3 E[|X_t-\hat X_t|^2]=3E[|Z-\hat Z|^2]+3E\left(\int_0^t a\,\mathrm ds\right)^2+3E\left(\int_0^t \gamma\,\mathrm dB_s\right)^2\\ +6E\left(\int_0^t a\,\mathrm ds+\int_0^t \gamma\,\mathrm dB_s\right)\\ +6E\left((Z-\hat Z)\int_0^t a\,\mathrm ds\int_0^t \gamma\,\mathrm dB_s\right) $$ but am not sure how to proceed to get the second line. What do I do with the last two terms?

For the second arrow, I know this comes from the assumed Lipschitz continuity but am not able to fill in the steps between the lines.

Lastly, why do we need the statement marked with the $({\color{red}\ast})$? And why is the set $[0,T]$ intersected with the rationals?

Thank you in advance for any help with this.

Answer 1

1. For the first step you use the fact that $(a+b+c)^2≤3(a^2+b^2+c^2)$, which is a consequence of Cauchy-Schwarz inequality as pointed out by @LutzLehmann
2. For the second step you use Cauchy-Schwarz inequality and Ito's isometry.
3. For the third step you should use the Lipchitz continuity condition:

$$3tE\bigg(\int_0^t a^2 ds\bigg)=3tE\bigg(\int_0^t (b(s,X_s)-b(s,\hat{X}_s))^2 ds\bigg)$$ $$\leq 3tD^2E\bigg( \int_0^t (X_s-\hat{X}_s)^2 \bigg)ds$$

and using the same reasoning we have

$$E\bigg(\int_0^t \gamma^2 ds\bigg)\leq 3D^2E\bigg( \int_0^t (X_s-\hat{X}_s)^2 \bigg)ds.$$

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I am not one hundred percent sure about this last part, maybe some other use could tell us whether this is right or not.

$$E(|X(t,\omega)-\hat X(t,\omega)|^2)=0$$ By simplicity of notation let $Z_t=X(t,\omega)-\hat X(t,\omega)$.

The this implies that for each fixed $t\in[0,T]$ $$P\big(\{\omega:Z_t(\omega)=0\}\big)=1.$$

(this means that $X$ is a modification of $\hat X$).

We actually need to show that

$$P\big(\{\omega:Z_t(\omega)=0,\forall t\in[0,T]\}\big)=1.$$ (this means the processes are indistinguishable).

Start by taking an ordering of the rational numbers in $[0,T]$, $(r_1,r_2,\cdots)$.

Then for each fixed $r_n$ we have that $P\big(\{\omega:Z_{r_n}(\omega)=0\}\big)=1$, this means that for each $r_n$ there exists $\Omega_n$ with full measure such that $Z_{r_n}(\omega)=0$ for all $\omega\in\Omega_n$.

Now take $\Omega'=\bigcap_{n=1}^{\infty} \Omega_n$. Then we have that $P(\Omega')=1$, and for each $\omega\in\Omega'$, $Z_{r_n}(\omega)=0$, for all $n$. This means that $$P(Z_t=0,\forall t\in[0,T]\cap Q)=1$$ Then use the fact that the process is continuous and you are done.

Answer 2

Let $Z_t(\omega) := | X_t(\omega) - {\hat X}_t(\omega) |$ and I'll post an answer only for the last one:

\begin{equation} P\big( Z_t = 0 \text{ for all } t \in \mathbb{Q} \cap [0, T] \big) = 1 \tag{1} \end{equation}

as the others have already been well-answered above.

Here, the appearance of $\mathbb{Q}$ in (1) is because probability $P(.)$ is a measure defined on a $\sigma$-algebra that is closed under countable union and intersection, so uncountable operations cannot be directly applied.

Formally, let $(t_n)$ be the sequence of all rational numbers within $[0, T]$, so that $\bigcup_n \{t_n\} = Q\cap [0, T]$ and $E_n := \{\omega : Z_{t_n}(\omega) = 0\}$ for $n \in \mathbb{N}$. Then, since $P(E_{n}) = P(Z_{t_n} = 0) = 1$ by $E(|Z_{t_n}|^2) = 0$, we have $P(E_n^c) = 0$, for all $n$, hence \begin{equation} 0 \leq P\big({\textstyle\bigcup_n} E_{n}^c\big) \leq {\textstyle\sum_n} P(E_{n}^c) = 0 \end{equation} by sub-additivity. Therefore, by De Morgan's law, $P\big(\bigcap_n E_{n}\big) = 1 - P\big(\bigcup_n E_{n}^c\big) = 1$, where \begin{align} P\big({\textstyle \bigcap_n} E_{n}\big) = P\big( Z_{t_n} = 0 \text{ for all } n \big) &= P\big( Z_t = 0 \text{ for all } t \in \textstyle\bigcup_n \{t_n\} \big) \\ &=P\big( Z_t = 0 \text{ for all } t \in Q \cap [0, T] \big). \end{align}