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Consider a time homogeneous Ito diffusion satisfying a SDE,

\begin{equation}\label{1} dX_t=b(X_t)dt+\sigma(X_t)dB_t, X_s=x \end{equation}

$t\geq s$. The unique solution of the SDE is denoted by $X_t=X_t^{s,x}$. \begin{equation} X_{s+h}^{s,x}=x+\int_s^{s+h}b(X_u^{s,x})+\int_{s}^{s+h}\sigma(X_u^{s,x})dB_u \end{equation} \begin{equation} =x+\int_0^{h}b(X_{s+v}^{s,x})+\int_{0}^{h}\sigma(X_{s+v}^{s,x})d\tilde{B}_v \end{equation}

where $\tilde{B}_v=B_{s+v}-B_s,v\geq 0$. And, \begin{equation} X_{h}^{0,x}=x+\int_s^{h}b(X_v^{0,x})+\int_{0}^{h}\sigma(X_v^{0,x})dB_v \end{equation}

$\{\tilde{B}_v\}_{v\geq 0}$ and $\{B_v\}_{v\geq 0}$, have the same $P^0$ distribution, where $P^0$ is the probability distribution of $B_t$ starting at 0. I understood uptill here. I didn't understand the following claims.

It says that since both the version of the brownian motion have the same $P^0$ distribution, it follows by weak uniqueness of the solution of the SDE, \begin{equation} dX_t=b(X_t)dt+\sigma(X_t)dB_t, X_0=x \end{equation} that, $\{X_{s+h}^{s,x}\}_{h\geq0}$ and $\{X_{h}^{0,x}\}_{h\geq0}$ have the same $P^0$ distributions. I didn't understand how weak uniqueness in invoked here and what does it mean for both the $X$'s to have the same $P^0$.

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  • $\begingroup$ I don’t see why you would even a probabilistic argument here. The coefficients do not depend on time so time-invariance/homogeneity follows. $\endgroup$ – Calculon May 1 at 10:25
  • $\begingroup$ @Calculon It is not clear to me what it means for both the $X$ to have the same $P^0$ distribution. $\endgroup$ – user88923 May 1 at 10:30
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Let's recall the definition of weak uniqueness for an SDE of the form

$$dY_t = b(Y_t) \, dt + \sigma(Y_t) \, dB_t, \qquad Y_0 = x. \tag{1}$$

The SDE $(1)$ is said to have a unique weak solution if the following condition holds: If $(\Omega^{(i)},\mathcal{A}^{(i)},\mathbb{P}^{(i)})$, $i=1,2$, are probability spaces and $(Y_t^{(i})_{t \geq 0}$ are processes on $\Omega^{(i)}$ such that

  • $(Y_t^{(i)})_{t \geq 0}$ is adapated to a filtration $(\mathcal{F}_t^{(i)})_{t \geq 0}$ which is admissible for a Brownian motion $(B_t^{(i})_{t \geq 0}$ on $\Omega^{(i)}$,
  • $Y_t^{(i)} -x = \int_0^t b(Y_s^{(i)}) \, ds + \int_0^t \sigma(Y_s^{(i)}) \, dB_s^{(i)}$ $\mathbb{P}^{(i)}$-almost surely,

then $(Y_t^{(1)})_{t \geq 0}$ has the same finite dimensional distributions as $(Y_t^{(2)})_{t \geq 0}$, i.e. $$\mathbb{P}^{(1)}(Y_{t_1}^{(1)} \in A_1,\ldots,Y_{t_n}^{(1)} \in A_n) = \mathbb{P}^{(2)}(Y_{t_1}^{(2)} \in A_1,\ldots,Y_{t_n}^{(2)} \in A_n)$$ for any $t_1,\ldots,t_n \geq 0$, $n \in \mathbb{N}$ and measurable sets $A_i$.

Now in your case ,we are given a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and a Brownian motion $(B_t)_{t \geq 0}$ (started at $B_0=0$) on this probability space. On the one hand, we know that $Y_t^{(1)} := X_t^{0,x}$ is a solution to the SDE

$$dY_t = b(Y_t) \, dt + \sigma(Y_t) \, dB_t, \qquad Y_0 = x,$$

i.e. we have

$$Y_t^{(1)} -x = \int_0^t b(Y_r^{(1)}) \, dr + \int_0^t \sigma(Y_r^{(1)}) \, dB_r^{(1)}, \quad \mathbb{P}^{(1)}-\text{a.s.}$$

for $B_r^{(1)} := B_r$ and $\mathbb{P}^{(1)} := \mathbb{P}$. On the other hand, the process $Y_t^{(2)} := X_{s+t}^{s,x}$ satisfies

$$Y_t^{(2)}-x = \int_0^t b(Y_r^{(2)}) \, dr + \int_0^t \sigma(Y_r^{(2)}) \, dB_r^{(2)} \quad \mathbb{P}^{(2)}-\text{a.s.}$$

for $B_r^{(2)} := B_{s+r}-B_s$ and $\mathbb{P}^{(2)} := \mathbb{P}$. Since you are assuming that the SDE has a unique solution, it follows from the very definition (see above) that $(Y_t^{(1)})_{t \geq 0}$ and $(Y_t^{(2)})_{t \geq 0}$ have the same distribution on $(\Omega,\mathcal{A},\mathbb{P})$, i.e.

$$\mathbb{P}(Y_t^{(1)} \in A_1,\ldots,Y_{t_n}^{(1)} \in A_n) = \mathbb{P}(Y_{t_1}^{(2)} \in A_1,\ldots, Y_{t_n}^{(2)} \in A_n)$$

for any $t_i \geq 0$ and measurable sets $A_i$. By the very definition of $Y^{(1)}$ and $Y^{(2)}$ this means that $(X_t^{0,x})_{t \geq 0}$ and $(X_{t+h}^{s,x})_{t \geq 0}$ have the same distribution on $(\Omega,\mathcal{A},\mathbb{P})$.

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