Time homogeneity of Ito diffusion

Consider a time homogeneous Ito diffusion satisfying a SDE,

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

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

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

$$\{\tilde{B}_v\}_{v\geq 0}$$ 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, $$$$dX_t=b(X_t)dt+\sigma(X_t)dB_t, X_0=x$$$$ that, $$\{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$$.

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

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 $$(\Omega,\mathcal{A},\mathbb{P})$$.