Will taking one more step from a different starting point lead to the same end point?

25 Views Asked by At

I am dealing with a total variance which can be represented as: $$d(t)=\|v_{t+1}-v_{t}||_{TV}$$ The two distributions have the following relation: $$v_{t+1}=W_{t}\times v_{t}=W_{t}W_{t-1}...W_{0}\times v_{0}=W_{0,t}\times v_{0}$$ and thus $$v_{t} =W_{0,t-1}\times v_{0}$$ I want to use coupling to analyze the changing rate of $d(t)$ and it then turns out that the coupling should be defined as $(X_{t+1}, Y_{t})$ according to the theorem: $$\|u-v\|_{TV}\leq P\{X\neq Y\}$$ but under this definition the two chains $X_{t+1}$ and $Y_{t}$ seems never coupled since they are not synchronous. Upon fixing this asynchronicity, I wonder if there exists a new start point $v'_{0}$ that is different from $v_{0}$ and it satisfies the following property: $$v_{t} = W_{0,t}\times v'_{0}$$ and now the $d(t)$ is: $$d(t)=\|W_{0,t}\times v_{0}-W_{0,t}\times v'_{0}\|_{TV}$$ and thus the synchronous coupling is possible to be defined.

I am not sure if the thought above is reasonable and it seems it is equivalent to asking that will taking one more step from a different starting point to lead to the same endpoint.