I am trying to show that $X_t^{s,x} = X_t^{r, X_r^{s,x}}$ for $0 \leq s \leq r \leq t$, $x \in \mathbb{R}^n$ is a given initial condition for time $s$, for some SDE: \begin{equation*} d X(u)=b(X(u))d u+\sigma(X(u))d B(u). \end{equation*}

I have seen a couple of proofs that use the uniqueness of the solution to show this, but I feel like some of the proofs have some circular reasoning in it. I understand the general idea, that since the stochastic process $X_t$ is unique, then at some time $t$, if we can write two expressions for the process's random value, then the expressions must be equal. In Oksendal the proof is basically what I wrote above and is two lines: enter image description here

But I wanted a more 'formalized' version of this proof, but couldn't find one until I searched online and found this following one:

From the following link: https://www.math.tecnico.ulisboa.pt/~czaja/ISEM/10internetseminar200607.pdf, the proof of this is the following: enter image description here

I am wondering how to justify the step of 'replacing' $Z_u$ with $X_u^{s,x}$. Isn't this saying that $Z_u = X_u^{s,x}$, where $Z_u = X_u^{r, X_r^{s,x}}$, and so $X_u^{s,x} = X_u^{r, X_r^{s,x}}$. And if we are asserting that this is true, isn't this the same as what we are trying to prove: $X_t^{s,x} = X_t^{r, X_r^{s,x}}$. So I am not sure what I am missing here.

Any help would be greatly appreciated. Thanks!

  • $\begingroup$ Like me you know about two-liners that simply use pathwise uniqueness. What is the point of agonizing over that particulal proof ? $\endgroup$
    – Kurt G.
    Sep 25, 2022 at 9:52


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