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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.