2
$\begingroup$

Consider a sequence of stochastic processes $\{\tilde{x}^n\}$, $\tilde{x}^n = \tilde{x}^n_t(\omega)$, and Brownian motions $\{\tilde{w}^n\}$. Suppose that for each $\tilde{x}^n$ solves the stochastic differential equation $$\tilde{x}^n_t = x + \int_0^t\sigma^n(s,\tilde{x}^n_s)d\tilde{w}_s + \int_0^tb^n(s,\tilde{x}^n_s)ds$$ (assuming everything is progressively measurable, $\sigma^n$, $b^n$ continuous and bounded)
Skorokhod has proved a lemma (not gonna waste words on the details) which in turn gives me a sequence of pairs $\{x^n,w^n\}$, such that the finite dimensional distributions of $\{x^n,w^n\}$ coincide with those of $\{\tilde{x}^n,\tilde{w}^n\}$, but $\{x^n,w^n\}$ have the nice property that for all $t\geq 0$, $\{x^n_t,w^n_t\}$ converges in probability. Now I would like to show that $$x^n_t = x + \int_0^t\sigma^n(s,x^n_s)dw_s + \int_0^tb^n(s,x^n_s)ds$$ also holds for each $n$ almost surely for all $t$. To prove this, I need for each $n$, $t\geq 0$: $$E[|x^n_t - x - \int_0^t\sigma^n(s,x^n_s)dw_s - \int_0^tb^n(s,x^n_s)ds|^2] = 0$$ I know this should be very straight forward to prove by adding and subtracting the terms with $\tilde{x}$ in place of $x$ and the approximate dyadic (in time) sums for the integrals, then using the fact that they have the same finite dimensional distributions, but I somehow couldn't work it out. Actually I'm not even fully convinced that this is true at all. Could anyone show me the details of showing this? Thanks in advance. Any help is appreciated.

This is found in the proof of Theorem II.6.1 in Controlled Diffusion Processes by N.V.Krylov. I know results by Krylov are highly unlikely to be wrong, but he left out a lot of details in the proof, so I'm not very sure about this.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.