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Let $(X_t)_{t\ge0}$ be a process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$ such that $X_t-X_s$ is independent of $\mathcal F_s$ for all $t\ge s\ge0$.

Are we able to show that

  1. $X_u-X_t$ is independent of $\sigma(X_t-X_s,\mathcal F_s)$ for all $u\ge t\ge s$?
  2. $(X_t-X_s)_{t\ge s}$ is independent of $\mathcal F_s$ for all $s\ge0$?

(2.) is easy to show in the space case when $\mathcal F_t=\sigma(X_s,s\le t)$, but I struggle to obtain the claim in the general case.

It is clearly sufficient to show that for all $n\in\mathbb N$ and $s\le t_1<\cdots<t_n$, the vector $(X_{t_1},\ldots,X_{t_n})$ is independent of $\mathcal F_s$.

If we would be able to show (1.), the latter should follow from this answer. However, I also don't know how to show (1.) either ...

EDIT: It's not the same as (1.), but we should have the following: If $u\ge t\ge s\ge0$, then \begin{equation}\begin{split}\operatorname E\left[(X_u-X_t)(X_t-X_s)\mid\mathcal F_s\right]&=\operatorname E\left[\operatorname E\left[(X_u-X_t)(X_t-X_s)\mid\mathcal F_t\right]\mid\mathcal F_s\right]\\&=\operatorname E\left[\operatorname E\left[X_u-X_t\mid\mathcal F_t\right](X_t-X_s)\mid\mathcal F_s\right]\\&=\operatorname E\left[X_u-X_t\right]\operatorname E\left[X_t-X_s\mid\mathcal F_s\right]\\&=\operatorname E\left[X_u-X_t\right]\operatorname E\left[X_t-X_s\right],\end{split}\tag1\end{equation} where

  • the 1st equation holds by the tower property, since $t\ge s$;
  • the 2nd equation holds, since $X_t-X_s$ is $\mathcal F_t$-measurable;
  • the 3rd equation holds, since $X_u-X_t$ is independent of $\mathcal F_t$ and hence $\operatorname E\left[X_u-X_t\mid\mathcal F_t\right]=\operatorname E\left[X_u-X_t\right]$;
  • the 4th equation holds, since $X_t-X_s$ is independent of $\mathcal F_s$ and hence $\operatorname E\left[X_t-X_s\mid\mathcal F_s\right]=\operatorname E\left[X_t-X_s\right]$.

Am I missing something? If not, we should be able to conclude that $$\operatorname E\left[e^{{\rm i}\left\langle\left(X_{t_n}-X_{t_{n-1}},\:\ldots\:,\:X_{t_1}-X_{t_0}\right),\:y\right\rangle}\mid\mathcal F_s\right]=\prod_{i=1}^n\operatorname E\left[e^{{\rm i}(X_{t_i}-X_{t_{i-1}})y_i}\right]\tag2$$ for all $y\in\mathbb R^n$, $n\in\mathbb N$ and $s\le t_0<\cdots<t_n$. And this should immediately yield (2.). Is this correct?

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