I am now having difficulty in showing that $\sup_{t \geq 1} (W_t – W_1)$ is independent of $W_1$, where the $W_i$ is a Brownian motion.
I know that $W_t – W_1$ is independent of $W_1$ for all $t \geq 1$ due to the independent increment property. But, how can I show that the supremum is also independent of $W_1$?
I thought that it suffices to show that $\sup_{t \geq 1} (W_t – W_1)$ is $\sigma((W_t – W_1)_{t\geq1})$-measurable, because $\sigma((W_t – W_1)_{t\geq1})$ is independent of $\sigma(W_1)$. Thus, instead of any Borel set, it suffices to show that the set $\big(\sup_{t \geq 1} (W_t – W_1) > a \big)$ is in $\sigma((W_t – W_1)_{t\geq1})$ for any $a \in \mathbb{R}.$
First, let me define, as C, the set on which $(W_t-W_1)_{t\geq1}$ has continuous sample paths. Then, clearly $\mathbb{P}(C)=1$ due to the defining property of Brownian motion, where $\mathbb{P}$ is the underlying probability measure.
Now, I tried to prove in this way : $$\big(\sup_{t \geq 1} (W_t – W_1) > a \big) = \bigg(\big(\sup_{t \geq 1} (W_t – W_1) > a \big) \cap C \bigg) \sqcup \bigg(\big(\sup_{t \geq 1} (W_t – W_1) > a \big) \cap C^\complement \bigg)$$
For $\bigg(\big(\sup_{t \geq 1} (W_t – W_1) > a \big) \cap C \bigg),$ using continuity, I can take rationals so that the set $\big(\sup_{t \geq 1} (W_t – W_1) > a \big)$ can be expressed as a countable union of sets contained in $\sigma(W_t – W_1)_{t\geq1}.$
Now, it remains to show that $C$ and $\bigg(\big(\sup_{t \geq 1} (W_t – W_1) > a \big) \cap C^\complement \bigg)$ are both contained in $\sigma(W_t – W_1)_{t\geq1}.$
But, then, I feel that there appears some delicacy that I find it somewhat hard to deal with.
- How can I show rigorously that $C$ is contained in $\sigma(W_t – W_1)_{t\geq1}.$?
- Assuming that $C$ is in $\sigma(W_t – W_1)_{t\geq1},$ $C^\complement$ is also in $\sigma(W_t – W_1)_{t\geq1}, $ and the measure of $C^\complement$ is evidently zero. Then, I feel that I need to use the completeness of the underlying probability measure $\mathbb{P}$ to deduce that $\bigg(\big(\sup_{t \geq 1} (W_t – W_1) > a \big) \cap C^\complement \bigg)$ is contained in $\sigma(W_t – W_1)_{t\geq1}.$ If my guess were right, then, if needed at all, on which probability space(triple) is the completion taken?
Thank you so much for reading and anything correcting my logic will be enormously appreciated!
Best Answer
Actually, $\sup_{t\ge 1}{(W_t-W_1)}=+\infty$ a.s. and so it is trivially independent of $W_1$.
If you consider the supremum over a finite interval $[1,T]$, $B_T:=\sup_{t\in [1,T]}(W_t-W_1)$, then $\sigma(B_T)\subseteq \sigma(W_t:t\ge 1)\vee \mathcal{N}$, where $\mathcal{N}$ is the collection of $\mathsf{P}$-null sets, and $\sigma(W_1)$ is independent of the latter $\sigma$-field.