Suppose I have two 1D Brownian bridges $(B^{(1)}_t,t\in [0,1]),(B^{(2)}_t,t\in [0,1])$, one from $0$ to $0$ and one from $x$ to $y$ where $x,y \geq 0$. Is there a neat way to show that there exists a coupling so that $\vert B^{(1)}_t \vert \leq \vert B^{(2)}_t \vert$ almost--surely for all $t \in [0,1]$? I have a partial proof but this needs a variety of case distinctions so I wondered if there was a good way to show this.
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$\begingroup$ just to be sure, you mean 2D Browinan bridges both parametrized by $[0,1]$, right? $\endgroup$– Kostya_ICommented Oct 14 at 16:44
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2$\begingroup$ Pretty sure he means 1D Brownian bridges, but his first coordinate is time. $\endgroup$– Martin HairerCommented Oct 14 at 18:11
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It seems that the following coupling is a good candidate.
For simplicity I only treat the case $x=y=1$ and denote $W^{(1)} = B^{(1)}$ as well as $W^{(2)} = B^{(2)} - 1$ which are now two BB pinned at $0$ at $t=0$ and $t=1$.
Denote $\tau_- = \inf( t \in [0,1] \mid W^{(1)}_t = -1/2 )$ the first hitting time of $-1/2$ and similarly $\tau_+ = \sup( t \in [0,1] \mid W^{(1)}_t = -1/2 )$ the last hitting time of $-1/2$. Conditionally on the event where the latter two exist and on the random pair $(\tau_-,\tau_+)$ I believe that the process $W^{(1)}_t, t \in [\tau_-,\tau_+]$ is again a BB pinned at $-1/2$. I don't have a nice or clean proof of this right now so I'll ask the OP and other interested readers to confirm I am not blundering here. The intuition is that conditionally on $(\tau_-,\tau_+)$ and on the excursions of the BB outside $[\tau_-,\tau_+]$ in $]-1/2, +\infty [$, the trajectory inside $[\tau_-,\tau_+]$ should be independent and again a BB, by a kind of BB version of the strong Markov property (the proof is clear I believe at least for a random walk on $\mathbb Z$ with conditioned final state).
You can now use the reflection principle with reflection horizontal axis $y = -1/2$ and set $W^{(2)}_ t = - 1 - W^{(1)}_t$ for $t \in [\tau_-,\tau_+]$ and set $W^{(1)}_t = W^{(2)}_t$ elsewhere. On $ t \in [\tau_-,\tau_+]$ it holds $ | W^{(1)}_t | = | W^{(2)}_t + 1 | $ while elsewhere $ W^{(2)}_t = W^{(1)}_t \geq -1/2 $ so that the distance to $0$ is smaller than the distance to $-1$. In the end, $ | W^{(1)}_t | \leq | W^{(2)}_t + 1 | $ .
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1$\begingroup$ Thank you Mathias, this is quite elegant. I agree with you that the conditional on $(\tau_-,\tau_+)$ the process restricted to $t\in [\tau_-,\tau_+]$ is a Brownian bridge. $\endgroup$– DavidCommented Oct 15 at 12:34
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1$\begingroup$ This approach doesn't work when $y=0$ (equivalently $x=0$) because you're not sufficiently bounded away from $0$. In that case there is a different coupling: Run the two BB independently until $\tau = \inf\{t: \vert B^{(1)}_t \vert = \vert B^{(2)}_t \vert \}$. If $B^{(1)}_\tau = B^{(2)}_\tau$, then let them run together. If $B^{(1)}_\tau = - B^{(2)}_\tau$, set $B^{(1)}_t = -B^{(2)}_t$ for all $t \geq \tau$. This coupling has the desired property. $\endgroup$– DavidCommented Oct 15 at 12:39
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1$\begingroup$ Your coupling then also work to couple a BB from $x$ to $y$ with a BB from $x-y$ to $0$ (without loss of generality $x-y>0$). And in a second step using the coupling I described you can couple it to a BB from $0$ to $0$. $\endgroup$– DavidCommented Oct 15 at 12:42