Let $W$ be a one dimensional standard Brownian motion, and $\sigma: [0, \infty) \to \mathbb R$ a Borel function with $c < \sigma < C$ for some constants $c, C > 0$.


Does there exist some $M > 0$ such that, conditional on $W\_t > -1$ for all $0 \leq t \leq 1$ and $W\_1 > M$,


$\int\_0^1 \sigma(t) \, dW\_t \geq 0$, almost surely?


**Remark:** The statement would be true if the integrator had uniformly bounded variation, but I suspect it does not hold for Brownian motion which has unbounded variation.



Conditioning the stochastic integral on the driving Brownian motion being large
No such $M$ exists for the following $\sigma$. Partition $[0,1]$ into countably many intervals, with endpoints $t\_0=0,t\_1,t\_2,...$ and let $\sigma$ take value 1 on the odd ones and value 2 on the even ones. Given any $M$, the following event $A\_M$ has positive probability:


$A\_M$ requires that the increments $W\_{t\_k}-W\_{t\_{k-1}}$ are in $(9,10)$ for the first $M$ odd values of $k$, and in $(-6,-7)$ for the first $M$ even values of $k$, with $W\_t-W\_{t\_{k-1}}>-1$ for $t \in [t\_{k-1},t\_k]$ when $0<k<2M$.


Then given $A\_M$, we have $W\_{t\_{2M}}>2M$, so the event $W\_1>M$ holds with high probability, yet on $A\_M$, the stochastic integral considered is $<-2M$ if you integrate up to $t\_{2M}$, and the integral over $[t\_{2M},1]$ is likely to be $<2M$.


