# Conditioning the stochastic integral on the driving Brownian motion being large

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.

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 $$0.
Then given $$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$$.