# pr.probability – Convergence of conditioned stochastic integral

Let $$B_t$$ be a standard Brownian motion, $$f: (0, T) to mathbb R$$ a bounded Borel measurable function, and $$X_t$$ a process independent of $$B_t$$ with sample paths that almost surely start at 0, and are continuous and of bounded variation.

For every $$varepsilon > 0$$, denote by $$Gamma_{varepsilon}$$ the event

$$sup_{t in (0,T)} |B_t – X_t| leq varepsilon.$$

Denote by $$Y_t$$ and $$Z_t$$ the processes $$int_{0}^t f(t) dB_t$$ and $$int_{0}^t f(t) dX_t$$ respectively, where the latter is a path-by-path ordinary Lebesgue-Stiltjes integral.

We write also $$Y$$ and $$Z$$ for the processes above, considered as $$C((0, T))$$-valued random variables.

Question: Is it true that the conditioned random variables $$Y|Gamma_{varepsilon}$$ converge to $$Z$$ in law as $$varepsilon to 0$$?

Note: Here we equip $$C((0, T))$$ with the topology of uniform convergence.

Remarks:

1. The conditions on $$X_t$$ ensure that the event $$Gamma_{varepsilon}$$ has nonzero probability for every $$varepsilon > 0$$, so that conditioning on it is well defined.

2. I believe that this is true, and the strategy should be to approximate the integral with respect to $$B_t$$ with elementary processes, and use this same approximation to write the Lebesgue-Stiltjes integral, and then bound the difference between the increments on $$Gamma_{varepsilon}$$. However the technicalities to do with convergence in law on $$C(0, T)$$ elude me.