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.


  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.