Let $f: [0, 1] \to \mathbb R$ be a bounded, continuous function, and $W$ a standard Brownian motion.


Denote $Y := \int\_0^1 f(t) \, dW\_t$.


For each $\varepsilon > 0$, consider the conditioned random variable $Y\_\varepsilon := \varepsilon Y | \{W\_1 \geq \frac{1}{\epsilon}\}.$


**Question:** Is it true that $Y\_\varepsilon$ converges in law to the deterministic random variable $\int\_0^1 f(t) \, dt$ as $\varepsilon \to 0$?



A large noise limit
Let $\varphi$ be the standard normal density. Since
$P[W\_1 \ge x] =(1+o(1))\varphi(x)/x$ as $ x \to \infty$ by [1], we obtain for fixed $\delta>0$ that as $\epsilon \to 0$,
$$P[W\_1 \ge \epsilon^{-1}+\delta \,| \,W\_1 \ge \epsilon^{-1}] \le 2 \varphi(\epsilon^{-1}+\delta)/ \varphi(\epsilon^{-1}) \to 0 \,, $$
so in particular,


$(\*)$ given $W\_1 \ge \epsilon^{-1}$, we have $\epsilon W\_1 \to 1$ in probability as $\epsilon \to 0$.


Write $W\_t=tW\_1+B\_t$, where
$$\{B\_t: 0 \le t \le 1\}=\{W\_t-tW\_1 : 0 \le t \le 1\}$$ is a standard Brownian bridge in $[0,1]$, independent of $W\_1$. Then
$$Y = \int\_0^1 f(t) \, dW\_t = W\_1 \int\_0^1 f(t) \, dt + \int\_0^1 f(t) \, dB\_t \,,
$$
so by $(\*)$, given $W\_1 \ge \epsilon^{-1} \,,$ we have
$$ \epsilon Y = \epsilon W\_1 \int\_0^1 f(t) \, dt + \epsilon \int\_0^1 f(t) \, dB\_t \to \int\_0^1 f(t) \, dt
$$
as $\epsilon \to 0$ in probability (and hence also in law.)


[1] <https://en.wikipedia.org/wiki/Mills_ratio#cite_note-S-4>


