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 $\varepsilon \to 0$?
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.)
