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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 $\varepsilon \to 0$?

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  • $\begingroup$ If you condition on W_1 then you are integrating with respect to brownian bridge, and you get a normal distribution you can find explicity. $\endgroup$
    – mike
    Feb 27, 2022 at 7:15
  • $\begingroup$ Thats if i condition on $W_1$ being a certain fixed value though. $\endgroup$
    – Nate River
    Feb 27, 2022 at 9:45
  • $\begingroup$ Then of course you have to uncondition. I don't know how it works out bt the mean is certainly what you are looking for $\endgroup$
    – mike
    Feb 27, 2022 at 15:44
  • $\begingroup$ Since the conditional distribution of $Y_{\epsilon}|W_1$ is normal distribution $N(\epsilon W_1(\int_0^1f(t)dt), \epsilon^2[\int_0^1f^2(t)dt-(\int_0^1f(t)dt)^2])$, $Y_\epsilon \nrightarrow \int_0^1f(t)dt$. $\endgroup$
    – JGWang
    Mar 2, 2022 at 8:19

1 Answer 1

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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

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    $\begingroup$ Beautiful. The convergence you mention of $\epsilon W_1$ to $1$ is indeed what I had in mind, but didn’t see how to continue. $\endgroup$
    – Nate River
    Mar 23, 2022 at 8:23

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