# Endpoint of Brownian motion conditional on high maxima

Note: This question is closely related to an earlier question: A large noise limit.

Let $$W$$ be a standard one dimensional Brownian motion.

For every $$\varepsilon > 0$$, let $$A_\varepsilon$$ denote the event

$$\{\underset{0 \leq t \leq 1}{\text{max}} W_t \geq \frac{1}{\varepsilon}\} \;,$$

and let $$\mathbb P^\varepsilon$$ be the probability measure given by

$$P^\varepsilon (E) = \frac{\mathbb P(E \cap A_\varepsilon)}{\mathbb P(A_\varepsilon)} \;,$$

for all measurable events $$E$$.

We denote by $$\mathbb E_{\mathbb P^\varepsilon}$$ the expectation under $$\mathbb P^\varepsilon$$.

Question: Is it true that

$$\lim_{\varepsilon \to 0} \mathbb E_{\mathbb P^\varepsilon} \big [\lvert \varepsilon W_1 - 1 \rvert \big ]= 0?$$

• This conditional distribution is a well known consequence of the reflection principle, and easy to work with.
– mike
Jun 13, 2022 at 11:50


By the reflection principle (see e.g. Proposition 2, for $$M:=\max_{0\le t\le1}W_t$$, \begin{aligned} &P(M\ge1/\ep,W_1\in dx) \\ &=P(W_1\in2/\ep-dx)1(x<1/\ep)+P(W_1\in dx)1(x>1/\ep) \\ &=\vpi(2/\ep-x)1(x<1/\ep)dx+\vpi(x)1(x>1/\ep)dx, \end{aligned} where $$\vpi$$ is the standard normal pdf.

So, $$$$P(A_\ep)=P(M\ge1/\ep)=2G(1/\ep)\sim2\ep\vpi(1/\ep),$$$$ where $$1-G$$ is the standard normal cdf, and \begin{aligned} &E1(M\ge1/\ep)|\ep W_1-1| \\ &=\int_{\mathbb R}P(M\ge1/\ep,W_1\in dx)|\ep x-1| \\ &=2\big(\ep\vpi(1/\ep)-G(1/\ep)\big)=o(\ep\vpi(1/\ep)) \end{aligned} (as $$\ep\downarrow0$$). Hence, $$$$\de(\ep):= E_{P_\ep}|\ep W_1-1|=\frac{E1(M\ge1/\ep)|\ep W_1-1|}{P(A_\ep)}\to0. \tag{1}\label{1}$$$$

The expression for $$\de(\ep)$$ in \eqref{1} can be rewritten as follows: $$$$\de(\ep)=\frac\ep{r(1/\ep)}-1, \tag{2}\label{2}$$$$ where $$r:=G/\vpi$$ is the Mills ratio. Then one can use known asymptotic expansions of and bounds on the Mills ratio (see e.g. this paper) to get asymptotic expansions of and bounds on $$\de(\ep)$$. For instance, from Propositions 1.3 and the simplest two cases of formula (1.8) (with $$m=1,2$$) of that paper we get $$$$\de(\ep)=\ep^2 - 2 \ep^4 + 10 \ep^6 - 74 \ep^8 + 706 \ep^{10}+O(\ep^{12})$$$$ (as $$\ep\downarrow0$$) and $$$$\frac{\ep^2}{1 + 2 \ep^2}<\frac{\ep^2 + 7 \ep^4}{1 + 9 \ep^2 + 8 \ep^4} <\de(\ep)<\frac{\ep^2 + 3 \ep^4}{1 + 5 \ep^2}<\ep^2.$$$$ One may note that the difference between the upper and lower bounds $$\dfrac{\ep^2 + 3 \ep^4}{1 + 5 \ep^2}$$ and $$\dfrac{\ep^2 + 7 \ep^4}{1 + 9 \ep^2 + 8 \ep^4}$$ on $$\de(\ep)$$ is $$\dfrac{24 \ep^8}{1 + 14 \ep^2 + 53 \ep^4 + 40 \ep^6}$$, which is $$<0.00000024$$ if $$\ep=0.1$$.

• Thank you for your answer! Around the same time, Yuval Peres also wrote an answer on a different website. I will be adding his answer too shortly. Jun 13, 2022 at 14:43
• This seems such a remarkable result. Given that the maximum was greater than $1/\epsilon$, the endpoint is within $O(\epsilon)$ of $1/\epsilon$ with high probability via a simple Markov bound (the endpoint $W_1$ is within $100 \epsilon$ of $1/\epsilon$ with probability of more than $.99$). Is it that with overwhelming probability the first hitting time of the maximum is close to $1$ as $\epsilon$ shrinks? That would ensure that $W_1$ cannot fall too far below the maximum value attained (which is $\geq1/\epsilon$), but I struggle to see why $W_1$ can't be $O(1)$ above $1/\epsilon$. Jun 13, 2022 at 21:02
• @user196574 : Yes, $W_1$ must be close to the maximum $M$ with large (conditional) probability -- given what $M$ takes a large value. The entire trajectory $(W_t\colon t\in[0,1])$ will then be close to the linear function $(tW_1\colon t\in[0,1])$, again conditionally on $M$ (or, equivalently, $W_1$) taking a large value. The latter statement follows by (say) the independence of the Brownian bridge $(W_t-tW_1\colon t\in[0,1])$ from $W_1$. Jun 13, 2022 at 21:18
• @NateRiver : Right, conditioning on $M$ taking large values is almost the same as conditioning on $W_1$ taking large values, because, by the first display in my answer, $P(W_1>b-h|M>b)\to1$ if $b\to\infty$ and $bh\to\infty$, so that (say) $h=\tfrac{\ln b}b$ will do here. Also, trivially, $P(M>b|W_1>b)=1$. Aug 15, 2022 at 16:00
• @NateRiver : Write $\varepsilon W_t-t=A_t+tB$, where $A_t:=\varepsilon(W_t-tW_1)$ and $B:=\varepsilon W_1-1$. We showed that $E(|B|\,| M\ge 1/\varepsilon)\to0$. Clearly, $E(\sup_{t\in[0,1]}|A_t|\,|W_1\ge1/\varepsilon)=E\sup_{t\in[0,1]}|A_t|\to0$. I thought that from here it should be rather easy to get $E(\sup_{t\in[0,1]}|A_t|\,| M\ge 1/\varepsilon)\to0$, since the conditioning on $M\ge1/\varepsilon$ is almost the same as the conditioning on $W_1\ge1/\varepsilon$, but I have not checked those details. Dec 1, 2022 at 17:30

Iosif Pinelis has already posted an answer, but here is an alternate answer communicated to me by Yuval Peres on a different website. Any typoes/mistakes are most definitely mine.

Write

$$\tau = \text{min}\{t > 0 \, : \, W_t \geq \frac{1}{\varepsilon}\}.$$

By the reflection principle, we have

$$\mathbb P(\tau \leq 1) = \mathbb P(A_\varepsilon) = 2 \Phi (-\frac{1}{\varepsilon}),$$

where $$\Phi(x) := \int_{-\infty}^x (2\pi)^{-\frac{1}{2}} e^{-\frac{t^2}{2}} \, dt$$ denotes the CDF of the standard normal distribution.

Using the strong Markov property at time $$\tau$$, we have that $$|W_1 - W_\tau|$$ is a standard half normal random variable with parameter $$\sigma = 1 - \tau$$, independent of $$\mathcal F_\tau$$.

Thus we compute

$$\mathbb E \big [|W_1 - \frac{1}{\varepsilon}| \, \big | \, \tau \leq 1 \big ]$$ $$= \mathbb E \big [|W_1 - W_\tau| \, \big | \, \tau \leq 1 \big ]$$ $$= \frac{\mathbb E[\mathbf 1_{\{\tau \leq 1\}} |W_1 - W_\tau|]}{\mathbb P(\tau \leq 1)}$$ $$= \frac{\mathbb E[\mathbf 1_{\{\tau \leq 1\}} \mathbb E[ |W_1 - W_\tau| \big | \sigma(\tau) ]]}{\mathbb P(\tau \leq 1)}$$ $$= \frac{\mathbb E \big [\mathbf 1_{\{\tau \leq 1\}} \sqrt{\frac{2}{\pi} (1 - \tau)} \big ]}{\mathbb P(\tau \leq 1)}$$ $$\leq \sqrt \frac{2}{\pi}.$$

Thus

$$\mathbb E_{\mathbb P^\varepsilon} [|\varepsilon W_1 - 1|] = \mathbb E[|\varepsilon W_1 - 1 | \big | A_\varepsilon] \leq \varepsilon \sqrt{\frac{2}{\pi}}$$

which tends to $$0$$, as desired.

In fact, the bound above may be further improved as observed by Yuval Peres.

We compute, for any $$0 < \delta < 1$$,

$$\mathbb P[\tau \leq 1 - \delta \, | \, \tau \leq 1] = \Phi \left ( -\frac{1}{\sqrt{ 1 - \delta} . \varepsilon } \right ) \big /\Phi \left ( -\frac{1}{\varepsilon } \right )$$

$$\leq \frac{\text{exp} \big (\frac{1}{2 \varepsilon^2} \big )}{\text{exp} \big (\frac{1}{2 \varepsilon^2 (1 - \delta)} \big )}$$|

where the inequality above is due to the fact that $$\Phi(-r) e^{\frac{r^2}{2}}$$ is decreasing in $$r$$.

Thus,

$$\mathbb E \left [ \frac{\sqrt{1 - \tau}}{\varepsilon} \, \big | \, \tau \leq 1 \right ] = \int_0^\infty \mathbb P(\sqrt{1 - \tau} \geq r\varepsilon \, | \, \tau \leq 1) \, dr$$

$$\leq \int_0^\infty \text{exp}\big (-\frac{r^2}{2} \big )\, dr = \sqrt{\frac{\pi}{2}}$$

And so finally,

$$\mathbb E_{P^\varepsilon} [|\varepsilon W_1 -1|] \leq \sqrt{\frac{2}{\pi}}\sqrt{\frac{\pi}{2}} \varepsilon^2 = \varepsilon^2$$

and this is sharp asymptotically.

• that's nice, can you please provide a link to the website? Jun 13, 2022 at 16:43
• Ah, it is a private mathematics group on Facebook. Here is the link to the group if you are interested: facebook.com/groups/1923323131245618 Jun 13, 2022 at 16:47
• I don't have a Facebook account, so thanks for posting; nice question and nice answers too. Jun 13, 2022 at 16:56
• Thanks, and you're welcome! Jun 13, 2022 at 17:02
• We can also get sharp bounds (sharp of any order) and asymptotic expansions for the conditional expectation in question by using (i) the exact expression for this expectation and (ii) sharp bounds and asymptotic expansions for the MIlls ratio. This remark is now added to my answer. Jun 13, 2022 at 17:36