Endpoint of Brownian motion conditional on high maxima

Question

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

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

Answer 1

$\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\de}{\delta}$Yes, this is true:

By the reflection principle (see e.g. Proposition 2, for $M:=\max_{0\le t\le1}W_t$, \begin{equation} \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} \end{equation} where $\vpi$ is the standard normal pdf.

So, \begin{equation} P(A_\ep)=P(M\ge1/\ep)=2G(1/\ep)\sim2\ep\vpi(1/\ep), \end{equation} where $1-G$ is the standard normal cdf, and \begin{equation} \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} \end{equation} (as $\ep\downarrow0$). Hence, \begin{equation} \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} \end{equation}

---

The expression for $\de(\ep)$ in \eqref{1} can be rewritten as follows: \begin{equation} \de(\ep)=\frac\ep{r(1/\ep)}-1, \tag{2}\label{2} \end{equation} 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 \begin{equation} \de(\ep)=\ep^2 - 2 \ep^4 + 10 \ep^6 - 74 \ep^8 + 706 \ep^{10}+O(\ep^{12}) \end{equation} (as $\ep\downarrow0$) and \begin{equation} \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. \end{equation} 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 $\ep=0.1$.

Answer 2

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.