*Note: This question is closely related to an earlier question: [A large noise limit](https://mathoverflow.net/questions/417015/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


$$\lim\_{\varepsilon \to 0} \mathbb E\_{\mathbb P^\varepsilon} \big [\lvert \varepsilon W\_1 - 1 \rvert \big ]= 0?$$



Endpoint of Brownian motion conditional on high maxima
$\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\de}{\delta}$Yes, this is true:


By the reflection principle (see e.g. [Proposition 2](https://ocw.mit.edu/courses/15-070j-advanced-stochastic-processes-fall-2013/aca1518a09539a09ddd37428ab0d0268_MIT15_070JF13_Lec7.pdf), 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](https://eudml.org/doc/122588)) 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 $\ep=0.1$.


