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
$$\lim_{\varepsilon \to 0} \mathbb E_{\mathbb P^\varepsilon} \big [\lvert \varepsilon W_1 - 1 \rvert \big ]= 0?$$
 A: $\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\de}{\delta}$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 $\ep=0.1$.
A: 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.
