Note: Here all processes take values in $[0, 1]$.


Let $W$ be a standard one dimensional Brownian motion, and $\sigma > 0$ a constant.


Let $X$ be the solution to the SDE


$$dX\_t = \sigma X\_t \, dW\_t$$


with $X\_0 = 1$ a.s.


For every $\varepsilon > 0$, let $A\_\varepsilon$ be the event $\{\text{max}\_{0 \leq t \leq 1} X\_t \geq \frac{1}{\varepsilon}\}$, and let $\mathbb P^\varepsilon$ be the probability measure defined by


$$\mathbb P^\varepsilon (E) =\frac{ \mathbb P(E \cap A\_\varepsilon)}{\mathbb P(A\_\varepsilon)}.$$


for all events $E$.


Define also for each $\varepsilon > 0$ the process $Y^\varepsilon$ by


$$Y^\varepsilon\_t := X\_t^{-C(\varepsilon)}.$$


where


$$C(\varepsilon) := \frac{\log \frac{1}{\varepsilon} +\frac{\sigma^2}{2}}{\sigma}.$$


**Question:** Is it true that


$$\lim\_{\varepsilon \to 0} \mathbb E\_{\mathbb P^\varepsilon} \left [\int\_{0}^1 \lvert Y\_t^\varepsilon - e^t \rvert \, dt \right ] = 0?$$


Where $\mathbb E\_{\mathbb P^\varepsilon}$ denotes the expectation under $\mathbb P^\varepsilon$.


*Remark:* It may be useful to adapt the method given by Yuval Peres in the answer [here](https://mathoverflow.net/questions/417015/a-large-noise-limit), however I do not know how to deal with the additional integral of $\sigma X\_t$ against the Brownian bridge.



Another large noise limit
$\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}\newcommand\num{\operatorname{num}}\newcommand\den{\operatorname{den}}\newcommand{\R}{\mathbb R}
\newcommand{\vpi}{\varphi}$The conjecture is not true in general.


The limit depends on $\si$. In particular, let us show that the limit in question is, not $0$, but $\infty$ if
\begin{equation\*}
 \si>2 + \sqrt3; \tag{-2}\label{-2}
\end{equation\*}
(also see the heuristics at the end of this answer).


Indeed, let $P:=\mathbb P$, $P\_\ep:=\mathbb P\_\ep$, $E\_\ep:=\mathbb E\_{P\_\ep}$,
\begin{equation\*}
 m:=\ln\frac1\ep\to\infty,\quad l:=m+\si^2/2,\quad\mu:=-\frac\si2, \quad r:=\frac m\si, 
\end{equation\*}
\begin{equation\*}
 M\_t:=\max\_{s\in[0,t]}(W\_s+\mu s),
\end{equation\*}
\begin{equation\*}
 B\_t:=\{M\_t\ge r\}. 
\end{equation\*}


Note that $(X\_t)$ is a geometric Brownian motion, so that
\begin{equation\*}
 X\_t=\exp(\si W\_t-\si^2 t/2), 
\end{equation\*}
whence
\begin{equation\*}
 Y\_t:=Y^\ep\_t=X\_t^{-C(\ep)}=e^{\si l t/2}e^{-l(W\_t+\mu t)} \tag{-1}\label{-1}
\end{equation\*}
and
\begin{equation\*}
 A\_\ep=\{M\_1\ge r\}\supseteq B\_t; 
\end{equation\*}
here and in the sequel, $t\in(0,1)$.


It follows that
\begin{equation\*}
 E\_\ep Y\_t\ge\frac\num\den, \tag{0}\label{0}
\end{equation\*}
where
\begin{equation\*}
 \num:=Ee^{-l(W\_t+\mu t)}1\_{B\_t},\quad \den:=P(A\_\ep). 
\end{equation\*}


Formula 1.4.8(1) on p. 256 in Handbook of Brownian Motion - Facts and Formulae, Second Edition, by Borodin and Salminen can be rewritten as
\begin{equation\*}
 P(M\_t<u,W\_t+\mu\_t\in dz) \\ 
 =\vpi\Big(\frac{z-\mu t}{\sqrt t}\Big)\frac{dz}{\sqrt t}
 -e^{2\mu u}\vpi\Big(\frac{z-2u-\mu t}{\sqrt t}\Big)\frac{dz}{\sqrt t} 
 \tag{1}\label{1}
\end{equation\*}
for $z<u$, where $\vpi$ is the standard normal pdf.


Using \eqref{1} (and noting that $M\_t\ge W\_t+\mu\_t$), one can find
\begin{equation\*}
\begin{gathered}
 \num=\int\_\R P(M\_t\ge r,W\_t+\mu\_t\in dz)e^{-lz} \\ 
 =
 \frac{1}{2} \left(\text{erf}\left(\frac{\si t \left(2 m+\si ^2+\si \right)-2 m}{2 \sqrt{2} \si 
 \sqrt{t}}\right)+1\right) \\ 
 \times \exp \left(\frac{1}{8} \left(\frac{4 m^2 (\si t-4)}{\si }+4 m (\si +1)
 (\si t-2)+(\si +2) \si ^3 t\right)\right) \\ 
 +\frac{1}{2} e^{\frac{1}{8} t \left(2 m+\si ^2\right)
 (2 m+\si (\si +2))} \text{erfc}\left(\frac{\si t \left(2 m+\si ^2+\si \right)+2 m}{2
 \sqrt{2} \si \sqrt{t}}\right)
\end{gathered}
\end{equation\*}
and
\begin{equation\*}
\begin{gathered}
 \den=P(M\_1\ge r) \\ 
 =
1-\frac{1}{2} e^{-m}
 \left(\text{erfc}\left(\frac{\frac{\si }{2}-\frac{m}{\si }}{\sqrt{2}}\right)-2\right)-\frac{1}{2}
 \text{erfc}\left(-\frac{\frac{m}{\si }+\frac{\si }{2}}{\sqrt{2}}\right). 
\end{gathered}
\end{equation\*}


If now $\si>2+\sqrt3$, then the interval $I\_\si:=(\frac{4\si-1}{\si^2},\min(1,\frac4\si))$ is nonempty and contained in the interval $(0,1)$. Moreover, for any $\si>2+\sqrt3$ and any $t\in I\_\si$, we have $\frac\num\den\to\infty$ (as $m\to\infty$), and hence, by \eqref{0}, $E\_\ep Y\_t\to\infty$. Thus, by Fatou's lemma, the limit in question is $\infty$. $\quad\Box$




---


Let me offer two competing heuristics to explain this result:


**Heuristic I: The large-deviation effect:** The large-deviation event $A\_\ep=\{M\_1\ge r\}=\{M\_1\ge m/\si\}$ (with $m\to\infty$) implies that $W\_t\approx mt/\si$. (In this case, this follows, for instance, from the independence of $W\_1$ from the Brownian bridge $B\_\cdot$, where $B\_t:=W\_t-tW\_1$ for $t\in[0,1]$.) So, on the event $A\_\ep$ we have $X\_t\approx\exp((m-\si^2/2)t)$ and hence
\begin{equation\*}
 Y\_t\approx\exp\Big(-\frac{m^2}\si\,(1+o(1))t\Big), \tag{2}\label{2}
\end{equation\*}
so that we may expect $\int\_0^1 Y\_t\,dt$ to be somewhat small on the event $A\_\ep$, on the order of $\si/m^2$. The smaller $\si$ is, the more pronounced this effect should be. I think we will indeed have $E\_\ep\int\_0^1 Y\_t\,dt\to0$ if \eqref{-2} does not hold, but I have not checked all the details here.


**Heuristic II: The counterbalancing effect of a re-weighting exponential factor:** However, if $\si$ is large enough, then the large-deviation effect of Heuristic I may be overshadowed by the factor $e^{-lW\_t}$ in the representation of $Y\_t$ in \eqref{-1}. Indeed, this exponential factor can be very large for negative values of $W\_t$ and negligible for positive values of $W\_t$, since $l\sim m\to\infty$. So, even though negative values of $W\_t$ are somewhat suppressed by the large-deviation condition $M\_1\ge m/\si$, this suppression may be counterbalanced by the re-weighting exponential factor $e^{-lW\_t}$, which greatly "favors" negative values of $W\_t$. This counterbalancing effect will be more successful when the large-deviation effect is less strong, that is, when the spread/diffusion coefficient $\si$ of the Brownian motion is large enough. In this case,
the conditional expectation of $Y\_t$ given $A\_\ep$ may resemble much more
the unconditional expectation of $Y\_t$, which is
\begin{equation\*}
 e^{l^2t/(2+o(1))}=e^{m^2t/(2+o(1))}, 
\end{equation\*}
which is very, very large (as $m\to\infty$).


Heuristic II is absent in the [previous setting](https://mathoverflow.net/questions/417015/a-large-noise-limit), where we do not have a very influential re-weighting exponential factor, such as the just considered factor $e^{-lW\_t}$.


