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

$$Y^\varepsilon_t := X_t^{-C(\varepsilon)}.$$


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

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, however I do not know how to deal with the additional integral of $\sigma X_t$ against the Brownian bridge.

  • $\begingroup$ Can you disclose how you arrived at this conjecture, in particular, to $e^t$? Are you sure your expression for $C(\varepsilon)$ is correct? $\endgroup$ Commented Apr 20, 2022 at 20:36
  • $\begingroup$ Thank you for your detailed answer. I arrived at the conjecture by guessing that conditional on $\max_{0 \leq t \leq 1} W_t \geq M$, then formally $\frac{1}{M} dW_t$ behaves as $dt$ as $M \to \infty$. (The condition on the corresponding max for $X$ translates directly to a condition of this form on $W$.) Thus the SDE $dX_t = \sigma X_t dW_t$ formally becomes $dX_t = M \sigma X_t dt$, which admits solution $X = e^{M \sigma t}$. The normalisation $C(\varepsilon)$ is then simply $M \sigma$. $\endgroup$
    – Nate River
    Commented Apr 21, 2022 at 15:58
  • $\begingroup$ As for motivation, this arose in trying to derive asymptotic bounds on the price of short maturity Asian options, modelled as geometric Brownian motion. $\endgroup$
    – Nate River
    Commented Apr 21, 2022 at 15:58
  • $\begingroup$ It would be quite enlightening to know intuitively why the current conjecture fails, while the simpler one in the linked post holds true. $\endgroup$
    – Nate River
    Commented Apr 21, 2022 at 16:05
  • 1
    $\begingroup$ I will try to add such heuristics later, on why one conjecture holds and the other fails. I think this can also be discerned by analyzing the sources and contributions of the various terms in the expressions for $\text{num}$ and $\text{den}$ in my answer, even though I have not done such work. $\endgroup$ Commented Apr 21, 2022 at 17:20

1 Answer 1


$\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}\newcommand\num{\operatorname{num}}\newcommand\den{\operatorname{den}}\newcommand{\R}{\mathbb R} \newcommand{\vpi}{\varphi}$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, where we do not have a very influential re-weighting exponential factor, such as the just considered factor $e^{-lW_t}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.