# Another large noise limit

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

• Can you disclose how you arrived at this conjecture, in particular, to $e^t$? Are you sure your expression for $C(\varepsilon)$ is correct? Apr 20, 2022 at 20:36
• 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 $M \sigma$. Apr 21, 2022 at 15:58
• As for motivation, this arose in trying to derive asymptotic bounds on the price of short maturity Asian options, modelled as geometric Brownian motion. Apr 21, 2022 at 15:58
• It would be quite enlightening to know intuitively why the current conjecture fails, while the simpler one in the linked post holds true. Apr 21, 2022 at 16:05
• 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. Apr 21, 2022 at 17:20


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 $$\begin{equation*} P(M_t for $$z, 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 $$e^{-lW_t}$$.