| Russian Mathematical Surveys |
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This article is cited in 1 scientific paper (total in 1 paper) Brief Communications On the Davis–Monroe problem P. A. Yaskov Steklov Mathematical Institute of Russian Academy of Sciences, Moscow
Russian version:
Uspekhi Matematicheskikh Nauk, 2022, Volume 77, Issue 6(468), DOI: https://doi.org/10.4213/rm10088 
Bibliographic databases:
    In this paper we solve the Davis and Monroe problem [2] about the values of $\varepsilon>0$ for which $\mu_\varepsilon$, the distribution of the Brownian motion with nonlinear drift $B_\varepsilon=B+\varepsilon F$ considered as a random element of $C[0,1]$, is equivalent to (is singular with respect to) the Wiener measure $\mu$ on $C[0,1]$, where $B=(B(t))_{t\in[0,1]}$ is the standard Brownian motion, $F(t)=\sqrt{(t-\tau)^+}$, $t\in[0,1]$, and $\tau$ is a random moment of time that is uniformly distributed on $[0,1]$ and independent of $B$. This problem is closely related to the theory of Gaussian multiplicative chaos. Note that by the Cameron–Martin theorem (see [4], Theorem 1.38) this kind of power drift is critical: if $F(t)\equiv t^\alpha$ and $\alpha>0$, then for all $\varepsilon>0$ the above measures $\mu_\varepsilon$ are equivalent to $\mu$ ($\mu_\varepsilon\sim \mu$) when $\alpha>1/2$ and singular with respect to $\mu$ ($\mu_\varepsilon\perp \mu$) when $\alpha\leqslant 1/2$. The case $\alpha=1/2$ becomes more delicate when the drift occurs at a random moment of time $\tau$. As shown in [2], if $F(t)\equiv\sqrt{(t-\tau)^+}$ and $\tau$ has the above properties, then $\mu_\varepsilon\ll\mu$ (that is, $\mu_\varepsilon$ is absolutely continuous with respect to $\mu$) for $\varepsilon<2$ and $\mu_\varepsilon\perp\mu$ for $\varepsilon>\sqrt{8}$ . Our main result, Theorem 1 below, covers the unexplored case of $\varepsilon\in [2,\sqrt{8}\,]$. Theorem 1. The following relations hold: $\mu_\varepsilon\sim\mu$ for $0<\varepsilon<\sqrt{8}$ and $\mu_\varepsilon\perp\mu$ for $\varepsilon\geqslant\sqrt{8}$. To prove Theorem 1 in the case when $\varepsilon\geqslant\sqrt{8}$ we use the following observation: $\varlimsup M_n(Z)\leqslant 0$ almost surely at $Z=B$ and $\varlimsup M_n(Z)>0$ almost surely at $Z=B_\varepsilon$ for $\varepsilon\geqslant\sqrt{8}$ , where the upper limit is taken as $n\to\infty$,
$$
\begin{equation*}
M_n(Z)=\max_{k=0,\dots,4^n-2^n-1}\biggl\{\frac1{\sqrt{\ln(4^n-k)}} \int_{(k+1)4^{-n}}^1\frac{dZ(t)}{\sqrt{t-k\,4^{-n}}}- \sqrt{2\ln(4^n-k)}\,\biggr\},
\end{equation*}
\notag
$$
and the integral is understood in the Riemann–Stieltjes sense.
To prove Theorem 1 in the case when $\varepsilon<\sqrt{8}$ we consider the process $B_{\varepsilon,\delta}=B+\varepsilon F_\delta$ with values in the measurable space $(C[0,1],\mathcal C)$ and its distribution $\mu_{\varepsilon,\delta}$, where $\delta>0$, $F_\delta(t)=\sqrt{(t-\tau)^++\delta}-\sqrt{\delta}$ for $t\in[0,1]$, and $\mathcal C$ is the Borel $\sigma$-algebra in the space $C[0,1]$ with the uniform metric. Then it is obvious that $B_{\varepsilon,\delta}\to B_{\varepsilon}$ almost surely in the uniform metric as $\delta\to0$. The rest of the proof that $\mu_\varepsilon\ll \mu$ comes down to verifying that $\mu_{\varepsilon,\delta}\ll\mu$ for all $\delta>0$ and
$$
\begin{equation}
\frac{d\mu_{\varepsilon,\delta}}{d\mu} \text{ converges in }L_1(C[0,1],\mathcal C,\mu)\text{ as }\delta\to0.
\end{equation}
\tag{1}
$$
Further, let $(\Omega,\mathcal F,\mathsf P)=(C[0,1],\mathcal C,\mu)$ be a probability space with the Brownian motion $B$ defined by $B(\omega)\equiv \omega$, $\omega\in\Omega$. According to the Cameron–Martin–Girsanov formula, we have $\mu_{\varepsilon,\delta}\ll\mu$ and $\mu_{\varepsilon,\delta}/d\mu=I(\varepsilon X_\delta/2)$, where $X_\delta(t):=\displaystyle\int_t^1 \frac{dB(s)}{\sqrt{s-t+\delta}}$ , $t\in [0,1]$, and $I(X):=\displaystyle\int_0^1\mathcal E(X(t))\,dt$ for the Gaussian process $X=(X(t))_{t\in[0,1]}$ on $(\Omega,\mathcal F,\mathsf P)$ with values in $C[0,1]$ and $\mathcal E(\xi):=e^{\xi-\mathsf E\xi^2/2}$, for the Gaussian variable $\xi$ (here $\mathsf E$ is the expectation with respect to the measure $\mathsf P$). The model of our interest corresponds to $\delta=0$. By setting $\delta=0$, under the integral sign in $I(\varepsilon X_0/2)$ we obtain formally a ‘Gaussian’ process $X_0(t):=\displaystyle\int_t^1 \frac{dB(s)}{\sqrt{s-t}}$ , whose covariance function is non-negative and has the form $K(s,t)=-\ln|s-t|+g(s,t)$ for some continuous function $g$ on $[0,1]^2\setminus (1,1)$. There is no such process in the usual sense since $K(t,t)=\infty$ for all $t\in[0,1)$. However, $X_0$ can be defined as the family of Gaussian variables $\displaystyle\int_0^1X_0(t)\,\rho(dt)$ indexed by non-negative measures $\rho$ on $(0,1)$ with the standard Borel $\sigma$-algebra such that $\displaystyle\iint_{(0,1)^2}K(s,t)\,\rho(ds)\,\rho(dt)<\infty$. It is well known in the theory of Gaussian multiplicative chaos (see, for instance, [1]) that if $g$ in the representation of $K$ is continuous on $[0,1]^2$, then for $\varepsilon<\sqrt{8}$ the integral $I(\varepsilon X_0/2)$ can be defined as the limit of $I(\varepsilon X_{\delta,G}/2)$ in $L_1$ as $\delta\to0$, where $X_{\delta,G}(t)=\displaystyle\int_0^1 X_0(s)\,d\biggl(1- G\biggl(\dfrac{t-s}{\delta}\biggr)\biggr)$ and $G\colon\mathbb R\to[0,1]$ is a probability distribution function with support in $(0,1)$ and such that $\displaystyle\sup_{0\leqslant t\leqslant 1/2}\displaystyle\int_0^1 \ln\dfrac1{|t-s|}\,dG(s)<\infty$. The above limit does not depend on the particular choice of $G$. Formally, our case is not covered by the known results. Nevertheless, it can be analysed by modifying the method from [1] (see the proof of Theorem 4.1.1 in [3]). In this way we can establish (1), from which it follows that $\mu_{\varepsilon}\ll\mu$ and $d\mu_\varepsilon/d\mu=I(\varepsilon X_0/2)$ is the limit of $I(\varepsilon X_{\delta}/2)$ in $L_1$ for $\varepsilon<\sqrt{8}$. To prove that $\mu\ll \mu_{\varepsilon}$ we only need to check that $\mu(I=0)=0$ for $I=I(\varepsilon X_0/2)$. This follows from the estimate below:
$$
\begin{equation}
I\geqslant \frac{I_1(\gamma)}2\inf_{t\in [0,1/2-\gamma]} \mathcal E\biggl(\frac{\varepsilon Z_0(t)}{2}\biggr)+ \frac{I_2}{2} \quad\text{for all } \gamma\in\biggl(0,\frac{1}{2}\biggr),
\end{equation}
\tag{2}
$$
where $I_1(\gamma)=2\displaystyle\int_{0}^{1/2-\gamma} \mathcal E\biggl(\dfrac{\varepsilon Y_0(t)}{2}\biggr)\,dt$, $\gamma\in\biggl(0,\dfrac{1}{2}\biggr)$, and $I_2=2\displaystyle\int_{1/2}^1 \mathcal E\biggl(\frac{\varepsilon X_0(t)}{2}\biggr)\,dt$ are independent (the integrals $I_1(\gamma)$ and $I_2$ are found by taking limits in $L_1$ similarly to $I$), and we have and
$$
\begin{equation*}
Z_0(t):=\int_{1/2}^1\frac{dB(s)}{\sqrt{s-t}}= \frac{B(1)}{\sqrt{1-t}}-\frac{B(1/2)}{\sqrt{1/2-t}}+ \frac{1}{2}\int_{1/2}^1\frac{B(s)\,ds}{(s-t)^{3/2}}
\end{equation*}
\notag
$$
for $t\in[0,1/2)$. In this case, as follows from a suitable change of variables under the integral sign and the properties of the Brownian motion, $I_1(0)$ and $I_2$ have the same distribution as $I$. In addition, since trajectories of the Gaussian process $Z_0$ are continuous almost surely, the infimum in (2) is positive almost surely for all $\gamma\in(0,1/2)$. In particular, all this allows us to show that
$$
\begin{equation*}
\begin{aligned} \, \mu(I=0)&\leqslant \mu\bigl(I_1(\gamma)=0 \ \forall\,\gamma\in(0,1/2);I_2=0\bigr) \\ &=\mu\bigl(I_1(0)=0;I_2=0\bigr)=[\mu(I=0)]^2, \end{aligned}
\end{equation*}
\notag
$$
that is, $\mu(I=0)=0$ or $1$. Since $\mathsf E I=\lim_{\delta\to 0}\mathsf E I(\varepsilon X_\delta/2)=1$, we have $\mu(I=0)=0$; hence $\mu\ll \mu_{\varepsilon}$.
Citation in format AMSBIB
\Bibitem{Yas22} |
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