Density near at $0$ for the integral of the positive part of the Brownian motion

Question

This question was asked recently on MO and then deleted by the owner, user Aalon. I think the question deserves to be answered, which is what I will try to do here. Aalon was reading this paper, where it is claimed (on page 32, in line 5) that \begin{equation} P\Big(\int_0^1(W_t)_-\,dt\le\varepsilon\Big)\le\varepsilon \tag{1} \end{equation} for $\varepsilon>0$, where $W_\cdot$ is the standard Wiener process. The "reason" for this claim given in the mentioned paper (on page 32, in line 4) was that \begin{equation} \int_0^1(W_t)_-\,dt\ge\Big|\int_0^1 W_t\,dt\Big|. \tag{2} \end{equation} Aalon had doubts about both (1) and (2).

Answer 1

Thank you for your work on this question! I am one of the authors of the cited paper and, yes, we stated a wrong claim (1) in the appendix of this paper (let me remark that the results in the main part remain unaffected as they do not rely on the exact bound). I am sorry for the confusion.

I want to add here how a true upper bound for the probability can be obtained.

Denote $(W_t)_{t\ge 0}$ a standard Brownian motion and $(W_t)^+$ its positive part and $(W_t)^-$ its negative part, respectively. Observe that $\int_0^1(W_t)^-\,dt\stackrel{d}{=}\int_0^1(W_t)^+\,dt$, such that \begin{align*}P\Big(\int_0^1(W_t)^-\,dt\le x\Big)=P\Big(\int_0^1(W_t)^+\,dt\le x\Big)\,,x>0\,.\end{align*} Let us bound the second probability. For any $\epsilon>0$, the inequality \begin{align*}\int_0^1(W_t)^+\,dt\ge \int_0^1W_t \cdot\mathbb{1}(W_t>\epsilon)\,dt\ge \epsilon \int_0^1\mathbb{1}(W_t>\epsilon)\,dt\end{align*} leads us to \begin{align*}P\Big(\int_0^1(W_t)^+\,dt\le x\Big)&\le P\Big(\epsilon\int_0^1\mathbb{1}(W_t>\epsilon)\,dt\le x\Big)\\ &=P\Big(1-\int_0^1\mathbb{1}(W_t\le \epsilon)\,dt\le x/\epsilon\Big)\\ &=P\Big(\int_0^1\mathbb{1}(W_t\le \epsilon)\,dt\ge 1-x/\epsilon\Big)\,. \end{align*} For the last expression, we can use a nice generalization by Takács of Lévy's arc-sine law, see https://projecteuclid.org/download/pdf_1/euclid.aoap/1034968240. Using (15) and (16) from this work, we obtain that \begin{align*} P\Big(\int_0^1\mathbb{1}(W_t\le \epsilon)\,dt\ge 1-x/\epsilon\Big)&=\frac{1}{\pi}\int_{1-x/\epsilon}^{1}\frac{\exp(-\epsilon^2/(2u))}{\sqrt{u(1-u)}} du +2\Phi(\epsilon)-1\,, \end{align*} with $\Phi$ the cdf of the standard normal distribution. Thereby, we obtain that \begin{align*}P\Big(\int_0^1(W_t)^+\,dt\le x\Big)&\le \frac{1}{\pi}\int_{1-x/\epsilon}^{1}\frac{\exp(-\epsilon^2/(2u))}{\sqrt{u(1-u)}} du +2\int_0^{\epsilon}\frac{\exp(-u^2/2)}{\sqrt{2\pi}}du\,, \end{align*} and elementary bounds give the upper bound \begin{align*}P\Big(\int_0^1(W_t)^+\,dt\le x\Big)&\le \frac{2}{\pi}\sqrt{\frac{x}{\epsilon}}\frac{1}{\sqrt{1-x/\epsilon}} +\frac{2\epsilon}{\sqrt{2\pi}}\,. \end{align*} Since we are especially interested in the probability for $x\to 0$, the bound is not too bad for this case (for larger values of $x$ it can be crude). Choosing $\epsilon=x^{1/3}$, we obtain the upper bound \begin{align*}P\big(\int_0^1(W_t)^+\,dt\le x\big)&\le \frac{2}{\pi}x^{1/3}\frac{1}{\sqrt{1-x^{2/3}}} +\frac{2x^{1/3}}{\sqrt{2\pi}}\,. \end{align*} Observe that $P\big(\int_0^1(W_t)^+\,dt\le x\big)=\mathcal{O}(x^{1/3})$, $x\to 0$, is in line with the conjecture in (261) of Janson (https://projecteuclid.org/download/pdfview_1/euclid.ps/1178804352), where one can find an expansion with a precise constant of the leading term.

Answer 2

I consulted Igor Borisov, who is an expert on these (and other) matters and, in particular, wrote a paper (Theory Probab. Appl. 25 (1980), no. 3, 454--465 (1981)) on the existence of a bounded density for integral functionals of stochastic processes.

I. Borisov has then provided the following proof, reproduced here with his kind permission. This proof shows that inequality (1) is quite wrong; then of course inequality (2) is quite wrong as well, even if understood in terms of stochastic majorization.

Instead of $(W_\cdot)_-$ let us use $(W_\cdot)_+$, since these two processes are equal in distribution. Let \begin{equation*} M:=\max_{0\le t\le1}W(t). \end{equation*} Then for any real $x,c>0$ \begin{align*} P\Big(\int_0^1(W_t)_+\,dt\le x\Big)& \ge P\Big(\int_0^1(W_t)_+\,dt\le x,\ M\le c\Big) \\ & \ge P\Big(c\int_0^1 I\{W_t\ge0\}\,dt\le x,\ M\le c\Big) \\ & \ge P\Big(\int_0^1 I\{W_t\ge0\}\,dt\le\frac xc\Big)-P(M>c), \end{align*} where $I\{\cdot\}$ denotes the indicator. By Lévy's arcsine law, for real $z>0$ \begin{equation*} P\Big(\int_0^1 I\{W_t\ge0\}\,dt\le z\Big)=\frac2\pi\,\arcsin\sqrt z \end{equation*} Also, by the well-known formula (2), \begin{equation*} P(M>c)=\sqrt{\frac2\pi}\int_c^\infty e^{-t^2/2}\,dt\sim\sqrt{\frac2\pi}\,\frac{e^{-c^2/2}}c \end{equation*} as $c\to\infty$. Letting now $c=\sqrt{2\ln\frac1x}$ with $x\downarrow0$, we see that \begin{equation*} P\Big(\int_0^1(W_t)_+\,dt\le x\Big)\ge c_0\sqrt x\Big/\Big(\ln\frac1x\Big)^{1/4} \end{equation*} for some real $c_0>0$ and all small enough $x>0$, so that indeed (1) fails to hold.