[This question](https://mathoverflow.net/questions/327167/lower-bound-for-integral-of-a-negativ-part-of-a-brownian-motion-with-respect-to) 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](https://arxiv.org/pdf/1408.3768.pdf), 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). 



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


