# Inequality on Brownian motions and Ito integrals

I know that Ito's integral isn't monotonic, i.e. if $$X \le Y$$ almost surely, then $$\int_0^t X_s \, dX_s \le \int_0^t Y_s \, dY_s$$ almost surely for $$X,Y$$ semi-martingales. However, is it true that $$\mathbb{P}\left(\int_0^t B_s \, dB_s \ge \int_0^t |B_s| \, dB_s\right) = 0$$ for any Brownian motion $$B$$?

My try is the following:

1. Prove that $$\int_0^t B_s - |B_s| \, dB_s$$ is a martingale in $$L^2$$ for any $$t \ge 0$$, because the co-variation is bounded, i.e. $$\mathbb{E}\left[\left|\langle \int_0^t B_s - |B_s| \, dB_s, \int_0^t B_s - |B_s| \, dB_s\rangle_t \right|\right] = \mathbb{E}\left[\left|\int_0^t B_s-|B_s|\, d\langle B,B\rangle_s\right|\right] < \infty$$ and since the co-variation of a Brownian motion is $$\langle B,B \rangle_s = s$$ we obtain $$\int_0^t B_s - |B_s| \, ds$$ which is bounded for any $$t$$ by $$-2B_s \le B_s-|B_s| \le 0$$ almost surely. Then, the co-variation is bounded from below by $$-2t\sup_{s\le t} B_s$$ and from above by $$0$$, and by Doob's inequality and taking the expected value of the co-variation we finally achieve $$\mathbb{E}[|\langle \int_0^t B_s - |B_s| \, dB_s, \int_0^t B_s - |B_s| \, dB_s\rangle_t|] \le 2t C \mathbb{E}[|B_t|] < +\infty$$
2. Since it is a martingale, we can use the Maximal inequality to bound the probability $$\mathbb{P}\left(\int_0^t B_s- |B_s| \, dB_s \ge \lambda \right) \le \frac{2}{\lambda} \mathbb{E}[|\int_0^t B_s- |B_s| \, dB_s|]$$ however from this we cannot infer that it goes to $$0$$ since we don't know if $$\mathbb{E}[|\int_0^t B_s- |B_s| \, dB_s|] < +\infty$$ nor that it goes to $$0$$.

Furthermore, I reckon that such method would suffice to show that both $$\mathbb{P}\left(\int_0^t B_s \, dB_s \ge \int_0^t |B_s| \, dB_s\right) = 0$$ and $$\mathbb{P}\left(\int_0^t B_s \, dB_s \le \int_0^t |B_s| \, dB_s\right) = 0$$ which can't be the case.

It can not be true, since the process $$M$$ defined by $$M_t = \int_0^t (|B_s|-B_s) ~\mathrm{d}B_s$$ is a martingale with expectation $$0$$. If it was non-negative almost surely, it would be null almost surely.
Informally, $$\mathrm{d}B_s$$ is not a measure, but when you integrate continuous adapted processes, you may approach it by random signed measures. That is why inequalities are not preserved by stochastic integration.
• I don't really get the sentence ''If it was non-negative almost surely, it would be null almost surely'', we don't know $M_t$ is non-negative or am I missing something? Apr 25, 2022 at 14:47
• You ask whether $\int_0^t B_s \mathrm{d}B_s \le \int_0^t |B_s| \mathrm{d}B_s$ almost surely or not. If the answer was yes, the difference would be non-negative almost surely. Apr 25, 2022 at 17:24