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:
- 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
$$\mathbb{E}[|\langle \int_0^t B_s – |B_s| \, dB_s, \int_0^t B_s – |B_s| \, dB_s\rangle_t|] \le 2t C \mathbb{E}[|B_t|] < +\infty$$ - 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.
Best Answer
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.