Let $B$ be a standard Brownian motion. I want to prove that the SDE
\begin{equation}
X_t = \int_0^t 1_{\{ X_s \geq 0 \}} \, dB_s
\end{equation} has no solution on any set-up.
A hint I was given was to consider $f(x) = -x^3 1_{\{ x < 0 \}}$ and then $f(X_t)$. If we apply Ito's formula to $f$ we get
\begin{equation}
-X_t^3 \, 1_{\{ X_t <0 \}}= f(X_t) = -3 \left( \int_0^t X_s^2 1_{\{ X_s < 0 \}} dX_s + \int_0^t X_s 1_{\{ X_s < 0\}} ds \right)
\end{equation}
and so by plugging-in $X_s$ from the SDE
\begin{equation}
X_t^3 \, 1_{\{ X_t <0 \}} = 3 \int_0^t X_s 1_{\{ X_s < 0\}} ds,
\end{equation}
since the integral over $X_s$ will have $X_s^2 1_{\{X_s < 0 \}} 1_{\{X_s \geq 0 \}}$ as the integrand and $B_s$ as the integrator. Having this I don't really know how to continue. I also don't know what ways there are to show that a SDE does not have a solution.
Best Answer
You didn't apply Itô's formula correctly; it should read
$$-X_t^3 1_{\{X_t<0\}} = -3 \left( \int_0^t X_s^2 1_{\{X_s<0\}} \, dX_s + \int_0^t X_s 1_{\{X_s<0\}} \, d\color{red}{\langle X \rangle_s} \right) \tag{1}$$
where $\langle X \rangle$ denotes the quadratic variation. From $dX_s = 1_{\{X_s>0\}} \, dB_s$, it follows that $$d\langle X \rangle_s = (dX_s)^2= 1_{\{X_s \geq 0\}} \, ds.$$ Plugging this into $(1)$ and using that $1_{\{X_s \geq 0\}} 1_{\{X_s<0\}}=0$, we get
$$-X_t^3 1_{\{X_t<0\}}=0.$$
Since $-X_t^3$ is strictly positive on $\{X_t<0\}$, this means that $\mathbb{P}(X_t<0)=0$. Hence, $$X_t = \int_0^t 1_{\{X_s \geq 0\}} \, dB_s = \int_0^t dB_s = B_t,$$
which contradicts $X_t \geq 0$ (...since Brownian motion takes negative values).