Consider the SDE $$dX_t=\text{sign}(X_t)dB_t$$ with $X_0=0$ and where $$\text{sign}(x)=\begin{cases}-1&\text{if }x\leq0\\1&\text{if }x>0\end{cases}.$$ I am asked to find a weak solution to this SDE. From the SDE, if $X_t$ is positive then it evolves as the Brownian motion $B$ and if non-positive then evolves as the reflection of the Brownian motion $B$. As $X_0=0$, the evolution will start as the reflection of $B$, and then evolve as the above, which makes me tempted to say that $X=-B$ is a weak solution.
However, in addition to not being sure how to show this, I am also asked to show that there is no strong solution – but I think $X=-B$ would be a strong solution as it would be adapted to the filtration generated by $B$?
Best Answer
Start with a Brownian motion $X$ carried by some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and define $W_t = \int_0^t \mathrm{sgn}(X_s)dX_s$. By Lévy's characterisation theorem, $W$ is a Brownian motion. Observe then that by associativity of the stochastic integral: $$\mathrm{sgn}(X_s) dW_s = \mathrm{sgn}(X_s)^2 dX_s = dX_s$$ showing that $(X,W)$ is a weak solution of your proposed SDE.