Solving SDE with sign function in drift term? Consider the following SDE with $X_0 = 1$,
$$
dX_t = X_t\operatorname{sign}(X_t) \, dt + X_t \, dW_t,
$$
where $\operatorname{sign}(x) = \mathbb{1}_{\{x \ge 0\}} -\mathbb{1}_{\{x < 0\}}$. How am I supposed to solve this SDE?
 A: We assume that $X_t = f(t,W_t)$ is an Ito process, so that we can assume the Ito formula to find a candidate for the solution of the equation, by matching the $dt$ and $dW_t$ drifts. Write $f \equiv f(t,x)$. We proceed to do some rough work in the next part.

Note that by the Ito formula, we have :
$$
dX_t = \left(\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f}{\partial x^2}\right) dt + \frac{\partial f}{\partial x} dW_t
$$
Therefore, we must find $f$, such that $\frac{\partial f}{\partial x} = f$ and $\frac{\partial f}{\partial t} + \frac{1}{2}\frac{\partial f}{\partial x^2} = f \mbox{sign}\{f\}$, along with $f(1,0) = 1$. Indeed, from the first condition we obtain that $f(t,x) = g(t)e^{x}$ for some $t$, then by the second condition we obtain that $g(t) = e^{\frac t2}$ fits, therefore we get $X_t = e^{\frac t2 + W_t}$ to fit in the boundary condition.
Note that $\mbox{sign}(X_t) = 1$ a.s. since $X_t =e^{R_t}$ for a real-valued random variable. Hence that particular term never contributes to the equation.

Of course, this is rough work since we did not rigorously solve the ODEs above, for example. But then we have a guess : one can easily verify now that the $X_t$ obtained from the rough work is indeed a strong solution using the Ito formula.
This solution is unique, since the coefficients satisfy the Lipschitz and linear growth conditions.
