Solving SDE with sign function in drift term?

Question

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?

Answer 1

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.

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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.

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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.