How do I use Ito's formula to solve this $ dX = \frac{1}{3} X^{1/3} dt + X^{2/3} dB $

Question

The title says it all. I tried many functions for Ito's formula but failed. $f(B,t)= B^{3}$ gives something close but that is not it. I'll appreciate any help.

Answer 1

It makes sense to look for solutions in the form $X_t = Y_t^3$ so our equation simplifies to $dX_t = \frac 13 Y_t dt + Y_t^2 dB_t$. We should also have that $dY_t = a_t dt + b_t dB_t$. Now applying Ito to find $dY_t^3$ gives

\begin{align*} dY_t^3 &= 3 Y_t^2 dY_t + 3 Y_t d \langle Y,Y \rangle_t = 3 Y_t^2 a_t dt + 3 Y_t^2 b_t dB_t + 3 Y_t b_t^2 dt \\ dX_t &= \frac 13 Y_t dt + Y_t^2 dB_t. \end{align*}

Comparing the $dB_t$ term gives that $b_t = \frac 13$, and then comparing the $dt$ terms gives $\frac 13 Y_t = 3Y_t^2 a_t + 3 Y_tb_t = 3Y_t^2 a_t + \frac 13 Y_t$ so we can simply set $a_t = 0$. Hence our solution is $Y_t = \frac 13 B_t$ and $X_t = \left(\frac 13 B_t\right)^3$.