What is the solution of the SDE $$dX_t = 3dt + 2\sqrt{X_t} dB_t$$ when $X_0=0$ and B is a one dimensional brownian motion?
I know what the solution is when the dimension of the brownian motion is the same as the dimension of the bessel process but how do you find the solution when the dimensions are not equal?
Skipping all technicalities.
Ito's formula gives:
$d f(X_t) = f'(X_t) d X_t + \frac 12 f''(X_t) d \langle X\rangle_t$
Here $$dX_t = adt + b \sqrt{X_t} d B_t,$$
(with $a=3,b=2$).
Here $d \langle X\rangle_t =b^2 X_t dt$. Now let's take $f(x) =\sqrt{t}$, and let $Y_t = f(X_t) = \sqrt{X_t}$. Then
\begin{align*} d Y_t &=d f(X_t)\\ & = \frac{1}{2\sqrt{X_t}}(a dt + 2\sqrt{X_t} d B_t) - \frac 18 \frac{1}{\sqrt{X_t^{3}}} b^2 X_t dt\\ & = d B_t -(\frac a2 - \frac {b^2}{ 8})\frac{1}{\sqrt{X_t}} dt\\ & = d B_t + \left (a - \frac{b^2}{4}\right) \frac{1}{2Y_t} dt. \end{align*}
Since Bessel process of dimension $r$ satisfies
$$ d \rho_t = d B_t + (r-1) \frac{1}{2\rho_t} dt,$$
It follows that $Y$ is Bessel process of dimension $a-\frac{b^2}{4}+1$. When $a=3$ and $b=2$, this means $Y$ is a Bessel process of dimension $3$, hence $X$ is the square of a Bessel process of dimension $3$.