Let the stochastic processes $U_t$ and $V_t$ be defined by the stochastic differential equations, we have: $$ dU_t = [ -aU_t - w_0 V_t]dt + BdW_t$$ $$dV_t = [-aV_t + w_0 U_t ] dt + Bd [W_t]_1 $$
where where $W_t, {W_t}_1$ denote two independent Wiener processes and $a, ω_0, B$ positive constants
To start my question I need to find $R_t^2 = U_t^2 +V_t^2 $ I then need to derive the stochastic differential equation for $R_t$, then I can begin with the rest of my question.
My lecture notes don't have any information on how to do this and I can't seem to find any online, does anyone have any suggestions on how to do this?
I would be grateful for any help you can provide.
Apply Ito's lemma to $U_t^2$ and $V_t^2$ :
$$d(U_t^2)=2U_tdU_t+d<U_t,U_t>$$ $$d(V_t^2)=2V_tdV_t+d<V_t,V_t>$$
We have $$d(U_t^2)=2U_tdU_t+B^2dt$$ $$d(V_t^2)=2V_tdV_t+B^2dt$$
$$d(U_t^2)=2U_t\left([ -aU_t - w_0 V_t]dt + BdW_t\right)+B^2dt$$ $$d(V_t^2)=2V_t\left([-aV_t + w_0 U_t ] dt + Bd [W_t]_1\right)+B^2dt$$ We add the two equations because we have $$dR_t^2=d(U_t^2)+d(V_t^2)$$
$$dR_t^2=-2a(U_t^2+V_t^2)dt+2B^2dt+2B(U_tdW_t+V_td [W_t]_1)$$
We introduce the process $$dX_t=\frac{U_tdW_t+V_td [W_t]_1}{\sqrt{U_t^2+V_t^2}}$$. We have $$d<X_t,X_t>=dt$$ because $W_t$ and $[W_t]_1$ are independent. By Levy's theorem, $X_t$ is a Brownian motion.
We can rewrite $$dR_t^2=-2a(U_t^2+V_t^2)dt+2B^2dt+\frac{2B}{\sqrt{U_t^2+V_t^2}}dX_t$$ We define $Y:=R_t^2$, $R_t=\sqrt{Y_t}$ Thus, $$dR_t^2=-2aR_t^2dt+2B^2dt+\frac{2B}{R_t}dX_t$$
$$dR_t=\frac{1}{2 \sqrt{Y_t}}dY_t+\frac{1}{2} \frac{-1}{4}Y_t^{-\frac{3}{2}}\frac{4B^2}{R_t^2}dt$$ $$dR_t=\frac{1}{2R_t}\left(-2aR_t^2dt+2B^2dt+\frac{2B}{R_t}dX_t\right)+\frac{1}{2} \frac{-1}{4}(R_t^2)^{-\frac{3}{2}}\frac{4B^2}{R_t^2}dt$$
Finally,
$$dR_t=\left(-aR_t+\frac{B^2}{R_t}-\frac{B^2}{2R_t^5}\right)dt+\frac{B}{R_t^2}dX_t$$