If $A_t = cos(X_t)$ and $B_t = sin(X_t)$ find the infinitesimal increment for $Y_t = A_t^2 + B_t^2$

Question

If $X_t$ is Brownian motion, I'm not sure how to apply Ito's lemma to get $d Y_t$ for $ Y_t = A_t^2 + B_t^2$ where $A_t = cos(X_t)$ and $B_t = sin(X_t)$ in particular, I get confused because $sin^2(x) + cos^2(x) = 1$ so shouldn't $d Y_t$ be just 0 and I'm not sure how to handle the correlation between $A_t$ and $B_t$ which are clearly not independent.

Answer 1

Your reasoning using the trigonometric identity is correct and so applying Ito's lemma is unnecessary here. However it is possible and gives the same result.

In the case of $f(x) = x^2$, Ito's lemma tells you that $$dA_t^2 = 2 A_t dA_t + d \langle A \rangle_t$$

This means we want to find an SDE satisfied by $A_t$. Ito's lemma applied with the function $\cos$ tells us that $$dA_t = - \sin(X_t) dX_t - \frac{1}{2} \cos(X_t) dt$$ where we use the fact that $\langle X \rangle_t = t$. Using this, we also find that $d \langle A \rangle_t = \sin(X_t)^2 dt$.

Hence substituting into the first equation gives $$dA_t^2 = - 2 \cos(X_t) \sin(X_t) dX_t - \cos(X_t)^2 dt + \sin(X_t)^2 dt$$

Similar calculations show that $$dB_t^2 = 2 \cos(X_t) \sin(X_t) dX_t - \sin(X_t)^2 dt + \cos(X_t)^2 dt.$$ Hence, $$dY_t = dA_t^2 + dB_t^2 = 0$$ as you predicted.