let $C = (C_t = C_0 e^{\alpha W_t})_{t\geq 0}$ where $C_0 \geq 0$ (constant) and $W_t$ is a standard Brownian motion
let $X_t$ be a stochastic process such that :
$dX_t = \mu X_t dt + \sigma X_t d W^*_t$
where $W^*_t$ is a Brownian motion independent of $W_t$
it is asked to find the SDE that the process $Y = (Y_t = X_tC_t)_{t \geq 0}$ solves, using Ito's lemma I ended up here :
$dY_t = \alpha Y_t dW_t + \sigma Y_t dW^*_t + (\frac{\alpha^2}{2}+\mu)Y_tdt + d \langle X, C\rangle _t$
how to simplify $ d \langle X, C\rangle _t$
Best Answer
By Ito's lemma, we can compute that $$dC_t = \alpha C_0 e^{\alpha W_t} dW_t + \frac{\alpha^2}{2}C_0 e^{\alpha W_t} dt$$
Since the covariation of a finite variation process and a semimartingale is $0$, we have that $$\langle X, C \rangle_t = \langle \int_0^\cdot \sigma X_s dW_s^* , \int_0^\cdot \alpha C_0 e^{\alpha W_s} dW_s \rangle = \int_0^t \alpha \sigma C_0 X_s e^{\alpha W_s} d\langle W_s^*, W_s \rangle = 0$$ where the final inequality follows since $W$ and $W^*$ are independent BMs.