Suppose I have two process $X_t:=\int_0^t f(X_s,s)dB_s$ and $Y_t:=\int_0^t g(Y_s,s)dB_s$. Let both $g,f$ be continuously differentiable functions. I am aware of the result that implies that the corresponding quadratic variations are given by $[X]_t=\int_0^tf^2(x,s)ds$ and $[Y]_t=\int_0^tg^2(x,s)ds$.
I am interested in a linear combination of a Bernouilli variable $W$ (known at $t=0$) and these processes, and want to calculate the quadratic variation of the process $WX_t + (1-W)Y_t$. I do not know much about $[X,Y]_t$ (the covariation). Is it still possible to compute the quadratic variation I am intersted in?
Thanks in advance.
Since either $W = 0$ or $W = 1$, $Z_t := W X_t + (1-W)Y_t$ will either equal $X_t$ or $Y_t$ for all $t$. Hence $[Z]_t = W [X]_t + (1-W)[Y]_t$.
This could be verified by checking that $Z_t^2 - [Z]_t$ is a local martingale, or by applying the result that $[A+B]_t = [A]_t+[B]_t + 2 [A,B]_t$ for any processes $A,B$ to $A=WX$ and $B=(1-W)Y$ (and using that $W(1-W)=0$ to get $[WX,(1-W)Y]_t=0$).