Covariation of Wiener processes decomposed into uncorrelated parts

Question

Suppose $W_1,W_2$ are independent brownian motions, and $p$ is a continuously differentiable $[-1,1]$-valued function. How do I compute the covariation $[W_1,W_3], [W_2,W3]$ for $W_3$ defined as $$W_3(t) = \int_{0}^{t}p(s)dW_1(s) + \int_{0}^{t} \sqrt{1-p^2(s)}dW_2(s)$$? Any hints are appreciated

Answer 1

Note that, for $j\in \{1,2\},$ $W_j(t)=\int_0^t 1\textrm{d} W_j(s)$ so, applying the defining properties of the stochastic integral, we find that

$$ \langle W_j,W_3\rangle_t=\int_0^t p(s)\textrm{d}\langle W_j,W_1\rangle_s+\int_0^t \sqrt{1-p^2(s)}\textrm{d}\langle W_j,W_2\rangle_s $$

Now, being independent Brownian Motions, we see that $\langle W_j,W_k\rangle_s=1_{j=k}s$

In conclusion,

$$ \langle W_1,W_3\rangle_t=\int_0^t p(s)\textrm{d}s\\ \langle W_2,W_3\rangle_t=\int_0^t\sqrt{1-p^2(s)}\textrm{d}s $$