Hey i have a fairly simple task and please just comment if it is done correctly. I have to show that if two processes
$dX(t)=a_X(t)dt+b_X(t)dW_t$
$dY(t)=a_Y(t)dt+b_Y(t)dW_t$
are equal then $a_X=a_Y$ and $b_X=b_Y$.
I calculate:
$d((X(t)-Y(t))^2=...=2(a_X(t)-a_Y(t))(X_t-Y_t)dt+(b_X-b_Y)^2dt+2(b_X-b_Y)(X_t-Y_t)dW_t$
$\sqrt{(X(t)-Y(t))^2}=X(t)-Y(t)$ so we since on the left we have $0$, then:
$2(a_X(t)-a_Y(t))=0$, so $a_X(t)=a_Y(t)$
$(b_X-b_Y)^2=0$ so $b_X=b_Y$
$a_X,a_Y: P(\{w:\int_0^T|a(s,w)|ds<\infty\})=1$ and $b_X,b_Y\in M_{[0,T]}^2=\left\{f:[0,T]\times\Omega\to\mathbb{R}:\text{f is adapted}, E\left(\int_0^Tf^2(t)dt\right)<\infty\right\}$
Is my proof correct or do you have any doubts?