Assume $X(t)$ and $Y(t)$ are two stochastic processes satisfying $$dX=\mu_1(X)dt+\sigma_1(X)dW_1$$ $$dY=\mu_2(Y)dt+\sigma_2(Y)dW_2$$ where $W_1$ and $W_2$ are independent standard Brownian motions.
Since $dX(t)dY(t)=0$, then $X(t)$ and $Y(t)$ are uncorrelated. My question is: if $X(t)$ and $Y(t)$ are uncorrelated, can we say they are also independent for any fixed $t$ ? (I don't need independence for $X(s)$ and $Y(t)$, only for the same $t$.) If the answer is no, is there any other condition I can check besides uncorrelated to guarantee independence?
My thought is: at any time t, $X$ and $Y$ are two random variables. For any two random variables $X$ and $Y$, if for any bounded measurable functions $f$ and $h$, $f(X)$ and $h(Y)$are uncorrelated, then $X$ and $Y$ are independent. But for stochastic processes, to verify $f(X)$ and $h(Y)$are uncorrelated, the only way I can think of is to use Ito's formula, which requires twice differentiability, which is more than bounded&measurable.
Thank you for any comments. Sorry if the answer is straightforward.
There seem to be two questions here.
First, "$X$ and $Y$ are solutions to the following SDEs, where the driving Brownian motions are independent. Are $X$ and $Y$ independent?" Second, "If two processes $X$ and $Y$ are uncorrelated in the sense that $X(t)$ and $Y(t)$ are uncorrelated, then are $X(t)$ and $Y(t)$ independent?" The answer to the second question seems obviously no. (I'm not saying DEW intended the second question. I'm just trying to make the first question clearer.)
As to the first, if each SDE has a strong solution, $X$ and $Y$ will be independent.
One way to make the second question more interesting would be to rephrase it as "If two continuous local martingales $X$ and $Y$ are uncorrelated, are they independent?" for which the example given in Karatzas and Shreve (Brownian Motion and Stochastic Calculus, Section 3.4.C) is apposite: Let $$X_t\triangleq \int_0^t1_{W_s\ge 0}\,\textrm{d}W_s\quad\text{and}\quad Y_t\triangleq \int_0^t1_{W_s<0}\,\textrm{d}W_s$$ where $W$ is a scalar Brownian motion. Then $$\operatorname{cov}\left(X_t,Y_t\right)=\operatorname{E}(X_tY_t)=0.$$ But $$\langle X\rangle_t+\langle Y\rangle_t=t,$$ so $\langle X\rangle$ and $\langle Y\rangle$, or more fundamentally $X$ and $Y$, are not independent.