I've been thinking about the class of SDEs which have the form: $$dx(t)=f(x)(dt+\rho dW_t)$$ Clearly when $\rho=0$ this is just an ODE: $$\frac{dx}{dt}=f(x) $$ My question is, how does the solution to the SDE relate to the solution of the ODE?
The effect of the noise is just to randomly speed up or slow down the passage of time. Numerically it's like stochastically perturbing the length of the time-step.
My intuition is that if $y(t)$ is the solution to the ODE then the solution to the SDE can be written in the form $x(t)=y(S(t,W_t))$, where $S$ is some function which controls how quickly time passes. I may be mistaken and I have no idea how to prove it. I'm new to the world of stochastic calculus!
Edit: I can give an example to make this clearer. We can have a geometric Brownian motion which is in the form I'm talking about. It would be: $$dx(t)=\mu x(dt+\rho dW_t).$$ Now when $\rho=0$, the solution is $y(t)=y(0)\exp(\mu t)$. In this case we also have an explicit form for the SDE given by $$x(t)=x(0)\exp\left(\mu\left(1-\frac{\mu\rho^2}{2}\right)+\mu W_t\right).$$ So for this example my intuition is correct, we can write $$x(t)=y\left(\left(1-\frac{\mu\rho^2}{2}\right)t+\rho W_t\right)$$ I want to know whether this can generalised to arbitrary functions $f$.