(I am not sure what would be the proper title for the question)
Let us have a 2D random process $(A(t), X(t))$ and take both $A$ and $X$ to be just reals. We can write a conditional expectation of $A$ with respect to $X = x$: $$ \langle a \rangle (x,t) = \bigl< A(t) \bigm| X(t) = x \bigr>. $$ I am interested in expressing the following quantity $$ \bigl< A(t + \Delta t) \bigm| X(t) = x \bigr>. $$ in terms of $\langle a \rangle (x,t)$. My current guess is that $$ \begin{multline} \bigl< A(t + \Delta t) \bigm| X(t) = x \bigr> = \langle a \rangle (x,t)\\[1ex] + \Delta t \frac{\partial}{\partial t} \langle a \rangle (x,t) + \frac{\partial}{\partial x} \Bigl< [X(t + \Delta t) - X(t)]\,A(t) \Bigm| X(t) = x \Bigr> + \ldots. \end{multline} $$
Is this true and if so, why? (Take $A$ and $X$ to be smooth enough or whatever is needed). Just a reference would suffice.
UPD Considering uniformly distributed $X$ would suffice. If you take $X(t)$ to be a molecule's position in a flow, the flow is in-compressible.
My current thoughts were, that by introducing $\Delta X = X(t+\Delta t) - X(t)$, I can go as $$ \begin{aligned} \bigl< A(t + \Delta t) \bigm| X(t) = x \bigr> &= \bigl< A(t + \Delta t) \bigm| X(t + \Delta t) - \Delta X = x \bigr>\\[1ex] &= \Bigl< \bigl< A(t + \Delta t) \bigm| X(t + \Delta t) = x + \Delta X \bigr> \Bigm| X(t) = x \Bigr>\\[1ex] &= \Bigl< \langle a \rangle (x + \Delta X, t+\Delta t) \Bigm| X(t) = x \Bigr>\\[1ex] &= \langle a \rangle (t+\Delta t) + \frac{\partial}{\partial x}\bigl< \Delta X \, A(t) \bigm| X(t) = x \bigr>, \end{aligned} $$ from which I got my guess. However, I am not sure why I can go from the first line to the second in the above.