I am fairly new to stochastic calculus and trying to get to grips with the concept and in particular how the notation is representing the time dependence. I summarise my understanding to date, which may help you to see what I might have missed:
Starting with the simple case of a stock price, this is usually written $S(W_t)$, and from Itô’s lemma in differential form we see both drift (i.e. deterministic) and diffusion (i.e. stochastic) components:
$$ dS(W_t) = \frac{\operatorname{dS}}{\operatorname{dW_t}} \operatorname{dt} + \frac12 \frac{\operatorname{d^2S}}{\operatorname{dW^2_t}} \operatorname{dW_t} $$
where $W_t$ is a stochastic (Weiner) process, indexed in time. This can be considered to come from Taylor expansion of $$ S(W_t +\delta W_t) - S(W_t) \; \approx\; S'(W_t) \;\,\delta W_t + \frac12 S''(W_t) \;\,\delta t $$ as $ \delta W_t \rightarrow 0 $ and taking ${dW_t}^2 = dt$
I find this description of Ito’s lemma immediately a little confusing, since it’s clear $S(W_t)$ has some time dependence (hence the $\operatorname{dt}$ term); however I’m happy to interpret this for now as coming through from $W_t$ - which we can consider to be $W(t)$.
Then we consider a derivative asset, $V(S(W_t))$ - e.g. a put option with $S$ as the underlying. Considering Wilmott (p129), which generalises to $dS(W_t) = A\operatorname{dt} + B\operatorname{dW_t}$, V is given the SDE: $$ \operatorname{dV}(S) = \frac{\operatorname{dV}}{\operatorname{dS}} \operatorname{dS}+ \frac12 B^2 \frac{\operatorname{d^2V}}{\operatorname{dS^2}}\operatorname{dt}$$
which by expanding $\operatorname{dS}$ from above, recovers the SDE in terms of $\operatorname{dW_t}$:
$$\operatorname{dV}(S) = \left( A\frac{\operatorname{dV}}{\operatorname{dS}} \operatorname{dS}+ \; \frac12 B^2 \frac{\operatorname{d^2V}}{\operatorname{dS^2}}\right)\operatorname{dt} \;+\; B \frac{\operatorname{dV}}{\operatorname{dS}} \operatorname{dW_t} $$
I can’t follow quite how these are derived; the Taylor expansion appears so be different from the dS case (e.g. what happened to A?), which makes me think I am missing something here.
Finally, we then find what seems to be referred to as the ‘2D case’, where the underlying is explicitly dependent on $t$ (in addition to the implicit dependence coming from $W(t)$.)
This gives a new SDE for the underlying: $$\operatorname{dS}(t,W_t) = \left( \frac{\partial S}{\partial t} + \frac12 \frac{\partial^2 S}{\partial W_t^2} \right)\operatorname{dt} + \frac{\partial S}{\partial W_t} \operatorname{dW_t} $$
and for the derivative function:
$$\operatorname{dV}(S(t,W_t)) = \left( \frac{\partial V}{\partial S} + \frac12 \frac{\partial^2 V}{\partial S^2} \right)\operatorname{dt} + \frac{\partial V}{\partial S} \operatorname{dW_t} $$
These, I can’t follow; it seems the process should match the above (only now using 2D Taylor expansions), but it seems different. Ican’t grasp the impact of adding ‘additional’ time-dependence, nor how to get from dS to dV.
So my question then, specifically, is what is the difference between: $S(W_t)$ and $S(t,W_t)$; and between $V(S(W_t))$ and $V(S(t,W_t))$ - and how do we derive the SDEs?
As an addendum, I’m not sure if there might also be the possibility of $V(S(W_t),t)$; if so - what would that mean/represent?