In non-Ito processes, does the identity $dW^{2}=dt$ always hold?

Question

Wiener Process $dB^2=dt$

From the above question, we see an answer wherein it is shown that $dW_{t}^{2}=dt$ through showing equivalence under integration for any square integrable function $g(W_{t})$. My question is, then, for a non-ito process which has a $dt$ term (for example, one of form $dX=a(N,t)dN+b(N,t)dt$ for Poisson process $N$), can one always make the replacement $dt=dW_{t}^{2}$ whenever $dt$ appears (after perhaps showing that the space can contain a Brownian motion)? Further, as it can be shown (via Ito's lemma) that $dW_{t}^{2}=2W_{t}dW_{t}+dt$, how exactly do these two separate identities exist if it seems that these are not necessarily always equal?

Answer 1

It seems that you are getting confused about what is getting squared. $dW_t^2 = dt$ means $dW_t dW_t = dt$ and is a heuristic to remind you that for a brownian motion you have that $[W]_t = t$ where $[\cdot]_t$ is the quadratic variation - so you consider $(dW_t)^2$. However in $dW_t^2 = 2W_t dW_t + dt$ it is really just $W_t$ that gets squared. In this case you are writing $d(W_t^2) = 2W_t dW_t + dt$.