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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?

BayesIsBae
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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 $d(W_t^2) = 2W_t dW_t + dt$.

Rhys Steele
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  • Ah, thanks for the correction! – BayesIsBae Jul 24 '19 at 23:40
  • Just to be sure of one thing, however: if it is the case that a Brownian motion can be supported in some probability space, can we then say $(dW_{t})^{2}=dt$? If not in general, under what condition can the substitution be made? – BayesIsBae Jul 24 '19 at 23:43
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    Sure, as long as you are working on some probability space that supports a Brownian motion and you are aware that what that means is that the quadratic variation of that Brownian motion is $t$. (just to completely avoid risk of confusion, what I have called Brownian motion others might sometimes call standard Brownian motion. You'll get a multiplicative factor if you change the variance) – Rhys Steele Jul 25 '19 at 08:03