Why time derivative of random process always zero?

Question

It seems that whenever I am applying Ito's Lemma and am doing the time derivative bit it always turns out $$\frac{d}{dt}V=0$$ Take an example from a book $$\frac{d}{dt}\left(\frac{B^{2}}{2}\right) = 0$$ I don't get it, simple Brownian motion is a function of time: $$B(t) = B(t-1) + Z \sqrt{t}$$ so how is this possible? Are we just pretending there is nothing beyond B in this composition?

Answer 1

Actually in Ito's Lemma you first recognize the expression as a function of $(x, t)$ and you differentiate the arguments accordingly, and then substitute the Brownian motion $W(t)$ back to the derivative by $x = W(t)$.

And although ordinary Brownian motion $W(t)$ has the distribution $\mathcal{N}(0, t)$ at every time point $t$, i.e. having the same distribution as $\sqrt{t}Z$ where $Z \sim \mathcal{N}(0,1)$, they are very different in nature. The path of $W(t)$ is rough and no-where differentiable, whereas $\sqrt{t}Z$ is a nice parabola with a random coefficient $Z$