# Why does the partial derivative, $X_t$, of an ABM $X(t)$ not involve standard Brownian motion $Z(t)$, even though $Z(t)$ varies with $t$?

Consider the arithmetic Brownian motion $X(t) = \alpha t + \sigma Z(t)$ and evaluating $dX(t)$ using Ito's lemma.

We have $\frac{\partial X}{\partial t} = \alpha$, which does not involve $Z(t)$, even though $Z(t)$ varies with $t$.

When using Ito's lemma, why is $Z(t)$ treated as an independent random variable from $t$ and therefore the partial with respect to $t$ treats $Z(t)$ as a constant?

Edit: I'm not sure why this is getting downvoted. What I have above is coming directly from my textbook "Models for Financial Economics" by Abraham Weishaus.

I'm looking for clarification on what the author writes in italics: "When using Ito's lemma, $Z(t)$ treated as an independent random variable from $t$ and therefore the partial with respect to $t$ treats $Z(t)$ as a constant" Your imagine from Ito's lemma is false. Although $Z(t)$ is a time-dependent process, we shouldn't apply $\frac{\partial Z(t)}{\partial t}$ in Ito's lemma. It is meaningless.
Let $f(t,x)=\alpha t+\sigma x\in \mathbb{C}^{1,2}\left([0,+\infty)\times \mathbb{R}\right)$. By application of Ito's lemma, we have $$df(t, Z(t))=\frac{\partial f}{\partial t}(t, Z(t))dt+\frac{\partial f}{\partial x}(t, Z(t))dZ(t)+\frac 12\frac{\partial^2 f}{\partial x^2}(t, Z(t))d[Z(t),Z(t)]\tag 1$$ thus $$dX(t)=\alpha dt+\sigma dZ_t\tag 2$$
• There may be some confusion about my question. I updated the original post with an imagine of my textbook. My notation $Z_t$ is to mean the partial derivative of $Z(t)$ with respect to $t$, not $Z$ as a function of $t$. I'm not sure why $\frac{\partial Z(t)}{\partial t}$ is meaningless, as you say. – user2521987 Dec 16 '16 at 1:40