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I am having trouble with this question:

$$\mbox{Let Y(t)}=\begin{cases} 0 & -1\le t \le 0\\ Z(t)-tZ(t) & 0 < t < 1 \\ 1-t & 1 \le t \le 2\end{cases}$$ Find the quadratic variations of $Y(t)$ over [-1,2].

Z(t) is standard brownian motion. The solution to this question says:

$$dY(t)=\begin{cases} 0 & -1\le t \le 0\\ dZ(t)-Z(2) dt & 0 < t < 1 \\ -dt & 1 \le t \le 2\end{cases}$$ $$[dY(t)]^2=\begin{cases} 0 & -1\le t \le 0\\ [dZ(t)]^2 & 0 < t < 1 \\ 0 & 1 \le t \le 2\end{cases}$$ ...by Itô's lemma because $dZ(t)dt=(dt)^2=0$.

So, the quadratic variation is $\int_{-1}^2 [dY(t)]^2 = \int_0^1 dt = 1$$

I do not understand why $dZ(t)dt=(dt)^2=0$. I understand how to apply Itô's lemma to stochastic differential equations, but I don't know much about it beyond that. Hopefully someone can clear this up in a relatively simple way.

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Well, what do you mean by $dZ(t)dt$? What do you mean by $(dt)^2$? The way it is typically defined in Ito calculus is $d[Z(t),t]$ and $d[t,t]$ where $[\cdot,\cdot]$ denotes the quadratic covariation. You just need to show that the quadratic variation of $f(t)=t$ is $0$ and that $[Z(t),t]=0$ and you're done.