# Why is twice ddifferential of running maximum zero?

I have saw the following statement in my notes. But I cannot see it clearly, if someone could explain?

Consider an Itô process given by $$\mathrm{d} Z_{t}=a_{t} \mathrm{~d} t+b_{t} \mathrm{~d} W_{t},$$ for some processes $$a, b$$ such that the two integrals associated with this expression are well-defined. Also, consider the running maximum and running minimum processes $$\bar{Z}$$ and $$\underline{Z}$$, which are defined by $$\bar{Z}_{t}=\max _{u \in[0, t]} Z_{u} \quad \text { and } \quad \underline{Z}_{t}=\min _{u \in[0, t]} Z_{u}$$

which imply that, $$\left(\mathrm{d} \bar{Z}_{t}\right)^{2}=\left(\mathrm{d} \underline{Z}_{t}\right)^{2}=\mathrm{d} \bar{Z}_{t} \mathrm{~d} \underline{Z}_{t}=\mathrm{d} \bar{Z}_{t} \mathrm{~d} t=\mathrm{d} \underline{Z}_{t} \mathrm{~d} t=\mathrm{d} \bar{Z}_{t} \mathrm{~d} W_{t}=\mathrm{d} \underline{Z}_{t} \mathrm{~d} W_{t}=0.$$

• Isn't there an ambiguity of notation with $Z$ and the running minimum process depending on $Z$ ? Commented Mar 29, 2022 at 15:59
• @SacAndSac Thank you for the comment, I have edited the post. Commented Mar 29, 2022 at 16:06

Note that $$\bar Z_t$$ is an increasing process, and so it has finite variation until it hits $$\infty$$. Since the integrals are well-defined, $$Z_t < \infty$$ for all $$t$$ and so $$\bar Z_t$$ has finite variation. Similarly, $$\underline{Z}_t$$ is a decreasing process and hence also has finite variation. Since the quadratic variation of any process of finite variation is $$0$$, the claims follow.
To see that any finite increasing process $$X$$ has finite variation on any interval $$[0,t]$$, observe that for any partition of the interval we have \begin{align*} \sum |X_{t_{i+1}}-X_{t_i}| = \sum (X_{t_{i+1}}-X_{t_i}) = X_t - X_0 <\infty \end{align*}