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

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

1 Answer 1

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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*}

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  • $\begingroup$ Thank you for the answer. Could you let me know why increasing processes have finite variations? $\endgroup$
    – David
    Commented Apr 19, 2022 at 2:22
  • $\begingroup$ @David I added the proof to the answer $\endgroup$ Commented Apr 19, 2022 at 17:10

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