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Let $t_1<t_2 \in [0,T]$, $f(t)\ge0$ and $Z_t$ an increasing process with $Z_0= 0$.

We have clearly that \begin{equation} \int_{[0,t_2[}f(s)dZ_s \ge \chi_{t_1<T}(Z_{t_2} - Z_{t_1})\min_{t_1\le s \le t_2}f(s) \end{equation}holds.

In a paper I'm reading the authors state based on this and using the fact that $Z_{t_1}\ge0$ it follows that \begin{equation} \int_{[0,t_2[}f(s)dZ_s \ge \chi_{t_1<T}Z_{t_2}(\min_{t_1 \le s \le t_2}(f(s) - f(t_2)) + f(t_2)) \end{equation} holds.

I'm confused how that can be, it seems like an error! Surely since $Z_{t_1}\ge0$ we must have $(Z_{t_2} - Z_{t_1}) \le Z_{t_2}?$ Would appreciate any insight into anything I may have glossed over here!

For reference I am talking about the beginning of page 264 of the following paper: Boetius, Frederik, and Michael Kohlmann. "Connections between optimal stopping and singular stochastic control." Stochastic Processes and their Applications 77.2 (1998): 253-281.

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  • $\begingroup$ I see how to match the hypothesis at the top of p. 264 with yours, but I don't see why the conclusion matches. The notation is sufficiently involved that I can buy it, but are you sure that what you have written matches their conclusion $J_f \ge \chi_{\sigma < T}(Z_\sigma^x - Z_\sigma^{x - \delta})f(\sigma) + \delta I_5(\delta) + \delta I_6(\delta)$? $\endgroup$
    – LSpice
    Commented Jul 25, 2020 at 20:33
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    $\begingroup$ Hi LSpice, to give you more context: When they are writing $J_f \ge \chi_{\sigma<T}(Z^x_\sigma - Z^{x-\delta}_\sigma)f(\sigma) + \delta I_5(\delta) + \delta I_6(\delta),$ by plugging in the terms we supposedly have $\chi_{\sigma^\delta<T}(X^*_{\sigma^\delta} - Z^{x-\delta}_\sigma)f(\sigma^\delta) + \int_{[0,\sigma^\delta[}f(s)d\xi^*_s \ge \chi_{\sigma^\delta<T}(X^*_{\sigma^\delta} - Z^{x-\delta}_{\sigma^\delta})f(\sigma^\delta) + \chi_{\sigma<T}\xi^*_{\sigma^\delta}f(\sigma^\delta) + \chi_{\sigma<T}\xi^*_{\sigma^\delta}\min_{\sigma \le s \le \sigma^\delta}(f(s) - f(\sigma^\delta))$ $\endgroup$
    – icewater
    Commented Jul 25, 2020 at 20:51
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    $\begingroup$ Yes I found it confusing too! I believe they are using it like so: $\chi_{A}=1$ if $A$ is true and $\chi_{A}=0$ otherwise. So yes $\chi_{t_1<T}Z_{t_2}$ is a product. $\endgroup$
    – icewater
    Commented Jul 25, 2020 at 21:01
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    $\begingroup$ Yes I agree with you. I am wringing my hands making sense of this hieroglyphic paper for a dissertation, no one seemed to understand my plight. Apparently it has 57 citations, I wonder if the 57+ authors cracked this issue. In any case thanks a lot for your time LSpice, I really appreciate it! $\endgroup$
    – icewater
    Commented Jul 25, 2020 at 21:18
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    $\begingroup$ Thank you LSpice, that is a great suggestion. I am using a workaround for the time being by just writing $\int_{[0,t_2[}f(s)d\xi^*_s \ge \xi^*_{t_2}\min_{0 \le s \le t_2}f(s)$ (since $\xi^*_0=0$). $\endgroup$
    – icewater
    Commented Jul 25, 2020 at 23:19

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