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Let $\overline{\widehat{Z}_i} = \frac{E_i\left[ \int_{t_i}^{t_{i+1}}\widehat{Z}_sds\right] }{\Delta t_i}$ with $\widehat{Z}$ a square integrable process, $\Delta t_i := t_{i+1} - t_i$, and $E_i$ denotes the conditional expectation w.r.t. $F_{t_i}$, with standard probability space/filtration.

Why is then $E_i\left[ \int_{t_i}^{t_{i+1}}(\widehat{Z}_s -\overline{\widehat{Z}_i})ds\right] =0$?

More details can be found in https://arxiv.org/pdf/2006.01496.pdf , Page 17, equation 5.8

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Breaking the integral into two terms, the first term is simply $E_i\left[ \int_{t_i}^{t_{i+1}}\widehat{Z}_sds\right]$.

The second term is $E_i \left[\int_{t_i}^{t_{i+1}} \overline{\widehat{Z}_i} ds\right]$. The term in the expectation is $\mathcal F_{t_i}$ measurable, and so the second term is just $\int_{t_i}^{t_{i+1}} \overline{\widehat{Z}_i} ds.$

The integrand being independent of $s$, this is just $\nabla t_i \overline{\widehat{Z}_i}$, which is $E_i\left[ \int_{t_i}^{t_{i+1}}\widehat{Z}_sds\right]$.

So the terms cancel and we get $0$ as required.

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    $\begingroup$ Thanks, Nate. You are absolutely right. $\endgroup$
    – freshst4r
    Commented Sep 11, 2021 at 14:41

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