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



Stochastic Integral + conditional expectation
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.


