# Stochastic Integral + conditional expectation

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

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.