$B_s$ is brownian motion. Because $\int_{t_{i-1}}^{t_i} e^{-μ({t_i} -s)}\sigma dB_s $ has a Brownian component, it is normally distributed with the mean zero according to Taylor & Karin 1998 's Introduction to Stochastic Modelling.
The above expectation is derived from the ornstein uhlenbeck model: $$dx_{t}= μ(θ-x_t)dt+\sigma B_t $$
and if $x_{t_i}$ given $x_{t_{i-1}}$.
$$ e^{μ{t_i}}x_{t_i}=x_{t_{i-1}}+ \int_{t_{i-1}}^{t_i} μe^{μs}θds + \int_{t_{i-1}}^{t_i} e^{μs}\sigma B_s $$
$$\iff x_{t_i}=x_{t_{i-1}}e^{-μ{t_i}}+ θ(1-e^{-μΔt}) + \int_{t_{i-1}}^{t_i} e^{-μ({t_i}-s)}\sigma B_s $$
where $Δt={t_i} - t_{i-1}$
Based on Leung & Li 2016's Optimal Mean Reversion Trading, the distribution of $x_t$ given $x_{t-1}$ under the Ornstein Uhlenbeck model is normally distributed with $$E[x_{t_i}|x_{t_{i-1}}]=x_{t_{i-1}}e^{-μΔt}+ θ(1-e^{-μΔt})$$
$$ Var(x_{t_i}|x_{t_{i-1}})= \frac{\sigma^2}{2μ}(1-e^{-2μΔt})$$
So $$E[x_{t_i}|x_{t_{i-1}}]=x_{t_{i-1}}e^{-μΔt}+ θ(1-e^{-μΔt})+E[\int_{t-1}^t e^{-μ(t-s)}\sigma dB_s |x_{t_{i-1}}]=x_{t_{i-1}}e^{-μΔt}+ θ(1-e^{-μΔt})$$ so that means $$E[\int_{t_{i-1}}^{t_i} e^{-μ({t_i}-s)}\sigma B_s |x_{t_{i-1}}]=0$$
How do I prove/derive this? Thank you!
Let $M_t := \int_0^t e^{-\mu(t_i-s)}\sigma dB_s$ and note that $M_t$ is a martingale because it is a stochastic intgerlad with a locally bounded integrand. By linearity of the stochastic integral, $\int_{t_{i-1}}^{t_i} e^{-\mu(t_i-s)}\sigma dB_s = M_{t_i}-M_{t_{i-1}}$ and hence \begin{align*} \mathbb{E}[\int_{t_{i-1}}^{t_i} e^{-\mu(t_i-s)}\sigma dB_s|x_{t_{i-1}}] = \mathbb{E}[M_{t_i}-M_{t_{i-1}}|x_{t_{i-1}}] = M_{t_{i-1}}-M_{t_{i-1}} = 0. \end{align*} There's a bit swept under the rug here (like the fact that $x_t$ is a Markov process and generates the same filtration as $B_t$), but that's basically it.