Conditional expectation of the solution to BSDE

Question

Let $B_s \in \mathbb{R}$ be the standard Brownian motion for $s\in [0,T]$ on the complete probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $\mathcal{B}^s_t$ be the $\sigma$-algebra $\sigma(B_\tau-B_t;\tau \in [t,s])$ for $s\in [t,T]$. Suppose $0<t_1<t_2<T$, $a_T$ is $\mathcal{B}^T_{t_2}$-measurable and $b_s$ is a $\mathcal{B}^T_{t_2}$-measurable process. Therefore, we can find a pair of $\{\mathcal{B}^s_{t_2}\}_{s\in [t_2,T]}$-progressively measurable process $(Y_s,Z_s)$ such that $$ Y_s = a_T +\int^T_s b_\tau d\tau - \int^T_s Z_\tau dB_\tau$$ for $s\in [t_2,T]$. We can also express $$Y_s = \mathbb{E}(a_T + \int^T_s b_\tau d\tau |B^s_{t_2}).$$ Is it true that $Y_s=\mathbb{E}(Y_s|\mathcal{B}^s_{t_2})=\mathbb{E}(Y_s|\mathcal{B}^s_{t_1}) = \mathbb{E}(a_T + \int^T_s b_\tau d\tau |B^s_{t_1})$?

Answer 1

\begin{align*} Y_s = \mathbb{E}[Y_s|\mathcal{B}_{t_2}^s] = \mathbb{E}\Big[\mathbb{E}[Y_s|\mathcal{B}_{t_2}^s]\Big|\mathcal{B}_{t_1}^s\Big](\text{tower property}) = \mathbb{E}[Y_s|\mathcal{B}_{t_1}^s](\text{as }Y_s = \mathbb{E}[Y_s|\mathcal{B}_{t_2}^s]) \end{align*}