I am starting to study the field of Backward Stochastic Differential Equations (BSDE) and have a conceptual question on numerical techniques to solve them. BSDE are of the form: $$Y_t=F((B_s)_{0\leq s\leq T})+\int_t^Tf(s,Y_s,Z_s)\text{d}s-\int_t^TZ_s\text{d}B_s$$ where $Y_T=F((B_s)_{0\leq s\leq T})$ is the terminal condition, $B_t$ a Brownian Motion, $Z_t$ a $\mathcal{F}_t$-adapted process which ensures the solution to the equation is adapted (where the filtration is normally the natural filtration of $B_t$) and $f(\cdot)$ is the so-called generator of $Y_t$. By adaptedness of $Z$ we have equivalently: $$Y_t=\mathbb{E}[Y_T|\mathcal{F}_t]+\int_t^T\mathbb{E}[f(s,Y_s,Z_s)|\mathcal{F}_t]\text{d}s$$ Thus numerical schemes to solve this kind of BSDE normally involve conditional expectations. Discretizing, we get something like: $$Y_{t_i}^\pi=\mathbb{E}[Y_{t_{i+1}}^\pi|\mathcal{F}_{t_i}^\pi]+f(t_i,Y_{t_i}^\pi,Z_{t_i}^\pi)\Delta_{i+1}$$ where the Brownian Motion $B$ have been discretized into a random walk $B^\pi$; $Y^\pi$ and $Z^\pi$ are discretized versions of $Y$ and $Z$; $\pi$ is some partition of $[t,T]$; $\mathcal{F}^\pi_{t_i}$ is the filtration generated by the random walk $B^\pi$; and $\Delta_{i+1}=t_{i+1}-t_i$.

Now, my question is on the calculation of the conditional expectation of the form $\mathbb{E}[Y_{t_{i+1}}^\pi|\mathcal{F}_{t_i}^\pi]$. In Ma, Protter, San Martín and Torres (2002), we have the following statement at the bottom of page 305:

We remark that the conditional expectation w.r.t. to the discrete filtration $(\mathscr{F}^{(n)})$ can be computed explicitly as follows. We assume $\Gamma$ is an $\mathscr{F}_{t_{k+1}}^{(n)}$-mesurable random variable and we take the $2^k$ atoms corresponding to the trajectories of the martingale [i.e. random walk] $M^{(n)}$ in $\mathscr{F}_{t_k}^{(n)}$. Each atom in $\mathscr{F}_{t_k}^{(n)}$ splits into two atoms of $\mathscr{F}_{t_{k+1}}^{(n)}$ [by definition of a random walk of course]. Then we have: $$\mathbb{E}[\Gamma|\mathscr{F}_{t_k}^{(n)}](\omega)=\frac{1}{2}(a+b)$$ where $a$, $b$ are the values of $\Gamma$ in the two atoms of $\mathscr{F}_{t_{k+1}}^{(n)}$ coming from the corresponding atom in $\mathscr{F}_{t_k}^{(n)}$ containing $\omega$.

I am not sure I understand this paragraph and thus how the conditional expectations are computed. The way I see, this is how we do it:

  1. Simulate the random walk $B^\pi$ on $t=t_1$, $t_2$, $\dots$ , $t_n=T$ therefore obtaining a trajectory $\omega$;
  2. Compute the terminal value $Y_T^\pi=Y_{t_n}^\pi=F((B_{t_i}^\pi)_{1\leq i\leq n})$ from the trajectory $\omega$;
  3. $\Gamma$ above is interpreted as $Y_{t_n}^\pi$. We know the trajectory $\omega$ of $B^\pi$ therefore we know a) the value of $B_{t_{n-1}}^\pi$, b) the value of $B_{t_n}^\pi$, and c) $Y_{t_n}^\pi \in \{a,b\}$. From a) and b) we know whether the random walk went up or down between $t_{n-1}$ and $t_n$ thus if for example $Y_{t_n}^\pi=a$ we can easily compute $b$. Once $b$ has been derived we compute: $$\mathbb{E}[Y_{t_n}^\pi|\mathcal{F}_{t_{n-1}}^\pi](\omega)=\frac{1}{2}(a+b)$$

Is my understanding of the computation of the conditional expectation, in particular step 3, accurate?


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