When I was reading the paper from Peng, I saw an equation which I had no idea about how to get it. The details are shown below:
For a BSDE :
$$y_t = \xi + \int_{t}^{T}g_0(s)ds - \int_{t}^{T}z_sdB_s$$
Which has a fixed $\xi \in L^2(\mathscr{F}_T)$ and $g_0(\cdot)$ satisfying $E(\int_{0}^{T}|g_0(t)|dt)^2 < \infty$. There exists a unique pair of process $(y., z.)\in L_{\mathscr{F}}^2(0,T;R^{1+d})$ satisfies the BSDE shown before.
Now, we consider the case where $\xi$ And $g_0(\cdot)$ are both bounded. Since we have
$$y_t = E^{\mathscr{F}_t}[\xi + \int_t^Tg_0(s)ds]$$
Thus the process is also bounded. We can apply Itô’s formula to $|y_s|^2e^{\beta s}$ for $s \in [t,T]$:
$$|y_t|^2e^{\beta t} + \int_t^T[\beta|y_s|^2+|z_s|^2]e^{\beta s} ds = |\xi|^2e^{\beta T} + \int_t^T2y_sg_0(s)e^{\beta s}ds - \int_t^T e^{\beta s}2y_sz_sdB_s$$
Here comes the problem. How did we get this equation? In other words, how did the Itô’s formula deal with $|y_s|^2e^{\beta s}$?
Note that your BSDE implies the following dynamics for $y$: $$dy_s = g_0(s) ds - z_s dB_s$$ with terminal condition $y_T = \xi$. This implies that the dynamics for $y_s^2$ are given by: $$d(y_s^2) = 2y_s dy_s + d \langle y \rangle_s= (2y_sg_0(s) +z_s^2) ds -2y_sz_s dB_s$$ Now, let $M_s = y_s^2 e^{\beta s}$. By Itô's product rule: $$\begin{align*} dM_s &= e^{\beta s} d(y_s^2) + \beta e^{\beta s} y_s^2 ds \\ &= (2y_sg_0(s) +z_s^2)e^{\beta s}ds + \beta e^{\beta s} y_s^2 ds - 2e^{\beta s}y_sz_s dB_s \end{align*}$$ Collecting terms and re-writing this in integral form, we get: $$\xi^2 e^{\beta T} - y_t e^{\beta t} = M_T - M_t = \int_t^T (\beta y_s^2 + z_s^2)e^{\beta s}ds + \int_t^T 2y_s g_0(s) ds -\int_t^T 2e^{\beta s}y_sz_s dB_s$$ After re-arranging once again, this agrees (up to some sign differences) with the result you proposed.