Consider the following BSDE: $$\begin{cases} dY_t &= -f(t, \boldsymbol{x}_t, Y_t, \boldsymbol{z}_t) ~ dt + \boldsymbol{z}_t^\intercal ~ d\boldsymbol{w}_t,\hspace{0.64cm} t\in[0, T] \\ Y_T &= \max \left( \displaystyle\prod_{i=1}^\mathrm{d} X_{i, T}^{\alpha_i} - K, 0 \right) \end{cases}$$ where $$\boldsymbol{x}_t = (X_{1,t}, \dots, X_{d,t}),$$ $$\boldsymbol{w}_t = (W_{1,t}, \dots, W_{d,t}),$$ $$f(t, \boldsymbol{x}_t, Y_t, \boldsymbol{z}_t) = -r Y_t,$$ and $$dX_{i,t} = (r- q_i) X_{i,t} ~ dt + X_{i,t} \sum_{j=1}^\mathrm{m} \bar{\sigma}_{ij} ~ dW_{j}, \ i=1, \dots, d.$$
Sufficient conditions for this BSDE to have a unique solution are: (i) $f$ is Lipschitz continuous on third and fourth argument, (ii) $f(\cdot, \cdot, 0, \boldsymbol{0})$ is predictable and $f(\cdot, \cdot, 0, \boldsymbol{0}) \in H_T^2(\mathbb{R})$, and (iii) $Y_T = \max \left( \displaystyle\prod_{i=1}^\mathrm{d} X_{i, T}^{\alpha_i} - K, 0 \right) \in L_T^2 (\mathbb{R}).$
I have proved conditions (i) and (ii) (please check the proof below):
(i) Take arbitrary $(y_1, \boldsymbol{z}_1), (y_2, \boldsymbol{z}_2) \in H_T^2 (\mathbb{R}) \times H_T^2 (\mathbb{R}^m)$. See that \begin{align*} \lvert f(t, \boldsymbol{x}_t, y_1, \boldsymbol{z}_1) - f(t, \boldsymbol{x}_t, y_2, \boldsymbol{z}_2) \rvert &= \lvert -ry_1 - ( - ry_2) \rvert \\ &= \lvert-r (y_1 - y_2) \rvert \\ &= r \lvert y_1 - y_2 \rvert \\ &\leq r (\lvert y_1 - y_2 \rvert + \lvert \boldsymbol{z}_1 - \boldsymbol{z}_2 \rvert). \end{align*} Thus $f$ is Lipschitz continuous on third and fourth argument.
(ii) $f(\cdot, \cdot, 0, \boldsymbol{0}) = -r (0) = 0$ and $\mathbb{E} \left( \int_0^T |f(\cdot, \cdot, 0, \boldsymbol{0})|^2 dt. \right) = 0 < \infty.$
So far I have not been able to prove point (iii). However, I get the following: $$\mathbb{E} (X_{i,t}^2) = X_{i,0}^2 \exp \left( 2 (r-q_i) t + \sum_{j=1}^m \bar\sigma_{ij}^2 t \right),$$ $$\mathbb{E} (X_{i,T}^2) = X_{i,0}^2 \exp \left( 2 (r - q_i) T + \sum_{j=1}^m \bar\sigma_{ij}^2 T \right) < \infty.$$ I need help to complete the prove, i.e., to get $\mathbb{E} (|Y_T^2|) < \infty$. Thank you very much.