and Quenez in [@EPQ], requires more stringent conditions than the aforementioned intuitive ones. More explicitly, the main result in [@EPQ] concerning the Malliavin differentiability of the solution to the BSDE (essentially) involves the following conditions (see [@EPQ Proposition 5.3] for a precise statement):

-   $\xi$ is Malliavin differentiable[^4] and $\E[|\xi|^4]<+\infty.$

-   At any time $t\in[0,T]$, the r.v. $\omega \longmapsto f(t,\omega,Y_t,Z_t)$ is Malliavin differentiable[^5] with Malliavin derivative denoted by $D_\cdot f(t,Y_t,Z_t)$ such that there exists a predictable process $K^{\theta}:=(K_t^\theta)_{t\in [0,T]}$ with $\int_0^T \E[(\int_0^T|K_s^\theta|^2ds)^2] d\theta<+\infty$, and such that for any $(y_1,y_2,z_1,z_2) \in {\mathbb{R}}^4$ it holds for a.e. $\theta \in [0,T]$ that: $$|D_\theta f(t,\omega,y_1,z_1) - D_\theta f(t,\omega,y_2,z_2)| \leq K^\theta_t(\omega) (|y_1-y_2|+|z_1-z_2|).$$

Roughly speaking, this means that $\xi$ and $\omega \longmapsto f(t,\omega,y,z)$ have to be Malliavin differentiable, but in order to prove that $Y$ and $Z$ are Malliavin differentiable, one needs to enforce an extra regularity conditions on each of the data: that is $\xi$ has a finite moment of order $4$, and the Malliavin derivative of the driver $f$ is Lipschitz continuous in $(y,z)$ with a sufficiently integrable stochastic Lipschitz constant $K$. Note that a careful reading of the proof allows one to conclude that the moment conditions on $\xi$ and $Df$ can actually 