Malliavin derivative of adapted processes Let $(\mathcal{F}_t)_{t\ge 0}$ be a filtration. A stochastic process $(X_t)_{t\ge 0}$ is adapted with respect to such a filtration, if $X_t$ is $\mathcal{F}_t$-measurable for all $t\ge 0$. Now consider two adapted processes $(u_t)_{t\ge 0}$ and $(v_s)_{s\ge 0}$.
Why these statements are (trivially?) true:

*

*If $s<t$ then $D_tv_s = 0$ where $D_t$ is the Malliavin derivative.

*If $s>t$ then $D_su_t = 0$.

The statemets appear in page 57 of Introduction to Malliavin Calculus of D. Nualart and E. Nualart. The authors mention these statements as trivial facts, without demonstration or further comment. I don't know what key observation I am overlooking, according to which both would be trivial or obvious statements.
 A: For $n \in \mathbb{N}$ let
$$
t_{n}^{k}= \frac{k}{2^n}, \quad k \in \{1, \dots, 2^n\}
$$
be the $(k+1)$-th element of the $n$-th order dyadic partition of the interval
$[0, T]$. Let
$$
I_{n}^{k} = (t_{n}^{k-1}, t_{n}^{k}], \Delta_{n}^{k}:= W_{t_{n}^{k}} - W_{t_{n}^{k-1}}
$$
be the k-th interval of the partition and the k-th increment of the Brownian
motion and define $\Delta_n = (\Delta_{n}^{1}, \dots, \Delta_{n}^{2^n})$. Take then a function $\varphi$ that is smooth and together with its derivatives is at most of polynomial growth.
If $k_n(t)$ is the only element $k \in \{ 1, ..., 2^n \}$ such that $t \in I_{n}^{k}$ for all $t \in (0,T]$, then the Malliavin derivative is
$$
D_tX := \frac{\partial \varphi}{\partial x_{n}^{k_n(t)}}(\Delta_n)
$$
I'll show the case $s > t$: if $X$ is $t$-adapted then it is independent from $\Delta_{n}^{k}$ for $k > k_n(t)$.
Indeed, if $t < t_{n}^{k_n(t)}$ then X is function only of $\Delta_{n}^{1}, ..., \Delta_{n}^{k_n(t)-1}$, while if $t = t_{n}^{k_n(t)}$ then it is function only of $\Delta_{n}^{1}, ..., \Delta_{n}^{k_n(t)}$. Hence since $k_n(t) < k_n(s)$ for n large enough so that s,t belong to disjoint dyadic intervals, from the definition
$$
\frac{\partial \varphi}{\partial x_{n}^{k_n(s)}}(\Delta_n) = 0
$$
