Let $X_t$ a random process such that its Malliavin derivative is well defined for all $t$. Then I have read that :

$D_s(\int_0^t \! X_u \, \mathrm{d}u)=\int_s^t \! D_s(X_u) \, \mathrm{d}u.$

What I don't understand is why in the second integral, the domain of integration is now $[s,t]$.

Thanks for your help !


I assume that $X_t$ is progressively measurable with respect to the corresponding Wiener process $W_t$ (otherwise the formula doesn't hold).

$X_t$ being $\sigma(W_s,s\le t)$-measurable implies that it does not depend on the "tail" $\{w(v),t<v\le T\}$ as a functional from $L_2(C([0,T],\mathbb{R}),\text{Wiener measure})$, $t<T\le\infty$. Thus, $D_hX_t$ is zero on the $\{h: \operatorname{supp}(h)\in(t,T]\}$ and $D_s(X_t)=0$ for $s>t$, therefore $$ D_s\left(\int_0^t X_udu\right)=\int_0^t D_s(X_u) du=\int_s^t D_s(X_u) du. $$


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