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I know that the Malliavin derivative of $B(t)$ is $1$ if $B$ is standard Brownian motion. But what about $B_1(t)+B_2(t)$ if $B_i$ are independent Brownian motions?

This is really a functional on the two dimensional Wiener space, so it should have a Malliavin derivative - but what is it and how to compute?

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If you assume that $\Omega=C_0([0,1];\mathbb R^2)$ and you let $H:=L^2[0,1]\oplus L^2[0,1]$ you can see that in Nualart's notation (see section 1.2. of The Malliavin Calculus and Related Topics) you have:

$$B_1(t)+B_2(t)=W(e_1\chi_{[0,t]})+W(e_2\chi_{[0,t]})$$ where $(e_1,e_2)$ is the canonical basis of $\mathbb R^2$ and $W$ is an isonormal gaussian process. Then by definition o the Malliavin derivative you have that $$D(B_1(t)+B_2(t))=e_1\chi_{[0,t]}+e_2\chi_{[0,t]}.$$

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