Cross-posted from: "https://math.stackexchange.com/questions/3917971/malliavin-derivative-of-stopped-brownian-motion"

I have a small question concerning the Malliavin derivatives. It could be rather simple, but I am unsure.

Let $B_t$ stand for the standard Brownian motion in $\mathbb{R}^d$. Denote $$T = \inf\{t| \|B_t\| = 1\}.$$ That is, $T$ is the first exit time from the unit ball.

I am interested in calculating the Malliavin derivative $DB_T$. The immediate suspect is $DB_T = 1_{[0,T]}$. I could not prove it though and I suspect it is not 100% correct.

Is the functional $B_T$ even Malliavin differentiable? If so, how to obtain the derivative?


1 Answer 1


Stopped Brownian motion is not Malliavin differentiable because if it was it would imply that $T$ is constant (see footnote pg.4 Locally Lipschitz BSDE driven by a continuous martingale path-derivative approach).

We would have that $W_{T}=\int_{0}^{\infty}1_{s\leq T}dW_{s}\in \mathbb{D}^{1,2}$ and $1_{s\leq T}\in \mathbb{D}^{1,2}$. However, by Nualart's proposition 1.2.6 (in "Malliavin and Related topics") we would get that $P[s\leq T]=0$ or $1$.

However, there is some work in Hitting times for Gaussian processes that still manages to get related estimates for hitting times using Malliavin calculus.


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