# Malliavin derivative of stopped Brownian motion

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?

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 $$1$$.