Malliavin derivative wrt time changed Brownian motion

Question

The Malliavin derivative $D^W_\alpha$, $\alpha \in \mathbb{R}$, with respect to a standard Brownian motion $W_t$ is $$ D^W_\alpha W_t = 1_{[0,t]}(\alpha). $$

What would be the Malliavin derivative with respect to a time-changed Brownian motion $B_{T(t)} = \int_0^t \sigma_u dW_u$ with $\sigma_u$ a positive process and $T(t) = \int_0^t \sigma^2_u du$?

My thought would be $$ D^B_\alpha B_{T(t)} = 1_{[0,T(t)]}(\alpha). $$

However I am not sure this is correct, and even if correct, how to go about showing this. Any pointers and/or references would be appreciated.

Thank you.

EDIT: It might be relevant to add that the process $\sigma_u$ above is independent of the standard Brownian motion $W_u$

Answer 1

Thanks to the formula $$B_{T(t)} = \int_0^t \sigma_u dW_u$$ that you already wrote, we can use the Malliavin derivative as done for SDEs, if you want a general theorem eg.theorem 2.2.1 in Nualart's related topics where he computes the Malliavin derivative for the general SDE

$$dX_{t}=B(t,X_{t})dt+A(t,X_{t})dW_{t}.$$

So in this particular case we just have

$$D_{a}X_{t}=\sigma(a),~~~~for~~~ a\leq t.$$