Can a stochastic integral be martingale with respect to two different filtrations?

Given a probability space $$(\Omega, \mathcal F, \mathbf P)$$ and a filtration $$\{\mathcal F_t\}_{t\ge0}$$ on it. Consider the following stochastic integral process $$X_t = \int_0^t H_s dW_s$$ where $$W$$ is a standard $$\{\mathcal F_t\}$$-Brownian motion and $$H$$ is an $$\{\mathcal F_t\}$$-adapted process satisfying suitable regular conditions so that the above stochastic integral process $$X$$ is well-defined and is a $$\{\mathcal F_t\}$$-martingale.

Now make a change of time $$t\to T(t)$$ with $$T:[0,\infty)\to [0,\infty)$$ a continuously differentiable increasing function. Let $$Y_t = X_{T(t)}.$$ Then clearly,

$$Y$$ is an $$\{\mathcal F_{T(t)}\}$$-martingale.

But on the other hand, $$$$Y_t = \int_0^{T(t)} H_s dW_s = \int_0^t H_{T(r)} dW_{T(r)}$$$$ by a change of variable $$s=T(r)$$. Since the quadratic variation $$\langle W_T, W_T\rangle = T$$, the martingale representation theorem (Theorem 3.4.2 in Karatzas & Shreve's book) yields that there is an extension $$(\tilde\Omega, \tilde{\mathcal F}, \tilde{\mathbf P}, \{\tilde{\mathcal F}_t\}_{t\ge0})$$ of $$(\Omega, \mathcal F, \mathbf P, \{\mathcal F_t\}_{t\ge0})$$ and a $$\{\tilde{\mathcal F}_t\}$$-Brownian motion $$B$$, such that $$W_{T(t)} = \int_0^t \sqrt{T'(r)} dB_r.$$ Hence, $$Y_t = \int_0^t H_{T(r)} \sqrt{T'(r)} dB_r.$$ This implies that

$$Y$$ is a $$\{\tilde{\mathcal F}_t\}$$-martingale.

Comparing the two claims, $$Y$$ is martingale under two filtrations, while these two filtrations $$\{\mathcal F_{T(t)}\}$$ and $$\{\tilde{\mathcal F}_t\}$$ seem to be irrelative. Is there anything weird?

• Is it obvious that $H_{T(r)}$ is adapted to $\tilde{\mathcal F}_r$? I think you need this to conclude $Y$ is an $\tilde{\mathcal F}_r$-martingale. Commented May 13, 2021 at 14:22
• @user6247850 Thank you very much. The second claim is incorrect in general. Inspired by your comment, I post an answer myself. Commented May 13, 2021 at 16:01

One thing I ignored is that the stochastic integral $$\int_0^t H_{T(r)} dW_{T(r)}$$ is understood in the sense that the $$\{\mathcal F_{T(t)}\}$$-adapted process $$H_T$$ is integrated with respect to the $$\{\mathcal F_{T(t)}\}$$-Brownian motion $$W_T$$. Both are in the filtration $$\{\mathcal F_{T(t)}\}$$. It cannot be understood in the original filtration $$\{\mathcal F_t\}$$, because we can say nothing about $$H_T$$ or $$W_T$$ in this filtration: $$H_T$$ need not be even adapted w.r.t. $$\{\mathcal F_t\}$$, and $$W_T$$ need not be an $$\{\mathcal F_t\}$$-martingale. So, we cannot proceed as the second claim to use the martingale representation to rewrite $$W_T$$ under the filtration $$\{\mathcal F_t\}$$ since it is not even a martingale.
The right way is in Theorem 8.5.7 in Oksendal's book. As stated in that theorem, if we define $$B_t = \int_0^{T(t)} \sqrt{(T^{-1})'(s)} dW_s,$$ then $$B$$ is an $$\{\mathcal F_{T(t)}\}$$-Brownian motion and a.s., $$\int_0^{T(t)} H_s dW_s = \int_0^t H_{T(r)} \sqrt{T'(r)} dB_r.$$ Hence, $$Y_t = \int_0^t H_{T(r)} \sqrt{T'(r)} dB_r$$ is still an $$\{\mathcal F_{T(t)}\}$$-martingale.