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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, \begin{equation} Y_t = \int_0^{T(t)} H_s dW_s = \int_0^t H_{T(r)} dW_{T(r)} \end{equation} 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?

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  • $\begingroup$ 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. $\endgroup$ Commented May 13, 2021 at 14:22
  • $\begingroup$ @user6247850 Thank you very much. The second claim is incorrect in general. Inspired by your comment, I post an answer myself. $\endgroup$
    – Dreamer
    Commented May 13, 2021 at 16:01

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The second claim is not correct.

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.

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