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3

First of all, your calculation is not correct. Itô's formula gives \begin{align*} M_T^{2m} &= 2m \cdot \int_0^T M_t^{2m-1} \, dM_t + m \cdot (2m-1) \cdot \int_0^t M_t^{2m-2} \, d \langle M \rangle_t \\ \Rightarrow \mathbb{E}(M_T^{2m}) &= m \cdot (2m-1) \cdot \mathbb{E} \left( \int_0^T M_t^{2m-2} \cdot X_t^2 \, dt \right)\tag{1} \end{align*} ...

1

First let me say that it was a little more tricky than I previously thought (you need more than only one $\theta$). As you have noticed $X_t$ is a continuous semi-martingale, so it has a decomposition into local martingale part $X^m$ plus a finite variation part $X^f$. Now by Itô on $Z^\theta_t=f_\theta(X_t,A_t)$ you get : ...

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When $\sigma$ and $B$ are independent, a solution is $$U=\sqrt{2\cdot\int_0^1\phi(t)^2\sigma_t^4\,\mathrm dt}.$$ When $\sigma$ is progressively measurable with respect to the filtration of $B$, the same solution might apply in the sense that $U\xi$ could be distributed like $\varepsilon_1$. As first steps in this direction, note that, for every real $x$, by ...

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This looks relevant. Kais Hamza and Fima C. Klebaner, “A Family of Non-Gaussian Martingales with Gaussian Marginals,” Journal of Applied Mathematics and Stochastic Analysis, vol. 2007, Article ID 92723, 19 pages, 2007. doi:10.1155/2007/92723 http://www.hindawi.com/journals/ijsa/2007/092723/abs/

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Name $Z=X/Y$ and use the Ito transformation formula. Or just expand for the infinitesimal time step: \begin{align*} Z_{t+dt}&=(t+dt)(B_t+dB_t)e^{−B_t-dB_t}\\ &=(t+dt)(B_t+dB_t)e^{B_t}(1-dB_t+\tfrac12dB_t^2+...)\\ &=tB_te^{−B_t}+te^{−B_t}(1-B_t)dB_t+e^{−B_t}(B_t+\tfrac12tB_t-t)dt+O(\sqrt{dt}^3) \end{align*} using the formal rule $(dB_t)^2=dt$. ...

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