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I don't know how to prove this :

let be $X_t = \int_{0}^{t}\sigma_{u}dW_{u}$ where $\sigma_{t}$ is a predictable process.

If $|\sigma_{t}| = c$ a.s. how can I prove that $X_{t}=c*\beta_{t}$ (equality in distribution) ? (obvious if there wouldn't be absolute value..)

Thank you

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Note that $X$ is a continuous martingale. Moreover, the quadratic variation is given by \begin{align*} \langle X_t, \, X_t\rangle = \int_0^t |\sigma_u|^2 du = c^2 t. \end{align*} That is, \begin{align*} \langle X_t/c, \, X_t/c\rangle = t. \end{align*} From Levy's characterization, $X/c$ is by law a Brownian motion, which we denote by $\beta$. Then, by law, \begin{align*} X_t = c\, \beta_t. \end{align*}

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  • $\begingroup$ TY ! I wouldn't have thought at Levy's characterization... $\endgroup$ – glork May 11 '15 at 16:42

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