3
$\begingroup$

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

$\endgroup$

1 Answer 1

2
$\begingroup$

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*}

$\endgroup$
1
  • $\begingroup$ TY ! I wouldn't have thought at Levy's characterization... $\endgroup$
    – glork
    May 11, 2015 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.