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Let $a \in R$,$B=(B^1,B^2)$ a brownian motion. $$X_t=e^{B_t^1}\left(\int_0^te^{-B_s^1}dB_s^2+a\int_0^te^{-B_s^1}ds\right)$$

Show there is a brownian motion $\beta$ such that $$X_t=\int_0^t \sqrt{1+X_s^2}d\beta_s+\int_0^t(a+X_s/2)ds$$

For the moment: $$U_t=e^{B_t^1}$$

By the Ito's formula: $$ U_t=1+\int_0^tU_sdB_s^1+(1/2)\int_0^tU_sds\\ dU_t=U_tdB_t^1+(1/2)U_tdt\\ dX_t=dB_t^2+X_tdB_t^1+(a+(1/2)X_t)dt $$ But how to show $$dB_t^2+X_tdB_t^1=\sqrt{1+X_t^2}d\beta_t?$$

Thank you

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1 Answer 1

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Define \begin{align*} \beta_t = \int_0^t \frac{dB_t^2 + X_t dB_t^1}{\sqrt{1+X_t^2}}. \end{align*} Then \begin{align*} \langle\beta, \beta\rangle_t = t. \end{align*} By Levy's characterization, $\{\beta_t \mid t \ge 0\}$ is a Brownian motion.

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  • $\begingroup$ ok, I understand. Thank you $\endgroup$
    – plouf
    Nov 13, 2015 at 17:37

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