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Let $T<\infty$, $(\Omega,\mathcal F,P)$ a probability space carrying a standard $d$-dimensional Brownian motion $(B_t)_{t\geq 0}$ and $(\mathcal F_t)_{t\geq 0}$ the natural $\sigma$-algebra filtration generated by $(B_t)_{t\geq 0}$, $\mathcal F=\mathcal F_T$. $(A_t)_{t\in[0,T]}$ is an absolutely continuous $R$-valued and strictly increasing $(\mathcal F_t)$-adapted processes satisfying $A_0=0$. Note that for a fixed $t\in[0,T]$, $A_t$ is a $\mathcal F_t$-stopping time.

If we define $\tilde B_t:=B_{A_t}$ and $\tilde{\mathcal F}_t:=\mathcal F_{A_t}$, then we have that $\tilde B_0=0$, $E[\tilde B_t]=0$ for each $t$, and $\tilde B_t-\tilde B_s$ is independent of $\tilde{\mathcal F}_s$, but $E[\tilde B^2_t]=E[A_t]$. So $(\tilde B_t)$ is not a Brownian motion under $P$.

I am wondering that if we can find an appropriate probability measure $\tilde P$ to make $\tilde B$ a standard Brownian motion under a new probability space $(\Omega,\tilde{\mathcal F},\tilde P)$?

My attempt: To make $(\tilde B_t)$ a Brownian motion, we only need to make it satisfy the Gaussian distribution $G$, so define the corresponding probability measure $\tilde P(\tilde B_t\in A):=G(A)$ for each $A\in\mathcal B(R^d)$.

I reffere to Theorem 3.4.6 in pp.174 of Brownian motion and stochastic calculus authored by Ioannis Katatzas and Steven Sherve, second edition. By this result, $(\tilde B_t)$ is a sandard Brownian motion if $(A_t)$ is a $\mathcal F_t$-martingale. Without this restriction, I can not obtain my desired result.

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  • $\begingroup$ the measure is $\mu([c,d]) = \frac{d-c}{A(c+d)-A(c)}$ ? $\endgroup$ – reuns Feb 25 '16 at 8:21
  • $\begingroup$ @user1952009 can you explain your idea more precisely? In my question, the Brownian motion is $R^d$ valued. $\endgroup$ – XIAO Lishun Feb 25 '16 at 8:34
  • $\begingroup$ 1) The fact that $A_t$ is a stopping time need not follow from adaptedness of $A$. 2) If $E[A_t] = t$, $B_t$ may well be a Brownian motion (and will be thanks to L&eacute;vy's theorem under some reasonable assumptions). $\endgroup$ – zhoraster Feb 26 '16 at 8:14
  • $\begingroup$ @zhoraster point (1) is right. For (2), the problem arise from it, $E[A_t]$ is unnecessarily equal to $t$. $\endgroup$ – XIAO Lishun Feb 26 '16 at 8:23

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