Let $M=(M_t)_{t\geq0}$ with $$M_t=\int_0^{\log\sqrt{1+2t}}e^s\text{d}B_s$$

where $(B_t)_{t\geq0}$ is a Standard Brownian Motion. Show that $M$ is also a Standard Brownian Motion and compute $\mathbb{E}[\int_0^tM^6_s\text{d}M_s]$ and $\mathbb{E}[\left(\int_0^tM_s\text{d}M_s\right)^3]$.

Can we use Levy's Characterization Theorem?

  • $\begingroup$ Indeed, the task is to prove that $(M_t)_{t\geqslant0}$ and $(M_t^2-t)_{t\geqslant0}$ are both martingales starting from $0$. $\endgroup$ – Did Dec 15 '12 at 10:30
  • $\begingroup$ The problem is that I don't know what to do with the term $\log\sqrt{1+2t}$ in the upper limit of the integral. $\endgroup$ – Nick Papadopoulos Dec 15 '12 at 10:38
  • $\begingroup$ Yes. Thus, you need to know $\mathrm dM_t$ where $M_t=K_{A(t)}$ for some (deterministic) increasing $t\mapsto A(t)$ and some (martingale) process $(K_t)_{t\geqslant0}$. Any idea? $\endgroup$ – Did Dec 15 '12 at 10:42
  • $\begingroup$ Well, I guess that would be $\text{d}M_t=\sqrt{1+2t}\text{d}B_{\sqrt{1+2t}}$. Then why is $M_t$ a martingale? $\endgroup$ – Nick Papadopoulos Dec 15 '12 at 10:48
  • $\begingroup$ No it is not. Please first find a formula in the more general setting I indicated (with a purpose, believe it or not) and only then apply this formula to your setting. $\endgroup$ – Did Dec 15 '12 at 10:50

From the exchanges in the comments, one reaches the conclusion that $M_t=K_{A(t)}$ with $$ K_t=\int_0^t\mathrm e^s\mathrm dB_s,\qquad A(t)=\log\sqrt{1+2t}. $$ Since $(K_t)_{t\geqslant0}$ is a martingale and the (deterministic) function $A$ is nondecreasing, this is enough to show that $(M_t)_{t\geqslant0}$ is a martingale. Furthermore, for every $0\leqslant s\leqslant t$, by Itô's isometry, the identity $$ (M_t-M_s)^2=\left(\int_{A(s)}^{A(t)}\mathrm e^u\mathrm dB_u\right)^2 $$ implies that $$ \mathbb E((M_t-M_s)^2)=\mathbb E\left(\int_{A(s)}^{A(t)}\mathrm e^{2u}\mathrm du\right)=\frac{\mathrm e^{2A(t)}-\mathrm e^{2A(s)}}2. $$ The function $A$ is tuned such that $\mathbb E((M_t-M_s)^2)=t-s$ hence $(M_t)_{t\geqslant0}$ is a Brownian motion. More generally, for every nonzero function $a$, $(M^a_t)_{t\geqslant0}$ is a Brownian motion, where $$ M^a_t=\int_0^{A^{-1}(t)}a(s)\mathrm dB_s,\qquad A(t)=\int_0^{t}a(s)^2\mathrm ds. $$ The two expectatons to be computed are direct since one can replace $(M_t)_{t\geqslant0}$ by $(B_t)_{t\geqslant0}$ without changing the result. For example, $$ X_t=\int_0^tB_t^6\mathrm dB_t $$ is an odd functional of $(B_s)_{0\leqslant s\leqslant t}$, hence $\mathbb E(X_t)=0$. Likewise, $$ Y_t=\int_0^tB_s\mathrm dB_s $$ is a centered gaussian random variable hence $\mathbb E(Y_t^3)=0$.

On time-changed Brownian motions, see this.

Edit: Since $A$ is invertible and continuous, the sigma-algebras $\mathcal F_t^B=\sigma(B_s;s\leqslant t)$ and $\mathcal F_t^M=\sigma(M_s;s\leqslant t)$ are such that $\mathcal F^M_t=\mathcal F^B_{A(t)}$. For every $s\leqslant t$, $M_t=M_s+\Delta$ where $$ \Delta=\int_{A(s)}^{A(t)}\mathrm e^u\mathrm dB_u. $$ The increments of $(B_u)_{u\geqslant A(s)}$ are independent of $\mathcal F^B_{A(s)}$ hence $\mathbb E(\Delta\mid\mathcal F^B_{A(s)})=0$, which shows that $\mathbb E(M_t\mid\mathcal F^M_s)=M_s$. Hence $(M_t)_{t\geqslant0}$ is a martingale.

  • $\begingroup$ I tried to show that $K_{A(t)}$ is a martingale using the definition but I failed. Could you give me a hint please? $\endgroup$ – Nick Papadopoulos Dec 15 '12 at 15:24
  • $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Dec 15 '12 at 18:32
  • $\begingroup$ Let $X_t=\int_0^te^s\text{d}B_s$. Observe that if we set:$$\tau_t=\inf\{s>0\ |\ \left<X,X\right>_s>t\}$$, then, by the Dambis/Dubins - Schwartz Theorem, $X_{\tau_t}$ is a standard Brownian Motion and $X_t=B_{<M,M>_t}$ for all $t\geq0$. In this case, $\tau_t=\inf\{s>0\ |\ s>\log(\sqrt{1+2t})\}$. $\endgroup$ – Nick Papadopoulos Dec 21 '12 at 19:16

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