Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to Itô’s representation, any $\xi \in L_2(\Omega, F_T , P)$ has a unique representation: $ \xi = E(\xi) + \int_0^T H_s dBs$ where $(H_s)$ is an adapted process belonging to $L_2$. $B$ is a standard Brownian Motion.

Question: Find $(H_s)$ for $\cos(B_T)$.

share|cite|improve this question
For some other examples, such as $M_T=B_T^3$, I can find the martingale process $M_t=E(B_T^3|F_t)$, then apply Ito's formula to find $H_t$ easily. But for $cos(B_T)$, I can't use this method. – XXX11235 Dec 5 '12 at 9:11
up vote 3 down vote accepted

Recall that for every $a$ in $\mathbb C$, $M_t=\mathrm e^{aB_t-a^2t/2}$ defines a martingale $(M_t)_{t\geqslant0}$ starting from $M_0=1$ and such that $\mathrm dM_t=aM_t\mathrm dB_t$. Hence, for every $T\geqslant0$, $$ M_T=1+\int_0^TaM_t\,\mathrm dB_t. $$ If $a=\mathrm i$, this yields $$ \mathrm e^{\mathrm iB_T}=\mathrm e^{-T/2}+\mathrm e^{-T/2}\int_0^T(\mathrm i\mathrm e^{\mathrm iB_t})\,\mathrm e^{t/2}\,\mathrm dB_t. $$ Keeping only the real part, one gets $$ \cos B_T=\mathrm e^{-T/2}-\mathrm e^{-T/2}\int_0^T\sin(B_t)\,\mathrm e^{t/2}\,\mathrm dB_t=\mathbb E(\cos B_T)+\int_0^TH_t\,\mathrm dB_t, $$ where $$ H_t=-\sin(B_t)\,\mathrm e^{(t-T)/2}. $$ Integrability is not an issue here since $\|H_t\|_\infty\leqslant1$ for every $t\leqslant T$.

Note: As regards your comment, writing $B_T=B_t+(B_T-B_t)$ yields the identity $$ \mathbb E(\cos B_T\mid \mathcal F_t)=\mathbb E(\cos(B_T-B_t))\cos B_t=\mathrm e^{(t-T)/2}\cos B_t, $$ which suggests to use the martingale $N_t=\mathrm e^{(t-T)/2}\cos B_t$. Then $$ \cos B_T=N_T=N_0+\int_0^T\mathrm dN_t, $$ Since $N_0=\mathrm e^{-T/2}$ and, by Itô's formula, $\mathrm dN_t=-\mathrm e^{(t-T)/2}\sin(B_t)\mathrm dB_t$, the identification of the solution $(H_t)_{0\leqslant t\leqslant T}$ follows directly. To sum up, $H_t$ solves the identity $$ H_t\mathrm dB_t=\mathrm dM_t,\qquad M_t=\mathbb E(\xi\mid\mathcal F_t). $$

share|cite|improve this answer
Thanks a lot!!!! – XXX11235 Dec 5 '12 at 21:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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