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I need to show $Y_t$ is a martingale by showing $\mathbb{E}[Y_t| F_s] = \mathbb{E}[e^{\frac{t}{2}} \sin(B_t)| F_s] = Y_s$, $s \leq t$, where $B_t$ is a Brownian motion.

I think the approach is to show

$$e^{\frac{t}{2}} \sin(B_t) = e^{\frac{t - s +s }{2}} \sin((B_t - B_s) + B_s)$$

But when I expand out using $\sin((B_t - B_s) + B_s)$ and expand out further using other trig identities, I end up back at $sin(B_t)$.

I am stuck how to to expand this out to show this is a martingale. The hints given in the question $e^{iA} = \cos(A) + i\sin(A)$ and $\cos(A + B) = \cos(A)\cos B - \sin(A)\sin(B)$, but I don't know how to apply these. Any help is appreciated.

Thanks

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1 Answer 1

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Recall that a BM $(W_t)_{t\geq 0}$ is Markov and $$E[e^{i\xi W_t}|W_s]=E[e^{i\xi (W_t-W_s)}]e^{i\xi W_s}=e^{i\xi W_s-\xi^2(t-s)/2}$$ So $$\begin{aligned}E[Y_t|\mathscr{F}_s]&=e^{t/2}E[\sin (W_t)|\mathscr{F}_s]=\\ &=e^{t/2}\frac{E[e^{iW_t}|\mathscr{F}_s]-E[e^{-iW_t}|\mathscr{F}_s]}{2i}=\\ &=e^{t/2}\frac{e^{iW_s-(t-s)/2}-e^{-iW_s-(t-s)/2}}{2i}=\\ &=e^{s/2}\frac{e^{iW_s}-e^{-iW_s}}{2i}=\\ &=e^{s/2}\sin(W_s)=\\ &=Y_s\end{aligned}$$

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