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How could show that the expectation of the following Ito process is zero. i.e. $$\mathbb{E}[\sin(5B_t)]=0,$$ where $B_t$ is a standard Brownian motion, $t\leq T$.

Here I want to use Ito's formula and Fubini's theorem as $$\sin(5B_t)=0+\int_{0}^{t}5\cos(5B_s)dB_s-\frac{25}{2}\int_{0}^{t}\sin(5B_s)ds.$$

Then $$\mathbb{E} \sin(5B_t)=5\mathbb{E}\int_{0}^{t}\cos(5B_s)dB_s-\frac{25}{2}\mathbb{E}\int_{0}^{t}\sin(5B_s)ds.$$

After this, I'm stuck. Anyone help?

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    The distribution of $B_t$ is symmetric, so no need to use Ito formula. – 2017-02-23
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    Going to be easier to say that $B_t$ is a normally distributed random variable with mean $0$ and variance $t$, now $\sin(nx)$ is an odd function and therefore the expected value vanishes – 2017-02-23

3 Answers 3

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Recall that $B_t$ is normally distributed with mean $0$ and variance $t$. Then by the "law of the unconscious statistician", $$\mathbb E(\sin(5B_t))=\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}\sin(5x)e^{-x^2/2t}dx.$$ The last integral converges and is equal to zero, since you are integrating an uneven function.

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$$\mathbb E[\sin(5B_t)]=\int\sin(5y)p(t,y)\mathrm ds=0.$$ Where $p(t,y)$ is the law of $B_t$, which is an even function and $\sin$ is odd.

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$Y$ is standard normally distributed, as well as $-Y$. Use the fact that $sin(-x)=-sin(x)$ $$E(sin(5B_t))=E(sin(5\sqrt{t}Y))=E(sin(-5\sqrt{t}Y))=-E(sin(5\sqrt{t}Y))$$

Then you can conclude