# Properities of sine(and cosine) of Brownian Motion

Suppose $$(B_t, \mathcal{F}_t)_{t \geq 0}$$ is a classical Brownian Motion (with the canonical filtration) and consider the process $$X_t=\sin(B_t)$$.

What properties does our new process share with a Brownian motion? Concretely: does it follow that $$(X_t, \mathcal{F}_t)_{t \geq 0}$$ is also a continous martingale?

It is definetively the case that $$X_t$$ also has stationary, independent increments, so that: $$\mathbb{E}[X_T|X_t] = \mathbb{E}[X_t + (X_T-X_t)|X_t] = X_t + \mathbb{E}[(X_T-X_t)|X_t] = X_t + \mathbb{E}[X_T-X_t] = X_t$$

Do you agree? Continuity also seems not difficult. On the other hand, the previous quick proof does not hold for $$\cos(B_t)$$ since $$\mathbb{E}[X_T] \neq \mathbb{E}[X_t] \neq 0$$.

Does this mean $$\cos(B_t)$$ is really not a martingale? I am using the results on this page by the way, for the expected values of $$\sin(B_t)$$ and $$\cos(B_t)$$.

• Are you familiar with Itô's lemma? Commented Apr 27, 2022 at 23:03
• A little bit, yes. I just took a look at the lemma again. If we prove that both process are Ito processes then we are done, right? @JoseAvilez Commented Apr 27, 2022 at 23:09
• Not quite, as you probably want to show that $X$ is not a martingale (hint: it isn't). Your calculation of conditional expectation is suspect... Commented Apr 27, 2022 at 23:10

Continuity is immediate as the composition of continuous functions is again continuous. However, $$\sin (B_t)$$ and $$\cos (B_t)$$ are not martingales.

Let $$f(x) = \sin x$$, so that $$f_x(x) = \cos x$$ and $$f_{xx}(x) = -\sin x$$. Then, by Itô's lemma, $$X_t$$ satisfies the following stochastic differential equation:

$$dX_t = f_x(B_t) dB_t + \frac{1}{2}f_{xx}(B_t) dt = \cos (B_t)dB_t - \frac{1}{2} \sin (B_t)dt$$

In particular, $$\sin B_t$$ has a non-zero drift term, so that it cannot be a martingale. You may argue similarly for $$\cos B_t$$,

• Thank you! Can I ask you one more thing: is $\sqrt{t} \sin(B_t)$ a martingale? I am reading about the Lemma, and maybe the $\sin$ cancels out there. Commented Apr 27, 2022 at 23:22
• Although here quantstart.com/articles/Itos-Lemma the lemma is stated for $C^2$ functions, and the square root is not $C^2$ in $\mathbb{R}$.. Commented Apr 27, 2022 at 23:24
• I see. However, we would then have that $f(x,t)= \sqrt{t}\sin(x)$, and using the notation of the page quantstart.com/articles/Itos-Lemma, $dW(t)^2$ is just $dB(t)^2$ which is $dt$. Adding them both does not give 0, so it still has a drift. Could you tell me what am I doing wrong? Commented Apr 27, 2022 at 23:41
• @Sarah The comment section is perhaps not the best place for us to have this conversation. Please post it as a separate question showing your application of Itô's lemma and I'll be able to comment. Commented Apr 27, 2022 at 23:42
• Ok! Give me a second. Thank you for the discussion btw. Commented Apr 27, 2022 at 23:43