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 $\cos(B_t)$.

  • $\begingroup$ Are you familiar with Itô's lemma? $\endgroup$ Commented Apr 27, 2022 at 23:03
  • $\begingroup$ 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 $\endgroup$
    – Barreto
    Commented Apr 27, 2022 at 23:09
  • $\begingroup$ 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... $\endgroup$ Commented Apr 27, 2022 at 23:10

1 Answer 1


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 $\cos B_t$,

  • $\begingroup$ 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. $\endgroup$
    – Barreto
    Commented Apr 27, 2022 at 23:22
  • $\begingroup$ Although here quantstart.com/articles/Itos-Lemma the lemma is stated for $C^2$ functions, and the square root is not $C^2$ in $\mathbb{R}$.. $\endgroup$
    – Barreto
    Commented Apr 27, 2022 at 23:24
  • $\begingroup$ 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? $\endgroup$
    – Barreto
    Commented Apr 27, 2022 at 23:41
  • $\begingroup$ @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. $\endgroup$ Commented Apr 27, 2022 at 23:42
  • $\begingroup$ Ok! Give me a second. Thank you for the discussion btw. $\endgroup$
    – Barreto
    Commented Apr 27, 2022 at 23:43

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