Ito integral representation of cosine of Brownian motion and expected value

I'm working through an exercise related to SDEs and I'm getting some conflicting results. Hoping someone can set me in the right direction. The problem:

Let $$X_T = \cos(B_T)$$ (where $$B$$ is Brownian motion or Wiener process). Find the process $$\mu$$ such that

$$X_T = \mathbb{E}[X_T] + \int_0^T \mu_s \, dB_s$$

and calculate $$\mathbb{E}[X_T]$$.

So, we can use a change of variable and consider that $$X_T = e^{\frac{1}{2}t} \cos B_t$$ and apply Ito:

$$dX_t = -e^{\frac{1}{2}t}\sin B_t \, dB_t \iff d(e^{\frac{1}{2}t}\cos B_t) = -e^{\frac{1}{2}t} \sin B_t \, dB_t$$, or

$$e^{\frac{1}{2}T} \cos B_t = \int_0^T-e^{\frac{1}{2}t} \sin B_t \, dB_t \iff \cos(B_t) = \int_0^T-e^{\frac{1}{2}(t-T)} \sin B_t \, dB_t$$

So, we have shown $$\cos(B_t)$$ is equivalent to just a stochastic integral, that is, it is a martingale with no drift and thus no expectation.

But, I've been searching around and it seems that a direct expectation on $$\cos(B_t)$$ yields $$e^{\frac{-t}{2}}$$?

When you rewrote the differential equation as an integral equation you forgot the initial condition. Please be more careful about notation (e.g. whether it is $t$ or $T$)

If we set $f(t,x) := e^{t/2} \cos(x)$, then $Y_t := f(t,B_t)$ satisfies by Itô's formula

$$dY_t = - e^{t/2} \sin(B_t) \, dB_t.$$

This is equivalent to saying

$$Y_T \color{red}{-Y_0} =- \int_0^T e^{t/2} \sin(B_t) \, dB_t.$$

Note that $Y_0 = e^0 \cos(0)=1$. Multiplying both sides with $e^{-T/2}$, we find

$$\cos(B_T) - e^{-T/2} = - \int_0^T e^{(t-T)/2} \sin(B_t) \, dB_t.$$

Hence,

$$X_T = \cos(B_T) = e^{-T/2} - \int_0^T e^{(t-T)/2} \sin(B_t) \, dB_t.$$

This shows, in particular, $\mathbb{E}(X_T) = e^{-T/2}$.

• @Archetupon You are welcome. – saz Apr 8 '17 at 8:21
• Probably this is irrelevant or obvious, but I found it interesting that the expected value of $X_T$ decreases over time and approaches 0 as $T \to \infty$. I guess the "extra noise" introduced by running the Brownian motion through $\cos$ dies out as we approach infinity and in the limit we again have that the expected value is 0. – baibo Apr 13 '20 at 15:32
• How do we first have the intuition to introduce exp(t/2) ? – TmSmth Nov 12 '20 at 19:56
• @TmSmth Have a look at this answer (there it's explained for $\sin$ rather than $\cos$, but the idea is the same). – saz Nov 12 '20 at 20:12