# Applying Ito's formula

This is probably an easy question but I am getting aquanted with Ito's formula and stuck on an exercise in my textbook.

Let $X_{t}=W_{t}-a t/2$ where $a$ is a real number and $W_{t}$ is brownian motion with $W_0=0$

I want to use the general Ito formula but unsure how to treat the derivatives when having both $W_{t}$ and $t$. Any advise in the right direction would be much appreciated.

This is what I mean by Ito's general formula:

$f(X) = f(X_0)+\int f^\prime(X)\,dX + \frac{1}{2}\int f^{\prime\prime}(X)\,d[X]$

We have $dX_t = dW_t - \frac{a}{2}dt$ and $d[X]_t = d[W]_t = dt$ since the deterministic piece $-at/2$ doesn't contribute to the quadratic variation. So, $$f(X_t) = f(X_0)+\int_0^t f'(X_s)\, dW_s - \frac{a}{2} \int_0^t f'(X_s) \,ds + \frac{1}{2} \int_0^t f''(X_s) \, ds \\ = f(X_0)+\int_0^t f'(X_s)\, dW_s + \frac{1}{2} \int_0^t \big[f''(X_s) - af'(X_s) \big]\,ds$$
• One quick question: Is $f'(X)= d(X)/dW+ d(X)/dt$? @Tom – simme Nov 12 '14 at 2:36
• No, the chain rule is replaced with the Ito rule: $$df(X_t) = \frac{\partial f}{\partial t}(t, X_t) dt + \frac{\partial f}{\partial x}(t, X_t) dX_t+ \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t, X_t) \cdot (dX_t)^2.$$ – m_gnacik Nov 12 '14 at 9:12