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, $$ f(X_t) = f(X_0)+\int_0^t f'(X_s)\, dW_s - \frac{a}{2} \int_0^t f'(X_s) \,ds + \frac{1}{2} \int_0^t f''(X_s) \, ds \\ = f(X_0)+\int_0^t f'(X_s)\, dW_s + \frac{1}{2} \int_0^t \big[f''(X_s) - af'(X_s) \big]\,ds $$

  • $\begingroup$ Now a big piece of the puzzle landed. Thank you! $\endgroup$ – simme Nov 12 '14 at 2:04
  • $\begingroup$ One quick question: Is $f'(X)= d(X)/dW+ d(X)/dt$? @Tom $\endgroup$ – simme Nov 12 '14 at 2:36
  • 1
    $\begingroup$ No, the chain rule is replaced with the Ito rule: $$ df(X_t) = \frac{\partial f}{\partial t}(t, X_t) dt + \frac{\partial f}{\partial x}(t, X_t) dX_t+ \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t, X_t) \cdot (dX_t)^2.$$ $\endgroup$ – m_gnacik Nov 12 '14 at 9:12
  • $\begingroup$ @simme check out math.stackexchange.com/a/2382314/466778 $\endgroup$ – encore Aug 9 '17 at 18:58

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