I am somewhat unsure how to go about showing this:

Use Ito's Lemma to show for any deterministic differentiable function, $f$: $$ \int_0^t f(s) dB(s) = f(t)B(t) - \int_0^t B(s)f'(s)ds $$ Where $B(t)$ is Brownian motion

I've used Ito's Lemma to show things are martingales, but I'm not sure how to go about doing this. I'm guessing it's something to do with the fact that f is deterministic, and Brownian motion is random, but am not really sure.


EDIT: Ito's lemma, as defined in my notes in integral form:

For $X_t = h(t, B(t))$ $$ X_t = X_0 + \int_0^th_B(s, B(s))dB(s) + \int_0^t(h_t(s, B(s)) + \frac{1}{2}h_{BB}(s, B(s))ds $$

  • $\begingroup$ Can you add a statement of what your version of Ito's lemma is? What you have there is a slight rearrangement of what I would take to be the statement of Ito's lemma applied to f(t)B(t). $\endgroup$ – Chris Janjigian Apr 4 '14 at 23:55
  • $\begingroup$ Sure thing. Edited the questing with it. Cheers $\endgroup$ – user140513 Apr 5 '14 at 0:12

By your definition $h(t,b) = f(t)b$ . Just plug right in. Basically the $X_0$ term goes away because $B_0 =0$ and the second derivative term goes away because h is linear in b. Then it's just a matter of isolating the Ito integral on the left hand side.

  • $\begingroup$ Oh okay. I was assuming $h(t,b)$ was just equal to $f$, so that's where I was going wrong. Cheers. $\endgroup$ – user140513 Apr 5 '14 at 4:46

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