I was reading a paper (link) from Richard Bass about Brownian functionals, and came across the following passage :

Let $X_t$ be a Brownian motion, $F_t$ its filtration and $g$ a real-valued function. We first assume that $g$ is bounded, has compact support, and is in $C^2$. By Clark's formula applied to the functional $g(X_1)$, $$ g(X_1)=\mathbb{E}g(X_1) + \int_0^1 \mathbb{E}[g'(X_1)|F_s]dX_s $$ (Another derivation of this representation is to use Ito’s lemma to take care of the case $g(x) = e^{iux}$ and then use linearity and a limiting process.)

My question is about the proposed alternative derivation of the given formula.

I tried to use Ito's lemma on $Y_t^u = e^{iuX_t}$ but didn't seem to be the right direction.

Can somebody give me some hints on how to proceed ?


Here are the steps I tried :

  1. First, observe that, conditional to $X_s$, $X_1$ is normally distributed with mean $X_s$ and variance $(1-s)$
  2. Then, we know $\mathbb{E}(e^{iuX_1}|F_s)$, which is given by the characteristic of a gaussian : $\mathbb{E}(e^{iuX_1}|F_s)=e^{iuX_s-\frac{1}{2}u^2(1-s)}$. We then have $\mathbb{E}[g'(X_1)|F_s]=g'(X_s)e^{-\frac{1}{2}u^2(1-s)}$
  3. Using Ito lemma, we have that $g(X_1)=1+\int_0^1 g'(X_s)dX_s + \frac{1}{2}\int_0^1 g''(X_s)d\langle X,X\rangle_s $

I'm still wondering if it's the right path to the proof !

  • $\begingroup$ What do you mean by not the right direction? You can't prove it for that choice of $g$ or don't know what to do next? $\endgroup$
    – LazyCat
    Sep 19, 2020 at 16:56
  • $\begingroup$ I meant for that particular choice of $g$. I guess that, since $g$ can be decomposed as a sum of Fourrier series, I should be able to arrive to the result. $\endgroup$
    – Aguelmame
    Sep 21, 2020 at 20:19
  • $\begingroup$ I've searched around a bit, take a look at page 25 of the following notes: math.wisc.edu/~kurtz/NualartLectureNotes.pdf I think, one can extract the details from there. If not - let me know I can take another look. $\endgroup$
    – LazyCat
    Sep 21, 2020 at 21:14
  • $\begingroup$ Hello, not sure to see the link ! The page goes through another route to prove Clark's formula, for general random variables. $\endgroup$
    – Aguelmame
    Sep 22, 2020 at 19:56


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy