In some notes, my professor writes the following for the price function of an geometric asian option:

\begin{align} \text{Price}(t)&=\tilde{\mathbb{E}}\left[\left(S(0)\exp\left(\frac{T}{2}\left(r-\frac{\sigma^2}{2}\right)+\frac{\sigma}{T}\int_0^T\tau d\tilde{W}(t)\right)-K\right)_{+} | \mathcal{F}_{W}(t)\right]\\ &=\tilde{\mathbb{E}}\left[\left((S(0)\exp\left(\frac{T}{2}\left(r-\frac{\sigma^2}{2}\right)+\frac{\sigma}{T}\int_0^T\tau d\tilde{W}(t)\right)-K\right)_{+}\right] \end{align}

Note that $\tilde{\mathbb{E}}$ and $\tilde{W}$ are the expectation in the risk neutral probability space. Does the above mean that the stochastic integral is $F_W(t)-$measurable? He just removes the filtration from the expectation. How can I show/motivate that the stochastic integral is $F_W(t)-$measurable?

  • $\begingroup$ When you construct an Ito Integral as $$\int_{h=0}^{h=t}X_h(\omega)W_h(\omega)$$, the integrand $X_t(\omega)$ is adapted to the filtration generated by the integrator $W_t(\omega)$ by definition. So the integral is by definition adapted. See for example the first paragraph of these lecture notes here, where Ito Integral is constructed from scratch. $\endgroup$ Feb 10 at 8:58
  • $\begingroup$ Should it not be $dW_h(\omega)?$ In my case, the $X_h(\omega)$ is just a deterministic constant, Which should be adapted to $F_{\tilde{W}}(t)$, but since $\tilde{W}(t)=W(t)+\text{constant},$ it's also $F_w(t)-$measurable right? $\endgroup$
    – Parseval
    Feb 10 at 9:16
  • 2
    $\begingroup$ Yes, it's a typo, should be $dW_h(\omega)$. You are correct that $\tilde{W}(t)=W(t)+K$ is also $F_W(t)$ measurable. $\endgroup$ Feb 10 at 9:20

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