# Show that stochastic integral is $F_W(t)-$measurable

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?

• 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. Feb 10 at 8:58
• 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? Feb 10 at 9:16
• 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. Feb 10 at 9:20