When a Stochastic Integral Is Measurable with Respect to Brownian Information

Summary

The document asks whether a stochastic integral appearing in a geometric Asian option valuation is measurable with respect to the Brownian filtration at time t. It points to a professor's expression in which a conditional expectation given that filtration is subsequently written without the conditioning, and asks whether the integral's measurability justifies that step. The setting uses risk-neutral expectation and Brownian motion, with an integral extending to maturity.

The source is a question rather than a worked explanation, so it provides no proof or answer. The issue depends on the relationship between the integral's time horizon and the filtration time: an integral involving Brownian increments through maturity is generally not known at an earlier time. Removing conditioning requires justification, such as the random variable already being measurable at the conditioning time or the conditional expectation being otherwise shown to equal the unconditional one. The notation and time limits would need careful checking before drawing a conclusion about the option price.

Key ideas

• The document questions whether a stochastic integral is measurable with respect to a Brownian filtration at time t.
• Dropping a conditional expectation requires a measurability or other probabilistic justification.
• An integral extending beyond time t may depend on Brownian increments not yet included in the filtration.
• The source poses the issue but does not supply a proof or resolution.

Tags

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


# 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?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.