Conditional Expectations for Brownian Motion Integrals

Summary

The document asks about a conditional expectation involving the integral of Brownian motion over an interval and the Brownian value at an intermediate time. It applies the tower property, splits the integral at that time, and conditions the later segment on the intermediate Brownian value. The remaining question is why the exponential involving the earlier segment can be moved outside the inner conditional expectation.

The key principle is measurability: given the Brownian motion’s information at the intermediate time, the earlier path integral is already known, so its exponential is fixed under that inner conditioning. The document itself poses the question but does not give this explanation or evaluate the resulting expectation. Its scope is a conditioning step for Brownian motion, without a broader pricing model or application.

Key ideas

• The tower property allows conditioning first on the Brownian value at an intermediate time.
• Splitting the path integral at that time separates the earlier path from the future increment segment.
• The earlier integral is measurable with respect to the information available at the intermediate time.
• A quantity measurable under the conditioning can be factored out of the inner conditional expectation.

Tags

# How to show the following conditional expectation relation holds for a Brownian motion?


# How to show the following conditional expectation relation holds for a Brownian motion?












Suppose that $B_k$ stands for a standard Brownian motion process.

\begin{equation} \mathbb{E}\Big(e^{-w\int_{t}^{S}B_k dk\, -uB_T}\Big| B_t = x\Big) \end{equation} where $w$ and $u$ are constants, and $t< T < S$. Then, using the tower property of the conditional expectation, we have \begin{equation} \mathbb{E}\Big(e^{-w\int_{t}^{S}B_k dk\, -uB_T}\Big| B_t = x\Big) = \mathbb{E}\Big(\mathbb{E}\Big(e^{-w\int_{t}^{S}B_k dk\, -uB_T}\Big|B_T\Big)\Big| B_t = x\Big) = \mathbb{E}\Big(\mathbb{E}\Big(e^{-w\int_{t}^{T}B_k dk\, -w\int_{T}^{S}B_k dk\, -uB_T}\Big|B_T\Big)\Big| B_t = x\Big) = \mathbb{E}\Big(e^{-w\int_{t}^{T}B_k dk\, - uB_T}\mathbb{E}\Big( e^{-w\int_{T}^{S}B_k dk}\Big|B_T\Big)\Big| B_t = x\Big) \end{equation} My question here is that in the last equation, why we can take this expression $e^{-w\int_{t}^{T}B_k dk}$ out of the inner expectation operator given $B_T$.

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