Using Stochastic Fubini to Rewrite an Integrated Brownian Motion

Summary

The document derives an alternate form for the time integral of Brownian motion, a step that arises in the short-rate Merton model. Representing Brownian motion at each time as the accumulation of its increments turns the time integral into an integral over a triangular region: increments at time u contribute from u through the terminal time. Reversing the order of integration therefore weights each Brownian increment by the remaining time interval.

The answer also gives an equivalent expression via integration by parts, involving terminal Brownian motion and a stochastic integral weighted by time. These identities clarify the geometry behind the change in integration order. The source provides a compact formula but does not explain the conditions that justify stochastic Fubini or the assumptions on the integration limits; its initial question’s lower limit t is also shifted to zero in the displayed answer.

Key ideas

• Brownian motion can be represented as the integral of its increments from the initial time.
• Reversing the integration order over the resulting triangular domain gives each increment a remaining-time weight.
• The integrated Brownian motion also equals terminal time multiplied by terminal Brownian motion minus a time-weighted stochastic integral.
• A rigorous derivation requires conditions that justify exchanging time and stochastic integration.

Tags

# How to understand the following brownian integral using Fubini's method?


# How to understand the following brownian integral using Fubini's method?












I am a little bit stucked with the following integral process, using Fubini's method, this is an intermediate step of short rate Merton Model.

$\int_{t}^{T} W(s)ds=\int_{0}^{\hat {T}}ds\int_{0}^{s}dW(u)\\=\int_{0}^{\hat {T}}dW(u)\int_{u}^{\hat {T}}ds\\=\int_{0}^{\hat {T}}(\hat{T}-u)dW(u)$

My more specific question is how did the change of integration variables proceed, as the process described by above integration is not very intuitive to me.

Many thanks!

## Answer by Michael Mark (score 2, accepted)

https://quant.stackexchange.com/a/27929

\begin{align*} \int_0^T W(t)\, dt &{}= \int_0^T\!\!\int_0^t dW(u)\,dt \\ &{}= \int_0^T\!\!\int_u^T dt\, dW(u) \\&{}= \int_0^T (T - u)\,dW(u) \\&{}= TW(T) - \int_0^T u\, dW(u) \end{align*}

however i am not sure if it is what you are asking for

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