Reversing Integration Limits in a Brownian Motion Integral

Summary

The note explains how to reverse the order of integration in a time integral of Brownian motion, expressing the result as a stochastic integral with the differentials in the opposite order. The original integration region has time variable s between zero and t, while u runs from zero to s. It forms a triangle in the (u, s) plane.

After reversing the order, u runs from zero to t, and for each fixed u, s runs from u to t. These bounds describe the same triangular region, viewed by fixing the other variable first. The explanation relies on ordinary geometric reasoning about integration domains and applies the stochastic Fubini theorem to the displayed integral. It clarifies the limits rather than developing conditions for when stochastic Fubini is valid, so readers should consult those conditions when applying the result in other settings.

Key ideas

• The original integration domain is described by 0 ≤ s ≤ t and 0 ≤ u ≤ s.
• Reversing the integration order keeps the same domain but changes which variable is fixed first.
• With u as the outer variable, its range is zero to t.
• For each fixed u, the inner variable s ranges from u to t.

Tags

# Limits of integration when applying stochastic Fubini theorem to Brownian motion


# Limits of integration when applying stochastic Fubini theorem to Brownian motion












I'm looking at the solution below from Quantuple, it's a nice, succinct solution but I'm confused about how the limits of the integrals in the second line come from. Could someone please elaborate on that part?

Integral of Brownian Motion w.r.t Time

Thanks

## Answer by LocalVolatility (score 6, accepted)

https://quant.stackexchange.com/a/43731

The equality that you are asking about is

$$ \int_0^t \int_0^s \mathrm{d}W_u \mathrm{d}s = \int_0^t \int_u^t \mathrm{d}s \mathrm{d}W_u. $$

When applying Fubini, you need to make sure that the domain that you are integrating over doesn't change. On the left-hand side, both $s$ takes values in $[0, t]$ and for any given $s$, $u \in [0, s]$. See the below plot.

Now when you reverse the order of integration, you let $u$ go from $[0, t]$. To integrate over the same area, for any given $u$, you need to let $s \in [u, t]$. See the second plot, which is essentially just the first one mirrored along the diagonal.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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