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



1 Answer 1


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.

enter image description here

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.

enter image description here


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