I am confused with a definition in Shreve's Stochastic Caclulus for Finance 2 book.

In page 129, Theorem 4.2.2, the Ito isometry theorem. It states that The Ito integral defined before satisfies $$\mathbb E I^2(t)=\mathbb E\int_0^t \Delta^2(u)du $$

I just can not find the definition of $u$. It looks to me it come out from nowhere...Could anybody just explain to me how $\Delta^2(u)$, $u$, and $du$ be defined here? And possibly tell me where can I find it in the book?

PS: I noticed that in Page 96 it uses letter $u$, but still...no explanation and I am not sure they are the same $u$.

Thank you!

  • 2
    $\begingroup$ I don't have the book, but it looks to me that $u$ is just a dummy variable $\endgroup$
    – Brenton
    Jun 18, 2015 at 19:07

1 Answer 1


As Brenton stated in the comments, the $u$ could be pretty much anything, because it goes away after you apply the bounds on the definite integral.

Perhaps a bit clearer would have been to designate it $t'$ instead of $u$:

$$\mathbb E I^2(t)=\mathbb E\int_0^t \Delta^2(t') dt'.$$

This makes it more semantically clear that you're integrating over time.


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