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I understand for some function $k: [0, 1] \to [0, c]$, the Itô integral is

$$\int_{0}^{1} k(r) \, \mathrm{d}B(r) = \lim_{n\to\infty} \sum_{i=1}^{n} k(r_{i-1}) [B(r_i) - B(r_{i-1})] $$

for a standard Brownian motion $B$. Does this mean that

\begin{align*} \int_{0}^{1} k(r) \, \mathrm{d}V(r) &= \lim_{n\to\infty} \sum_{i=1}^{n} k(r_{i-1}) \left[\left \{ B(r_i)- r_iB(1) \right \} - \left \{ B(r_{i-1})- r_{i-1}B(1) \right\} \right] \\ &= \lim_{n\to\infty} \sum_{i=1}^{n} k(r_{i-1}) \left[B(r_i) - B(r_{i-1})- B(1)(r_i - r_{i-1})\right] \end{align*}

for a standard Brownian bridge process $V$? I am new to Ito's calculus, and am checking my understanding. If you can confirm and/or provide a reference that would be much appreciated!

kpr62
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    The first formula is not correct. You should have $$ \int_{0}^{1} k(r) , \mathrm{d}B(r) = \lim_{n\to\infty} \sum_{i=1}^{n} \color{red}{k(r_{i-1})} [B(r_i) - B(r_{i-1})] $$ It is worth emphasizing that this change does make difference. (In terms of a fair stock market, this means that you can't predict the future!) – Sangchul Lee Apr 28 '23 at 04:10
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    To piggy-back a little on Sangchul Lee's comment: It's worth noting that the integral you've written is used sometimes (mostly in physics, it seems), but it is decidedly not the Ito integral. The naming convention for it seems all over the place, but some people even call it the "anti-Ito integral". – Brian Moehring Apr 28 '23 at 04:31
  • Edited to be more consistent with the financial realm. Is the Brownian bridge process integral correct? – kpr62 Apr 28 '23 at 15:52
  • First, I don't know if that definition is appropriate. In my experience, we'd first need to compute the SDE the Brownian bridge satisfies and then use that. – Brian Moehring Apr 28 '23 at 18:58

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