The problem
For a smooth function $g(t)$, the Riemann-Stieltjes integral
$$ \int_0^T f(t)\,dg(t) $$does not depend on whether the Riemann sum uses left endpoints, right endpoints, or midpoints.
What changes for Brownian motion
If we try to define
$$ \int_0^T B_t\,dB_t $$by sums of the form
$$ \sum_{i=0}^{N-1} B_{\xi_i}\bigl(B_{t_{i+1}}-B_{t_i}\bigr), $$then different choices of $\xi_i$ lead to different limits.
Consequence
For Brownian motion, stochastic integration is not automatic. We must choose a specific discretization rule.
Source Links
- STA447
- Phase 7 Atomic Reading Order
- Phase 7 — Itô Integral 定义与性质
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