In the book "Handbook of Stochastic Methods for Physics" by Crispin and Gardiner, I found the following calculation to show that stochastic integrals depend on the choice of partitioning points.
First, we define the stochastic integral $\int_{t_0}^t W(t') dW(t')$ as a limit of the partial sums: \begin{align} S_n = \sum_{i=1}^n W(\tau_i) [W(t_i)-W(t_{t-1})] \, , \end{align} where $\tau_i$ denotes a specific choice of intermediate points.
Then we can calculate \begin{align} \langle S_n \rangle &= \langle \sum_{i=1}^n W(\tau_i) [W(t_i)-W(t_{t-1})] \rangle \\ &=\sum_{i=1}^n [\text{min}(\tau_i ,t_i) -\text{min}(\tau_i,t_{i-1}) ] \\ &= \sum_{i=1}^n (\tau_i-t_{i-1}) \, . \end{align} I don't understand the first step here. Why is the expectation value equal to a difference of minima?
No, your second "insight" in the comments $⟨W(t)W(s)⟩=0$ for $s\ne t$ is wrong, I think you were thinking of the independence of increments, $⟨dW(t)dW(s)⟩=0$.
For the paths themselves you get for $t<s$ $$⟨W(t)W(s)⟩=⟨W(t)W(t)⟩+⟨W(t)(W(s)-W(t))⟩=t,$$ as the path on $[0,t]$ is independent of the (incremental) path on $[t,\infty)$. If you do not assume the order of arguments, then $⟨W(t)W(s)⟩=\min(s,t)$ results.