This problem is given as a motivating example for the Ito Integral in Mikosch's Elementary Stochastic Calculus with Finance in View...
Consider the Riemann-Stieltjes Sum $$S_n = \sum^n_{i=1} B_{t_{i-1}}\Delta_i B$$ where $$B=(B_t,t\geq 0)$$ is a Brownian Motion, $$\tau_n: 0=t_0<t_1<...<t_{n-1}<t_n=t$$ is a partition of $[0,t]$ and, for any function $f$ on $[0,t]$, $$\Delta f: \Delta_i f = f(t_i)-f(t_{i-1}), i=1,...,n,$$ are the corresponding increments of $f$ and $$\Delta_i = t_i - t_{i-1}, i=1,...,n.$$ The Riemann-Stieltjes sum $S_n$corresponds t the partition $\tau_n$ and the intermediate partition $(y_i)$ with $y_i=t_{i-1}$ is the left end point of the interval $[t_{i-1},t_i]$. It follows that $S_n$ can be written in the form $$S_n=\frac{1}{2}B^2_t - \frac{1}{2}\sum^n_{i=1}(\Delta_i B)^2.$$
I don't understand how the last equation for $S_n$ is derived since what I got is $$S_n=\frac{1}{2}B^2_{t_0}+B^2_{t_1}+B^2_{t_2}+...+B^{2}_{t_{n-1}}+\frac{1}{2}B^2_{t_n} - \frac{1}{2}\sum^n_{i=1}(\Delta_i B)^2$$ $$\Leftrightarrow S_n=B^2_{t_1}+B^2_{t_2}+...+B^{2}_{t_{n-1}}+\frac{1}{2}B^2_{t_n} - \frac{1}{2}\sum^n_{i=1}(\Delta_i B)^2.$$ My question is, where did $$B^2_{t_1}+B^2_{t_2}+...+B^{2}_{t_{n-1}}$$ go?
Oh, I am so sorry to be so stupid. It's just a matter of arithmetic.
$$S_n = \sum^n_{i=1} B_{t_{i-1}}\Delta_i B = \sum^n_{i=1} B_{t_{i-1}} (B_{t_i} - B_{t_{i-1}})=\sum^n_{i=1}(B_{t_{i-1}}B_{t_i}-B^2_{t_{i-1}}).$$ But by square of a binomial we get $$B_{t_{i-1}}B_{t_i}=\frac{1}{2}B^2_{t_i} + \frac{1}{2}B^2_{t_{i-1}}-\frac{1}{2}(\Delta_i B)^2.$$
So, $$S_n = \frac{1}{2}\sum^n_{i=1}(B^2_{t_i}-B^2_{t_{i-1}})-\frac{1}{2}\sum^n_{i=1}(\Delta_i B)^2$$$$=\frac{1}{2}(B^2_1-B^2_0+B^2_2-B^2_1+...+B^2_n-B^2_{n-1})+\frac{1}{2}\sum^n_{i=1}(\Delta_i B)^2.$$ Now, $B_{t_0}=B_0=0$ and $B_{t_n}=B_t$ the other $B_{t_i}, i=1,...,n-1$ cancel since they are of opposite signs. Hence, we end up with $$S_n=\frac{1}{2}B^2_{t}-\frac{1}{2}\sum^n_{i=1}(\Delta_i B)^2.$$