Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability.

Question

Let $\{t_i\}_{i=1}^n$ be a partition of $[0,t]$ and $W$ a standard Brownian motion. Write $W_i$ for $W_{t_i}$.
Show $$ \lim \sum W_{i} (W_{i+1}-W_i)=\frac12 W^2_t-\frac12 t $$ where the limit is in probability.

The proof is in our textbook (Kurtz, Stochastic Analysis). It goes as follow \begin{align} \lim \sum_{i=1}^n W_{i} (W_{i+1}-W_i) &= \lim \sum_{i=1}^n \left( W_i W_{i+1} - \frac12 W^2_{i+1}-\frac12 W^2_i \right)+\sum_{i=1}^n \left( \frac12 W^2_{i+1} - \frac12 W^2_i \right) \\ &=\frac12 W_t^2 - \lim \frac12 \sum_{i=1}^n \left( W_{i+1}-W_{i} \right)^2 \\ &=\frac12 W_t^2 - \frac12 t^2 \end{align}

How does the second equality follows?

Answer 1

The second sum in the RHS of the first equality is telescopic and gives the first term in the second line. For the first sum, note that $$W_iW_{i+1}-\frac 12W_{i+1}^2-\frac 12W_i^2=-\frac 12(W_{i+1}-W_i)^2,$$ hence a factor $\frac 12$ is missing.

Answer 2

Usually, for a Riemann integral, you say $$ \int f(t) dt = \lim_{|\delta t| \to 0} \sum f(t) \Delta t $$

Your textbook is showing that $\Delta W = W_{i+1}- W_i \approx \sqrt{\Delta t}$ since variance of random walk grows as the square-root of time.

First of all there's a telescoping sum: $$ \sum_{i=1}^n \left( \frac12 W^2_{i+1} - \frac12 W^2_i \right) = \frac12 W_t^2 $$

and then binomial formula $(x+y)^2 = x^2 + 2xy + y^2$:

$$ \lim \sum_{i=1}^n \left( W_i W_{i+1} - \frac12 W^2_{i+1}-\frac12 W^2_i \right) = \lim \sum_{i=1}^n \left( W_{i+1}-W_{i} \right)^2 = \frac12 t^2 $$

The word "lim" is very important. We can approximate the random walk over small time interval as binomial random variable $$ \mathbb{P}(\Delta W = \sqrt{\Delta t} )= \mathbb{P}(\Delta W = -\sqrt{\Delta t} )= \frac{1}{2}$$ and no matter what $(\Delta W)^2 = \Delta t$.