For a Wiener process $W_s$ and a partition $\{ S_0, S_1, \dots, S_n \}$ of $[0,t]$, I have been told that $$ \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} (W_{S_{i+1}} - W_{s_i})^2 = t $$ I am aware that the above summation is the expression for the 'quadratic variation' of $W_t$, but I am not sure why this is equal to $t$?
Can anyone please tell me how this result is derived?
(I will adapt an answer of mine in another post to answer this question)
Let $\pi=\{s_0,\dots,s_n\}$ denote the partition, we assume also $s_0=0$, $s_n=t$.
We define $$S_\pi(W):=\sum_{i=1}^n(W{_{s_i}}-W_{s_{i-1}})^2.$$ We want to show that $$\lim_{n\rightarrow\infty}S_{\pi}=t\ \ \ \ \text {in}\ \ \ L^2(\Omega),$$ which means $\lim_{n\rightarrow\infty}E[(S_\pi-t)^2]=0$ (where $\Omega$ denotes the measure space on which $W$ is defined).
Defining $Y_i:=(W_{s_{i}}-W_{s_{i-1}})^2-(s_i-s_{i-1})$ we can write $S_\pi-t=\sum_{i=1}^{n}Y_i$. We observe that the random variables $Y_1,...,Y_n$ are independent since each $Y_i$ is a function of $W_{s_{i}}-W_{s_{i-1}}$ and by the definition of Brownian motion its increments are independent. Moreover $$E[Y_i]=E[(W_{s_{i}}-W_{s_{i-1}})^2]-(s_i-s_{i-1})=\text{Var}(W_{s_{i}}-W_{s_{i-1}})-(s_i-s_{i-1})=0.$$ Thus $$E[(S_\pi-t)^2]=\sum_{i=1}^{n}E[Y_i^2].$$ We know that the distribution of $W_{s_{i}}-W_{s_{i-1}}$ is $N(0,s_{i}-s_{i-1})$ thus the distribution of $$Z:=\frac{W_{s_{i}}-W_{s_{i-1}}}{\sqrt{s_{i}-s_{i-1}}}$$ is $N(0,1)$. (Z doesn't depend on $i$ because we are interested only in its distribution and for every $i$ it is a standard normal). Using this fact we write $$\sum_{i=1}^{n}E[Y_i^2]=\sum_{i=1}^n E[\ [(W_{s_{i}}-W_{s_{i-1}})^2-(s_{i}-s_{i-1})]^2\ ]=\\ \sum_{i=1}^n (s_{i}-s_{i-1})^2\ E[\ [\left( \frac{W_{s_{i}}-W_{s_{i-1}}}{\sqrt{s_i-s_{i-1}}}\right)^2-1]^2\ ]=\\ \sum_{i=1}^n (s_{i}-s_{i-1})^2\ E[(Z^2-1)^2]=C\sum_{i=1}^n (s_{i}-s_{i-1})^2\leq \\ C \vert\vert\pi\vert\vert\sum_{i=1}^n s_{i}-s_{i-1}=Ct\vert\vert\pi\vert\vert,$$ where $C:=E[(Z^2-1)^2]$ is a constant and $\vert\vert\pi\vert\vert$ denotes the norm of the partition. Now if $n\rightarrow\infty$, then $\vert\vert\pi\vert\vert\rightarrow0$ and we have the result.