I'm a beginner in studying Brownian motion with some background in probability theory and I ran into some problems going through the textbook Brownian Motion by Schilling:
Problem Setup
Let's denote the random position of a particle at time $t$ in $[0, T]$ and we will model its movement (left and right) by iid distributed Bernoulli random variables, $\epsilon_k$, $k >= 1$, where
$P(\epsilon_k = 1) = P(\epsilon_k = 0) = 1/2$
so that $S_n = \epsilon_1 + ... + \epsilon_N$ and $N - S_N$ denote the number of right and left moves.
As a result, the position of the particle at $T = N\Delta t$ is
$X_T = S_N\Delta x - (N - S_N)\Delta x = \sum_{k=1}^N (2\epsilon_k-1)\Delta x$
Then, the book goes on to define $t = n\Delta t$ and say since the $\epsilon_k$ are iid random variables, the two increments $X_T - X_t$ and $X_t - X_0$ are independent. From there, we write $\sigma^2(t) := Var(X_t)$ and by Bienayme's identity we get
$Var(X_T) = Var(X_T - X_t) + Var(X_t - X_0) = \sigma^2(T - t) + \sigma^2(t)$
which means that the relationship is linear: $Var(X_T) = \sigma^2(T) = \sigma^2T$.
Questions
I don't understand why the book then says since $E(\epsilon_k) = 1/2$ and $Var(\epsilon_k) = 1/4 $ it can get by a direction calulation that
$Var(X_T) = N(\Delta x)^2$
and how it calculates $S_N^*$ and arrives at the conclusion that
$X_T = \sum_{k=1}^N (2\epsilon_k-1)\Delta x = (2S_N - N)\Delta x = S_N^*\sqrt T\sigma$, where $S_N^* = \frac{2S_N - N}{\sqrt N} = \frac{S_N - ES_N}{\sqrt {Var(S_N)}}$
and $S_N^*$ is the normalization (mean 0 and variance 1) of the random variable $S_N$.
Thank you for taking the time and helping me!