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Here, we introduce a construction of Brownian motion from a symmetric random walk. Divide the half-line $[0, \infty)$ to tiny subintervals of length $\delta$ as shown in Figure 11.30.

Each subinterval corresponds to a time slot of length $\delta$. Thus, the intervals are $(0,\delta]$, $(\delta, 2\delta]$, $(2\delta, 3\delta]$, $\cdots$. More generally, the $k$th interval is $\big((k-1)\delta, k\delta \big]$. We assume that in each time slot, we toss a fair coin. We define the random variables $X_i$ as follows. $X_i=\sqrt{\delta}$ if the $k$th coin toss results in heads, and $X_i=-\sqrt{\delta}$ if the $k$th coin toss results in tails. Thus, \begin{equation} \nonumber X_i = \left\{ \begin{array}{l l} \sqrt{\delta} & \quad \textrm{with probability }\frac{1}{2} \\ & \quad \\ -\sqrt{\delta} & \quad \textrm{with probability }\frac{1}{2} \end{array} \right. \end{equation} Moreover, the $X_i$'s are independent. Note that \begin{align*} E[X_i] &=0,\\ \textrm{Var}(X_i)&=\delta. \end{align*} Now, we would like to define the process $W(t)$ as follows. We let $W(0)=0$. At time $t=n \delta$, the value of $W(t)$ is given by \begin{align*} W(t)=W(n \delta)=\sum_{i=1}^{n} X_i. \end{align*} Since $W(t)$ is the sum of $n$ i.i.d. random variables, we know how to find $E[W(t)]$ and $\textrm{Var}(W(t))$. In particular, \begin{align*} E[W(t)]&=\sum_{i=1}^{n} E[X_i]\\ &=0,\\ \textrm{Var}(W(t))&=\sum_{i=1}^{n} \textrm{Var}(X_i)\\ &=n \textrm{Var}(X_1)\\ &=n \delta\\ &=t. \end{align*} For any $t \in (0,\infty)$, as $n$ goes to $\infty$, $\delta$ goes to $0$. By the central limit theorem, $W(t)$ will become a normal random variable, $$W(t) \sim N(0, t).$$ Since the coin tosses are independent, we conclude that $W(t)$ has independent increments. That is, for all $0 \leq t_1 \lt t_2 \lt t_3 \cdots \lt t_n$, the random variables \begin{align*} W(t_2)-W(t_1), \; W(t_3)-W(t_2), \; \cdots, \; W(t_n)-W(t_{n-1}) \end{align*} are independent. Remember that we say that a random process $X(t)$ has stationary increments if, for all $t_2>t_1\geq0$, and all $r>0$, the two random variables $X(t_2)-X(t_1)$ and $X(t_2+r)-X(t_1+r)$ have the same distributions. In other words, the distribution of the difference depends only on the length of the interval $(t_1,t_2]$, and not on the exact location of the interval on the real line. We now claim that the random process $W(t)$, defined above, has stationary increments. To see this, we argue as follows. For $0 \leq t_1 \lt t_2$, if we have $t_1=n_1 \delta$ and $t_2=n_2 \delta$, we obtain \begin{align*} W(t_1)&=W(n_1 \delta)=\sum_{i=1}^{n_1} X_i,\\ W(t_2)&=W(n_2 \delta)=\sum_{i=1}^{n_2} X_i. \end{align*} Then, we can write \begin{align*} W(t_2)-W(t_1)=\sum_{i=n_1+1}^{n_2} X_i. \end{align*} Therefore, we conclude \begin{align*} E[W(t_2)-W(t_1)]&=\sum_{i=n_1+1}^{n_2} E[X_i]\\ &=0,\\ \textrm{Var}(W(t_2)-W(t_1))&=\sum_{i=n_1+1}^{n_2} \textrm{Var}(X_i)\\ &=(n_2-n_1) \textrm{Var}(X_1)\\ &=(n_2-n_1) \delta\\ &=t_2-t_1. \end{align*}

Therefore, for any $0 \leq t_1 \lt t_2$, the distribution of $W(t_2)-W(t_1)$ only depends on the lengths of the interval $[t_1, t_2]$, i.e., how many coin tosses are in that interval. In particular, for any $0 \leq t_1 \lt t_2$, the distribution of $W(t_2)-W(t_1)$ converges to $N(0, t_2-t_1)$. Therefore, we conclude that $W(t)$ has stationary increments.

The above construction can be made more rigorous. The random process $W(t)$ is called the standard Brownian motion or the standard Wiener process. Brownian motion has continuous sample paths, i.e., $W(t)$ is a continuous function of $t$ (See Figure 11.29). However, it can be shown that it is nowhere differentiable.