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11.4.2 Definition and Some Properties

Here, we provide a more formal definition for Brownian Motion.
Standard Brownian Motion

A Gaussian random process $\{W(t), t \in [0, \infty) \}$ is called a (standard) Brownian motion or a (standard) Wiener process if
  1. W(0)=0;
  2. for all $0 \leq t_1 \lt t_2$, $W(t_2)-W(t_1) \sim N(0, t_2-t_1)$;
  3. 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;
  4. W(t) has continuous sample paths.
A more general process is obtained if we define $X(t)=\mu+\sigma W(t)$. In this case, $X(t)$ is a Brownian motion with $$E[X(t)]=\mu, \quad \textrm{Var}(X(t))=\sigma^2 t.$$ Nevertheless, since $X(t)$ is obtained by simply shifting and scaling $W(t)$, it suffices to study properties of the standard Brownian motion, $W(t)$.

Example
Let $W(t)$ be a standard Brownian motion. For all $s,t \in [0, \infty)$, find $$C_W(s,t)=\textrm{Cov}(W(s),W(t)).$$
  • Solution
    • Let's assume $s \leq t$. Then, we have \begin{align*} \textrm{Cov}\big(W(s),W(t)\big)&=\textrm{Cov}\big(W(s), W(s)+W(t)-W(s)\big)\\ &=\textrm{Cov}\big(W(s), W(s)\big)+\textrm{Cov}\big(W(s), W(t)-W(s)\big)\\ &=\textrm{Var}\big(W(s)\big)+\textrm{Cov}\big(W(s), W(t)-W(s)\big)\\ &=s+\textrm{Cov}\big(W(s), W(t)-W(s)\big). \end{align*} Brownian motion has independent increments, so the two random variables $W(s)=W(s)-W(0)$ and $W(t)-W(s)$ are independent. Therefore, $\textrm{Cov}\big(W(s), W(t)-W(s)\big)=0$. We conclude \begin{align*} \textrm{Cov}\big(W(s),W(t)\big)=s. \end{align*} Similarly, if $t \leq s$, we obtain \begin{align*} \textrm{Cov}\big(W(s),W(t)\big)=t. \end{align*} We conclude \begin{align*} \textrm{Cov}\big(W(s),W(t)\big)=\min(s,t), \quad \textrm{ for all }s,t. \end{align*}

If $W(t)$ is a standard Brownian motion, we have \begin{align*} \textrm{Cov}(W(s),W(t))=\min(s,t), \quad \textrm{ for all }s,t. \end{align*}

Example
Let $W(t)$ be a standard Brownian motion.
  1. Find $P(1 \lt W(1) \lt 2)$.
  2. Find $P(W(2) \lt 3 | W(1)=1)$.
  • Solution
      1. We have $W(1) \sim N(0,1)$. Thus, \begin{align*} P(1 \lt W(1) \lt 2)&=\Phi(2)-\Phi(1)\\ &\approx 0.136 \end{align*}
      2. Note that $W(2)=W(1)+W(2)-W(1)$. Also, note that $W(1)$ and $W(2)-W(1)$ are independent, and $$W(2)-W(1) \sim N(0,1).$$ We conclude that \begin{align*} W(2) | W(1)=1 \; \sim \; N(1,1). \end{align*} Thus, \begin{align*} P(W(2) \lt 3 | W(1)=1) &=\Phi\left(\frac{3-1}{1}\right)\\ &=\Phi(2) \approx 0.98 \end{align*}

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