From random walk to Brownian motion with drift

Suppose we divide a time interval of length $\Delta t$ into $n = \frac{\Delta t}{\Delta_n t}$ steps. Let $X$ be a random variable with initial value $X_0$ . At the first step, the value of the random variable can go up by $\Delta h$ or down by $\Delta h$ with probability $p$ and $q = 1 - p$ respectively. (This is a Bernoulli trial with probability of success $p$ ). Then at the first step we have $\Delta X_1 = X_1 - X_0$ and the expected value and variance are

$E[\Delta X_1] = p \Delta h - q \Delta h = (p - q) \Delta h$

and

$V[\Delta X_1] = E[(\Delta X_1)^2] - E^2[\Delta X_1]$

$= p(\Delta h)^2 + q(\Delta h)^2 - (p - q)^2 (\Delta h)^2$

$= (\Delta h)^2[p + q - (p - q)^2]$

$= 4pq(\Delta h)^2$

At the second step we similarly have $\Delta X_2 = X_2 - X_1$ with the same expected value and variance. And so on. This step-by-step process is a discrete-time random walk. Now let

$\Delta X_t = \Delta X_1 + \Delta X_2 + \cdots + \Delta X_n$

be the position after $n$ steps. Then $\Delta X_t$ is a binomial random variable with expected value

$E[\Delta X_t] = E[\Delta X_1 + \Delta X_2 + \cdots + \Delta X_n] = n(p - q)\Delta h = \Delta t (p - q) \frac{\Delta h}{\Delta_n t}$

and variance

$V[\Delta X_t] = V[\Delta X_1 + \Delta X_2 + \cdots + \Delta X_n] = n4pq(\Delta h)^2 = 4pq\Delta t \frac{(\Delta h)^2}{\Delta_n t}$

We now want to pass to continuous time by letting $n \rightarrow \infty$ or equivalently $\Delta_n t \rightarrow 0$ , and we want to make the expected value and variance independent of $\Delta h$ and $\Delta_n t$ in this limit. For this purpose, we need to define

$\Delta h = \sigma \sqrt{\Delta_n t}$

$p = \frac{1}{2} \bigg[1 + \frac{\alpha}{\sigma} \sqrt{\Delta_n t}\bigg]$

$q = \frac{1}{2} \bigg[1 - \frac{\alpha}{\sigma} \sqrt{\Delta_n t}\bigg]$

where $\alpha$ and $\sigma$ are two parameters whose role will become clear shortly. Substituting these three into the expressions for $E[\Delta X_t]$ and $V[\Delta X_t]$ and passing to the limit $\Delta_n t \rightarrow 0$ we get the expected value

$E[\Delta X_t] = \Delta t (p - q) \frac{\Delta h}{\Delta_n t} = \Delta t \bigg[\frac{\alpha}{\sigma} \sqrt{\Delta_n t} \bigg] \frac{\sigma \sqrt{\Delta_n t}}{\Delta_n t} = \alpha \Delta t$

and variance

$V[\Delta X_t] = 4pq\Delta t \frac{(\Delta h)^2}{\Delta_n t}$

$= 4 \times \frac{1}{4} \bigg[1 - \frac{\alpha^2}{\sigma^2} \Delta_n t \bigg] \Delta t \sigma^2 \frac{\Delta_n t}{\Delta_n t}$

$= \bigg[1 - \frac{\alpha^2}{\sigma^2} \Delta_n t \bigg] \Delta t \sigma^2$

$\rightarrow \sigma^2 \Delta t$

Thus, in the limit as $n \rightarrow \infty$ , both the mean and variance are independent of $\Delta h$ and $\Delta_n t$ . Note in particular that $\Delta h$ must depend on the square root of $\Delta_n t$ for this to be achieved. Were it not for this, the mean and variance would depend on $\Delta h$ and on the number of steps $n$ .

Now, as the number of steps $n$ becomes large, the binomial random variable $\Delta X_t$ becomes normally distributed, with

$\Delta X_t \sim \mathcal{N}(\alpha \Delta t, \sigma^2 \Delta t)$

Let $\epsilon_t \sim \mathcal{N}(0, 1)$ . Then we have

$\frac{\Delta X_t - \alpha \Delta t}{\sigma \sqrt{\Delta t}} \overset{d}{=} \epsilon_t$

$\Delta X_t \overset{d}{=} \alpha \Delta t + \sigma \epsilon_t \sqrt{\Delta t}$

Passing to differentials we then have the stochastic differential equation of Brownian motion with drift:

$dx = \alpha dt + \sigma dz$

where

$dz = \epsilon_t \sqrt{dt}$

is the (infinitesimal) increment of a Wiener process $z(t)$ (also known as Brownian motion), $\alpha$ is a drift parameter, and $\sigma$ is a variance parameter. Note that in the increment $dz$ of the Wiener process $z(t)$ we have $\epsilon_t \sim \mathcal{N}(0, 1)$ and $\epsilon_t$ is serially uncorrelated, i.e., $E[\epsilon_s \epsilon_t] = 0$ for $s \neq t$ . Thus, the values of $dz$ for any two different time intervals are independent, so $z(t)$ follows a Markov process with independent increments. We have $E[dz] = 0$ and $V[dz] = E[dz^2] = dt E[\epsilon_t^2] = dt$ . Therefore the variance of the change in a Wiener process grows linearly with the time horizon.

Note that the Wiener process does not have a time derivative in the conventional sense because with

$\Delta z = \epsilon_t \sqrt{\Delta t}$

we have

$\frac{\Delta z}{\Delta t} = \epsilon_t (\Delta t)^{-\frac{1}{2}}$

and this becomes infinite when we try to pass to the limit $\Delta t \rightarrow 0$ . This is why we need to use Itô calculus to deal with Brownian motion and related processes, rather than being able to use conventional calculus.

Brownian motion with drift is a simple generalisation of the Wiener process. Over any time interval $\Delta t$ , the change in $x$ , $\Delta x$ , is normally distributed with $E[\Delta x] = \alpha \Delta t$ and $V[\Delta x] = \sigma^2 \Delta t$ . From the above development starting with a discrete-time random walk, we can now see why $dx$ must depend on the square root of $dt$ , not just on $dt$ .