[[Brownian Motion]] is obtained as the continuous limit of a discrete [[Random Walk Model|Random Walk]]. The key is to scale the process so that as the number of steps increases, the [[Variance]] remains finite. We illustrate this with three cases. ## Constructing the Random Walk We begin by defining the [[Random Variable]] $z_t$ that represents the step taken at time $t$. Each step is assumed to be [[Independence and Identical Distribution|i.i.d.]] with the following properties: $ \mathbb E[z_t]=0, \quad \mathrm{Var}(z_t)=1$ The [[Covariance]] between two steps $z_t$​ and $z_s$​ is zero for different time points. This can be written using the Kronecker delta $\delta_{ts}$. $ \mathrm{Cov}(z_t, z_s) = \delta_{ts}, \quad \text{where}\quad \delta_{ts} = \begin{cases}0, & t \neq s\\ 1, & t=s \end{cases}$ Let $B_{\Delta t, T}$​ denote the random walk over a total time $T$ with a discrete time step $\Delta t$. The number of steps $n$ in the interval is given by: $ \Delta t = \frac{T}{n}, \quad B_{\Delta t, T}=\sum_{t=1}^n z_t$ ## Case 1: Rescaling The Time Step Only To move from the discrete Random Walk to a continuous-time interpretation, we let $n \to \infty$. As stated above, we require zero expectation and a finite variance in this limit to model Brownian Motion. **Expectation:** Since each $z_t$ has zero mean, the [[Expectation]] of the process $B_{\Delta t, T}$ is also zero. As this is not impacted by $n$, the expectation in the $\lim_{n \to \infty}$ stays zero. $ \mathbb E[B_{\Delta t,T}]=\mathbb E \left[\sum_{t=1}^n z_t\right] =n*\mathbb E[z_t]=0 $ **Variance:** As we are dealing with i.i.d. random variables, the [[Variance of Sum of Random Variables#Special Case of Independence|Variance of Sum of Random Variables]] is just the sum of individual variances. $ \mathrm{Var}(B_{\Delta t, T})=\mathrm{Var}\left(\sum_{t=1}^n z_t\right)=n \mathrm{Var}(z_t)=n $ As $n \to \infty$, the variance diverges. $ \lim_{n \to \infty} \mathrm{Var}(B_{\Delta t, T})= \infty$ This divergence occurs because the variance at each step remains constant while the number of steps increases without bound, making the process unsuitable for modeling Brownian motion. ## Case 2: Rescale Step Size with $\lambda = \frac{1}{n}$ To control the variance, we scale each $z_t$ by a factor of $\lambda$, where we set $\lambda = \frac{1}{n}$. We denote the scaled step as $\epsilon_t$. $ \epsilon_t=\lambda z_t \quad \text{where}\quad \lambda=\frac{1}{n} $ **Expectation:** The expectation of the process $B_{\Delta t,T}$ remains unchanged, as the scaling of zero expectation has no impact. $ \mathbb E[B_{\Delta t,T}]= \mathbb E\left[\sum_{t=1}^n \epsilon_t \right] = \lambda \: \mathbb E \left[\sum_{t=1}^n z_t \right]=n \lambda \:\mathbb E[z_t]=0 $ **Variance:** The variance is computed as: $ \mathrm{Var}(B_{\Delta t, T}) = \mathrm{Var}\left(\sum_{t=1}^n \epsilon_t \right) = \mathrm{Var}\left(\lambda \sum_{t=1}^n z_t \right)=\lambda^2n\mathrm{Var}(z_t) = \lambda^2n $ In the limit as $n \to \infty$, the variance of the process goes to $0$, because the inverse relationship of $\lambda=\frac{1}{n}$ is being squared (i.e. $\lambda^2$) and converges quicker to $0$ than $n$ converges to $\infty$. $ \lim_{n \to \infty} \mathrm{Var}(B_{\Delta t, T}) \to 0$ ## Case 3: Rescale Step Size with $\lambda=\sqrt{\Delta t}$ To achieve a variance that does not converge to $0$ or $\infty$, we scale each $z_t$ by a factor of $\lambda$, which we set to $\lambda = \sqrt{\Delta t}$. $ \epsilon_t=\lambda z_t \quad \text{where}\quad \lambda=\sqrt{\Delta t} $ **Expectation:** The expectation remains zero: $ \mathbb E[B_{\Delta t,T}]= \mathbb E\left[\sum_{t=1}^n \epsilon_t \right] = \lambda \: \mathbb E \left[\sum_{t=1}^n z_t \right]=n \lambda \:\mathbb E[z_t]=0 $ **Variance:** The variance of the process is: $ \mathrm{Var}(B_{\Delta t, T}) = \mathrm{Var}\left(\sum_{t=1}^n \epsilon_t\right) = \mathrm{Var}\left(\sqrt{\Delta t} \sum_{t=1}^n z_t\right) =\Delta t *n\mathrm{Var}(z_t) = \frac{T}{n}*n =T$ In this case, as $n \to \infty$ (or equivalently $\Delta t→0$), the variance remains constant at $T$. $ \lim_{n \to \infty} \equiv \lim_{\Delta t \to 0} \mathrm{Var}(B_{\Delta t, T}) \to T$ **Convergence in Distribution:** As we are dealing with a sum of many i.i.d. r.v's. that each have a finite variance of $\Delta t$, we can invoke [[Central Limit Theorem|CLT]]. Hence we can claim [[Modes of Convergence#Convergence in Distribution|Convergence in Distribution]] to a [[Gaussian Distribution|Gaussian]], where the mean and variance have already been determined above. $ \boxed{ \quad\lim_{n \to \infty} \equiv \lim_{\Delta t \to 0}=B_{\Delta t, T} \sim \mathcal N(0,T) \quad}$ > [!note:] > This convergence shows that, in the limit as $\Delta t \to 0$, the random walk process behaves like a Gaussian process with mean 0 and variance $T$. This is the key property of Brownian motion.