Deriving Normal Log Returns and Zero Covariance from Brownian Motion

# How to show that a Brownian motion is normally distributed and that the covariance is zero?


# How to show that a Brownian motion is normally distributed and that the covariance is zero?












I need help under standing this question. So i have the following given the logarithm of the price of a share of stock is given by

\begin{align*} p(t)=p(0)+\mu t+\sigma W(t), \quad t \in[0, T] \end{align*} where $p(0) \in \mathbb{R}$ is some fixed initial value, $\mu \in \mathbb{R}$ and $\sigma>0$ are constants, and $W(t)$ is a Brownian motion.

I know that the Brownian motion $W(t)$ has the properties

- $W(0)=0$

- $W$ has independent increments, i.e. if $0 \leq r<s \leq t<u,$ then \begin{align*} W(u)-W(t) \text { and } W(s)-W(r) \end{align*} are independent.

- The increments are normally distributed, i.e. \begin{align*} W(t)-W(s) \sim N(0, t-s) \end{align*} for all $0 \leq s \leq t$. Suppose that we have observed the price $p(t)$ at $n+1$ equidistant points \begin{align*} 0=t_{0}<t_{1}<\ldots<t_{n}=T \end{align*} with \begin{align*} t_{i}=\frac{i}{n} T, \quad i=0, \ldots, n \end{align*} I am given that the $n$ log-returns given as \begin{align*} r\left(t_{i}\right)=p\left(t_{i}\right)-p\left(t_{i-1}\right), \quad i=1, \ldots, n \end{align*}

I need to show the following

$r\left(t_{i}\right) \sim N\left(\mu \frac{T}{n}, \sigma^{2} \frac{T}{n}\right)$ and that $\operatorname{cov}\left(r\left(t_{i}\right), r\left(t_{i-1}\right)\right)=0$

## Answer by Pleb (score 1, accepted)

https://quant.stackexchange.com/a/60518

Normality of returns follows from the fact that the Brownian increments are normal distributed (property 3), $W(t_i)-W(t_{i-1}) \sim N(0,t_i-t_{i-1})$, and furthermore that $t_i-t_{i-1}=T(\frac{i}{n}-\frac{i-1}{n})=\frac{T}{n}$. See that:

\begin{align} r(t_i)&=p(t_i)-p(t_{i-1})\\ &=\left[p(0)+\mu\cdot t_i + \sigma W(t_i) \right] - \left[p(0)+\mu\cdot t_{i-1} + \sigma W(t_{i-1}) \right]\\ &= \mu (t_i-t_{i-1})+\sigma(W(t_i)-W(t_{i-1}))\\ &\overset{d}{=} \mu (t_i-t_{i-1}) + N\left(0,\sigma^2(t_i-t_{i-1})\right)\\ &=N\left(\mu\frac{T}{n},\sigma^2 \frac{T}{n}\right) \end{align} Now, the covariance of the returns follows directly from the independent increments property of Brownian motions. Under an alternative notation, $W(t_i)=W_{t_i}$, we have:

\begin{align} \small{Cov\left(r(t_i),r(t_{i-1})\right)} &= \small{\sigma^2 Cov\left(W_{t_i}-W_{t_{i-1}},\: W_{t_{i-1}}-W_{t_{i-2}} \right)}\\ &=\small{\sigma^2 \cdot (\mathbb{E}\left[(W_{t_{i}}-W_{t_{i-1}})\cdot (W_{t_{i-1}}-W_{t_{i-2}})\right]-\mathbb{E}\left[W_{t_{i}}-W_{t_{i-1}}\right]\cdot \mathbb{E}\left[W_{t_{i-1}}-W_{t_{i-2}}\right])}\\ &=\small{\sigma^2 \cdot (\mathbb{E}\left[(W_{t_{i}}-W_{t_{i-1}})\cdot (W_{t_{i-1}}-W_{t_{i-2}})\right] - 0)}\\ &=\small{\sigma^2 \cdot (\mathbb{E}\left[(W_{t_{i}}-W_{t_{i-1}})\right] \mathbb{E}\left[(W_{t_{i-1}}-W_{t_{i-2}})\right])}\\ &=\small{0}, \end{align} where we have used the independent increments property in the last equation and furthermore that the increments have expectation zero.