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Second-Order Properties of a Random Walk

If \(\{x_t\}_{t=1}^n\) is a random walk, then the population has the following properties.

\[ \mu_x = 0 \] and \[ cov(x_t, x_{t+k}) = t \sigma^2 \]

Why is \(cov(x_t, x_{t+k}) = t \sigma^2\)?

First, note that that since the terms in the white noise series are independent,

\[ cov ( w_i, w_j ) = \begin{cases} \sigma^2, & \text{if } ~ i=j \\ 0, & \text{otherwise} \end{cases} \]

Also, when random variables are independent, the covariance of a sum is the sum of the covariance.

Hence, \[\begin{align*} cov(x_t, x_{t+k}) &= cov ( \sum_{i=1}^t w_i, \sum_{j=1}^{t+K} w_j ) \\ &= \sum_{i=j} cov ( w_i, w_j ) \\ &= \sum_{i=1}^t \sigma^2 \\ &= t \sigma^2 \end{align*}\]

If \(k>0\) and \(t>0\), the correlation function is

\[ \rho_k = \frac{ cov(x_t, x_{t+k}) }{ \sqrt{var(x_t)} \sqrt{var(x_{t+k})} } = \frac{t \sigma^2}{\sqrt{t \sigma^2} \sqrt{(t+k) \sigma^2}} = \frac{1}{\sqrt{1+\frac{k}{t}}} \]

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