Autocovariance of a Random Walk Proof

Question

I'm currently looking at these notes online to try to understand the proof of the auto-covariance of a random walk (Slides 19 and 29 specifically) http://www.personal.psu.edu/asb17/old/sta4853/files/sta4853-2.pdf

Given a random walk we know at time $t$ the random process $Z_t = \delta t +\sum_{j=1}^{t} a_{j}$.

We know that the autocovariance is: $$\gamma(s,t) = Cov(Z_s,Z_t) = Cov(\sum_{j=1}^{s} a_{j},\sum_{j=1}^{t} a_{k})$$ $$= \sum_{j=1}^{s}\sum_{j=1}^{t} cov(a_j,a_k)$$ $$ = min(s,t) $$

I'm confused about the last line. Why does the summation equal to the minimum of s and t?

Answer 1

You're told at the top of slide $29$ that $\ a_t\sim\ \text{WN}(0, \sigma^2)\ $, so $\ Cov\left(a_j,a_k\right)=\sigma^2\delta_{jk}\ $. Therefore \begin{align} \sum_{j=1}^s\sum_{k=1}^t Cov\left(a_j,a_k\right) &=\sigma^2\sum_{j=1}^s\sum_{k=1}^t \delta_{jk}\\ &= \sigma^2\sum_{j=1}^s\mathbb{1}_{\{j\le t\}}\\ &= \sigma^2\min(s,t) \end{align}