The variance of a simple random walk/process

Question

I've been trying to wrap my head around this for the past day. Please help!

Let $\epsilon_i = \pm 1$ with equal probabilities independently for $i=1,...,N$. Then $Z_i = \epsilon_1 + ... + \epsilon_i$ is a random walk. $Z_i$ is a random walk process for $i = 1, ..., N$.

Why is the variance $var(\epsilon_i) = 1$ and $var(Z_i) = i$ ?

Answer 1

For the $\epsilon_i$ there are only two possibilities: $1$ and $-1$. As noted, the mean is $0$.

$\therefore var(\epsilon_i)=\frac12\left((1-0)^2+(-1-0)^2\right)=1 $

Answer 2

Since the $\{\epsilon_i\}$ are independent: $var(Z_i)=\sum_{j=1}^ivar(\epsilon_j) =i$.