The variance of a simple random walk/process 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 $var(Z_i) = i$ ? 
 A: we know the mean $E[\epsilon_j]$=0
so the variance $E[\epsilon_j^2] - (E[\epsilon_j])^2$ is $E[\epsilon_j^2]$ which is 1
due to independence 
$var(Z_i)=\sum_{j=1}^i var(\epsilon_j)=\sum_{j=1}^i 1 = i$
A: Since the $\{\epsilon_i\}$ are independent:
$var(Z_i)=\sum_{j=1}^ivar(\epsilon_j) =i$.
A: 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 $
A: I think there is a more intuitive way to see this, if you express the simple Random walk in terms of the Bernoulli random variable (which is considered the most basic random variable that exists).
Let's mark this Bernoulli Random Variable as $X_i$, where $X_i=1$ with probability $0.5$ and $X_i=0$ with probability $0.5$.
First note that $\epsilon_i$ can be actually expressed in terms of $X_i$:
$$\epsilon_i = 2X_i -1$$
You should double check the above expression to satisfy yourself that it holds true (if $X_i = 0$, indeed $\epsilon_i = -1$, and if $X_i = 1$, indeed $\epsilon_i = 1$).
Now it is super simple. $Z_i = \sum_{j=1}^{i} \left( 2X_i -1 \right) = 2* \sum_{j=1}^{i} (X_i) - i  $.
$$ Var(Z_i) = Var\left( 2* \sum_{j=1}^{i} (X_i) - i \right) = 4\sum_{j=1}^{i} Var(X_i)$$
Now $Var(X_i)$ for the Bernoulli random variable is just $p*(1-p)$. Since $p=0.5$, then $Var(X_i)=0.25$. Therefore:
$$4\sum_{j=1}^{i} Var(X_i) = \sum_{j=1}^{i} 4 * 0.25 = \sum_{j=1}^{i} 1 = i $$
