Expected value of product of two random variables If we have a process defined as:

$Y_t =  ε_t + Y_{t-1}$

How do you determine the expected value:

$E[Y_tY_{t-1}]$
Since $Y_t = Y_0 + \sum_{i=1}^t \epsilon_i$, if $Y_0 = 0$ we get $Y_t = \sum_{i=1}^t \epsilon_i$. Furthermore you said $\epsilon_i \sim N(0,1)$, and since under the assumption of independence of the $\epsilon_i$ the sum of (standard) normal random variables is again a normal random variable (with the means and variances added) we get $Y_t = \sum_{i=1}^t \epsilon_i \sim N(0,t)$. Writing out the expected value a bit, we get

$$\mathbb{E}[Y_t Y_{t-1}] = \mathbb{E}[(Y_{t-1} + \epsilon_t)Y_{t-1}] = \mathbb{E}[Y_{t-1}^2] + \mathbb{E}[\epsilon_t Y_{t-1}].$$

From $Y_{t-1} \sim N(0,t-1)$ it follows that $\mathrm{Var}(Y_{t-1}) = \mathbb{E}[Y_{t-1}^2] - \mathbb{E}[Y_{t-1}]^2 = \mathbb{E}[Y_{t-1}^2] = t - 1$. Finally, since $\epsilon_t$ and $Y_{t-1}$ are independent and symmetric around $0$, it follows that

$$\mathbb{E}[Y_t Y_{t-1}] = \mathbb{E}[Y_{t-1}^2] + \mathbb{E}[\epsilon_t Y_{t-1}] = (t - 1) + 0 = t - 1.$$

Without the assumption of independence of the $\epsilon_i$, however, this does not work.
$E[Y_{t}Y_{t-1}]=E[E[Y_{t}Y_{t-1}|\mathcal{F}_{t-1}]]=EY_{t-1}^{2}$