Showing Non-Stationarity of Random Walk with Drift

Question

So, I'm using the definition $Y_t= t\theta_0 + \sum e_t$ where $e_t$ is i.i.d with zero mean and constant variance. And I'm trying to show that $Cov(Y_t, Y_{t-k})=(t-k)\sigma_e^2$

So far I've got:

$Cov(Y_t, Y_{t-k})= Cov(t\theta_0 + \sum e_{t-1}, (t-k)\theta_0 + \sum e_{t-k-1}$)

$=Cov[(t\theta_0, (t-k)\theta_0)+(t\theta_0, \sum e_{t-k-1}) + (\sum e_{t-1}, (t-k)\theta_0)+(\sum e_{t-1}, \sum e_{t-k-1})]$

and then I think that all but the last covariances are 0, but then I'm just left with $Cov(\sum e_{t-1}, \sum e_{t-k-1})$ which looks like it should get me my desired result but what logic do I apply? Should I specify a specific lag or something so that I can do $Var(e_{t-k-1})$? Thank you!

Answer 1

$$ Cov(Y_t,Y_{t-k}) = Cov \Big(t \theta_{0} + \sum_{i=1}^{t} e_i, (t-k) \theta_{0} + \sum_{i=1}^{t-k} e_i \Big) \\ = Cov\Big( t \theta_{0}, (t-k) \theta_0\Big) + Cov\Big(t \theta_{0}, \sum_{i=1}^{t-k} e_i \Big) + Cov\Big(\sum_{i=1}^{t} e_i, (t-k) \theta_0\Big) + Cov\Big(\sum_{i=1}^{t} e_i, \sum_{i=1}^{t-k} e_i\Big) \\ = 0 + 0+0+ Cov\Big(\sum_{i=1}^{t} e_i, \sum_{i=1}^{t-k} e_i\Big) = \sum_{i=1}^{t-k} Var(e_i) = \sigma_{\epsilon}^{2}(t-k), $$ where one uses the bilinearity of covariance, the independence of the innovations $e_i$ and the fact that the covariance of two variables one of which is a constant equals zero.