Given the random walk model $y_t=y_{t-1}+v_{t}$ where $v_t$ are iid $N(0,\sigma^{2}_{v})$, I want to find an expression for cov$(y_t,y_{t-k})$.
I've had a look at past posts, but they all seem to show cov$(y_t,y_{t+h})=\text{var}(y_{t})=t\sigma^{2}_{v}$ for what I assume is $h>0$, and my result below is different.
My attempt:
Let $k>0$. Then \begin{align} \text{cov}(y_t,y_{t-k})=&\text{cov}(y_{t-1}+v_{t},y_{t-k})\\ =&\text{cov}(y_{t-1},y_{t-k})\\ &\vdots\\ =&\text{cov}(y_{t-k},y_{t-k})\\ =&\text{var}(y_{t-k})\\ =&(t-k)\sigma_{v}^{2} \end{align} where the second inequality holds because $y_{t-k}$ does not depend on the future value $v_{t}$.
So the covariance between lags $h$ apart does not depend on the time interval $h$, but ones $k$ apart do. How can that be?