How do you prove the result that for equation:
$$y_t = \beta_1 x_t + \beta_2 y_{t-1} + u_t$$
the beta parameters are biased when $u_t$ is autocorrelated? In other words, that$$ \text{Cov}(u_t, y_{t-1})$$ is not equal to 0 when $\text{E}(u_t, u_{t+j}) \neq 0$
Here is how I started, but got nowhere. First the definition of covariance:
$$\text{Cov}(u_t, y_{t-1}) = \text{E}[(y_{t-1}-\text{E}(y_{t-1})(u_t-\text{E}(u_t)]$$
Then figure out each component one by one. As a condition we have: $$\text{E}(u_t) = 0$$ As we know: $$y_{t-1} = \beta_1x_{t-1} + \beta_2y_{t-2} + u_{t-1}$$
The expected vale of which is (possibly where I go wrong). $$\text{E}(y_{t-1}) = \beta_1x_{t-1} + \beta_2y_{t-2}$$
Substituting the results into covariance equation:
$$\text{E}(u_{t-1})u_t = 0$$
So I possibly made a mistake somewhere. In addition I would like to inquire (perhaps relating to the issue) how is it that in these type of proofs for example:
$$\text{E} \left ( \frac{u_t}{1-b} \right )$$ is not equal to 0. Example in page 5 of this paper, onward from $\text{Cov}(u_t, y_t)=$: http://gauss.stat.su.se/gu/e/slides/Time%20Series/Simultaneous%20equation%20model.pdf (even more prepelexing, the same term appears to be 0 before in solving for $\text{E} \left ( y_t \right )$, same page.)
There you have it, a pdf works fine as an answer. I wouldn't mind a more intuitive way to understand the answer either.