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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.

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    $\begingroup$ Please clarify the title/the question. Your first display is just an equation, which cannot be "biased" in and of itself. Biased for what? $\endgroup$ Apr 6, 2016 at 10:38
  • $\begingroup$ @ChristophHanck Apologies, I mean that the Betas are biased (forgot the parameters from the equation, fixed now). $\endgroup$
    – Dole
    Apr 6, 2016 at 16:44

1 Answer 1

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Substitute the equation for $y_{t-1}$ in $\operatorname{Cov}(u_t,y_{t-1})$, so that $\operatorname{Cov}(u_t,y_{t-1}) = \operatorname{Cov}(u_t,x_{t-1}+y_{t-2}+u_{t-1})$.

If $\operatorname{Cov}(u_t,u_{t-1}) \neq 0$, you can see that the former can't be zero. But maybe I misunderstood your question.

There is a mistake when substituting into the covariance equation because $\operatorname{E}[(y_{t-1}-\operatorname{E}[y_{t-1}])(u_t-\operatorname{E}[u_t])] = \operatorname{E}[(y_{t-1}-\operatorname{E}[y_{t-1}])u_t] = \operatorname{E}[u_{t-1}u_t] \neq0$ and not $\operatorname{E}[u_{t-1}]u_t = 0$.

Regarding the second part of your question, $\operatorname{E}[\frac{u_t}{1-\beta_1}] = 0$ by linearity of the expected value operator.

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  • $\begingroup$ Thanks, indeed! (duh)... Although, I would still like to know what went wrong in the method I used. And also the 2nd question, regarding the expected value of ut $\endgroup$
    – Dole
    Apr 6, 2016 at 17:05
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    $\begingroup$ The problem was that you took $u_t$ out of the expected value operator. I edited my answer. $\endgroup$
    – Ale
    Apr 7, 2016 at 9:32
  • $\begingroup$ Actually the claim was that the expected value of the error divided by 1-b is not zero. In the paper there is actually an error with the parenthesis, and that explains it all (naturally I repeated the error). $\endgroup$
    – Dole
    Apr 7, 2016 at 9:42
  • $\begingroup$ Could you be more precise? On page 5 I see first that $\operatorname{E}[\frac{u_t}{1-\beta_1}] = 0$ (4); and then that $\operatorname{E}[\frac{u^2_t}{1-\beta_1}] = \frac{\sigma^2}{1-\beta_1} \neq 0$. Both are correct. $\endgroup$
    – Ale
    Apr 7, 2016 at 9:52
  • $\begingroup$ On page 5, 3rd equation after the COV sign: $E[\frac {u_t}{1-b}]u_t$. The error term is not in the expectation operator as it should. One last question, with your method it seems to be enough that $u_t$ is correlated with any error term $u_{t+k}$ (as you can keep substituting). With my method it's specifically $cov(u_t, u_{t-1})$, that has to be unequal to zero for bias. Which is correct, or where is the mistake in my analysis? $\endgroup$
    – Dole
    Apr 7, 2016 at 10:11

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