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I have two equations

$Y_i = \beta_0 + \beta_1X_i + \epsilon_i$

$X_i = Y_i + Z_i$

and additional information that $cov(\epsilon_i, Z_i) = 0$

And I need to prove that using the OLS in the first equation results in an inconsistent estimator of $\beta_1$.

I tried to do the following

$\hat{\beta_1} = \frac{\Sigma^N(Y_i - \bar{Y})(X_i - \bar{X})}{\Sigma^N(X_i - \bar{X})^2} = \frac{cov(X, Y)}{var(X)} = \frac{cov(\beta_0 + \beta_1X_i + \epsilon_i, Y_i + Z_i)}{var(Y_i + Z_i)}$

I know that when I use the formula $cov(x, y) = E[xy]-E[x]E[y]$ some terms will cancel out due to $cov(\epsilon_i, Z_i) = 0$ or $E[\epsilon]=0$ from the assumptions of the OLS. However, I am not sure hot to finalize the proof. Any ideas?

EDIT: I already expanded the numerator to

$\beta_1 cov(X, Y) + \beta_1 cov(X, Z) + cov(\epsilon, Y) + cov(\epsilon, Z)$

And after using $cov(x, y) = E[xy]-E[x]E[y]$ on the last two terms I end up with

$\beta_1 cov(X, Y) + \beta_1 cov(X, Z) + E[Y\epsilon]$

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  • $\begingroup$ Start out by expanding the covariance term in the numerator and see what you get... for starters, the constant term can be done away with, since its covariance with $Y_i$ and $Z_i$ = 0. $\endgroup$
    – jbowman
    Nov 30 '19 at 23:26
  • $\begingroup$ @jbowman yes, that was the idea I had when I said I knew some of the terms will cancel out. I edited the question to show what stays in the numerator. I do not know, however, if this is the end-step as there is nothing more to cancel out or there is still something to do. $\endgroup$
    – abu
    Nov 30 '19 at 23:33

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