# How to prove an OLS estimator is inconsistent

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

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

• 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. Nov 30 '19 at 23:26
• @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.
– abu
Nov 30 '19 at 23:33