$Y_i=\beta_0+\beta_1X_i+U_i$ is my regression model for an I.I.D. sample with N=1000 observations. Suppose $U_i\sim I.I.D.(0,\sigma^2)$ and Xi are also I.I.D for i=1,2,3......1000. Xi is independent of Ui. How to show that the estimator $\beta_1={{Y_3-Y_2}\over {X_3-X_2}}$ is a consistent estimator of the OLS estimator? Simplication gives $\tilde {\beta}={{\beta_0+\beta_1X_3+U_3-\beta_0-\beta_1X_2-U_2}\over {X_3-X_2}}={{\beta_1(X_3-X_2)+U_3-U_2}\over {X_3-X_2}}=\beta_1+{{U_3-U_2}\over {X_3-X_2}}$ What is the next step?
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$\begingroup$ The next step is to infer that what you have is not a consistent estimator. The reason is $\frac{U_3-U_2}{X_3-X_2}$ does not vary with the sample size and will always have the same variance. $\endgroup$– AmitNov 19, 2019 at 2:56
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Remember that consistency describes how the estimator behaves in the limit as N asymptotically approaches infinity. Assuming no errors in your math up to this point, you need to consider how your error terms $U_i$ behave asymptotically as well.
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$\begingroup$ I don't understand what happens to the error term as n goes to infinity. $\endgroup$ Nov 18, 2019 at 16:16
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$\begingroup$ Well, what does $U_i\sim I.I.D.(0,\sigma^2)$ mean in words? $\endgroup$– hehNov 18, 2019 at 16:17
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$\begingroup$ Oh, it should converge to 0. Are we using any weak/strong laws of large numbers? $\endgroup$ Nov 18, 2019 at 16:20
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$\begingroup$ Sort of but not really. In this case it follows directly from the specification you've given - by writing the error term out that way, you are supposing that this is true. The reason you suppose this is so that you can use that assumption later on, such as to resolve your initial question. :) $\endgroup$– hehNov 18, 2019 at 16:40
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1$\begingroup$ Given what you're asking, I would just recommend that you ensure you understand the whole process for showing consistency - at some point, you should be taking limits in probability (so-called "plim" operations) if you're being precise. But you haven't shown your full working of the problem here, so I'm just assuming you did those steps properly. $\endgroup$– hehNov 18, 2019 at 16:42