practical_1_derivations.pdf

practical_1_derivations.pdf - ECON2007 Practical 1 Notes...

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ECON2007 Practical 1 Notes Fall 2015 Term Original Author: Edmund Wright Amended by Giacomo Mason October 19, 2015 Contents 1 Assumption SLR.4: E [ u | x ] = 0 2 1.1 Law of Iterated Expectations (LIE) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 First implication of SLR.4: E [ u ] = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Second implication of SLR.4: E [ ux ] = 0 . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 R-squared 5 3 Proof of unbiasedness of OLS 6 3.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 3.2 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1
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1 Assumption SLR.4: E [ u | x ] = 0 Recall from the lectures the third assumption needed for unbiasedness, i.e. zero conditional mean (SLR.4 in Wooldridge): E [ u | x ] = 0. This assumption has two useful implications which we will use in several proofs: E [ u ] = 0 E [ ux ] = 0. To show these implications we need to use the Law of Iterated Expectations . 1.1 Law of Iterated Expectations (LIE) The Law of Iterated Expectations states the following: If we have two random variables a and b , then E ( a ) = E [ E ( a | b )] . For example, suppose we know that E ( a | b ) = 1+2 b , and we know that E ( b ) = 3. We can use these two facts with LIE to find E ( a ). LIE tells us that E ( a ) = E [ E ( a | b )]. If we plug in our formula for E ( a | b ) we then get E ( a ) = E [1 + 2 b ] = 1 + 2 E [ b ], and if we then plug in our value for E [ b ] we get E ( a ) = 1 + 2 · 3 = 7. 1.2 First implication of SLR.4: E [ u ] = 0 Proof Let’s prove this implication holds using LIE. If we substitute u for a in LIE, and x for b we get: E ( u ) = E [ E ( u | x )] . Now assume that SLR.4 holds, i.e. that E [ u | x ] = 0. Plugging this into the above equation gives us: E ( u ) = E [0] . The expected value of 0 is just 0. And so we end up with: E ( u ) = 0 And so we have shown that E [ u | x ] = 0 E [ u ] = 0. 2
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Strength It is stated in the lectures that E [ u ] = 0 is not a strong restriction. Why? Let’s see with an example. Suppose for example that the true relationship between y and x in the population can be described by the equation y = β 0 + β 1 x + u, where β 0 = 2 β 1 = 0 . 3 E [ u ] = 4 . It would appear at first glance that this population relationship must break SLR.4, as E [ u ] 6 = 0. But let’s manipulate the above equation a bit. First, add and subtract 4 from the RHS: y = β 0 + 4 + β 1 x + u - 4 . Obviously this does not change the relationship. Now define two new objects: β new 0 β 0 + 4 u new u - 4 . We can substitute these into our equation like so: y = β new 0 + β 1 x + u new , Notice that as u new u - 4 and E [ u ] = 4, we must have that E [ u new ] = 0. And so we have shown that a population relationship that initially appeared to break SLR.3 can be rewritten in a way that conforms to it.
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