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140_problems7_answers

140_problems7_answers - ECON 140 Fall 2008 11/6 Alex...

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ECON 140, Fall 2008 - 11/6 Alex Rothenberg Practice Problems: Serial Correlation and Panel Data Problem 1 Suppose we have the following autoregressive model for the error term (the unobservables) in a regression equation: ε t = ρ ε t - 1 + u t (1) u t iid N (0 , σ 2 u ) (2) This question asks you to investigate some of the properties of this process. 1. Fixing the value of ε 0 = 0, write out the first few terms of the sequence { ε t } in terms of the u t ’s. (Do this for t = 0 , 1 , 2 , 3 , 4) ANSWER: We write out the terms of the sequence as follows (using recursive substitution): ε 1 = ρ ε 0 + u 1 = u 1 ε 2 = ρ ε 1 + u 1 = ρu 1 + u 2 ε 3 = ρ ε 2 + u 3 = ρ 2 u 1 + ρu 2 + u 3 ε 4 = ρ ε 3 + u 4 = ρ 3 u 1 + ρ 2 u 2 + ρu 3 + u 4 2. Assuming E [ ε 0 ] = 0, what is the expected value of each term? ANSWER: Because each of the ε t ’s is just a linear combination of the u t ’s, which are mean zero random variables, each ε t will also be mean zero. For example: E [ ε 3 ] = ρ 2 E [ u 1 ] + ρ E [ u 2 ] + E [ u 3 ] = ( ρ 2 × 0) + ( ρ × 0) + 0 = 0 3. Assuming that Var( ε t ) = σ 2 for all t (homoskedasticity), express σ 2 as a function of σ 2 u . ANSWER: Returning to the recursion equation, we can apply the variance operator: Var( ε t ) = Var( ρ ε t - 1 + u t ) = ρ 2 Var( ε t - 1 ) + Var( u t ) 1
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ECON 140, Fall 2008 - 11/6 Alex Rothenberg Where the second line makes use of the independence between u t and ε t - 1 . Invoking the homoskedasticity assumption, we have: σ 2 = ρ 2 σ 2 + σ 2 u σ 2 = σ 2 u 1 - ρ 2 Note that this trick of solving for σ 2 and dividing by 1 - ρ 2 is not possible if ρ = 1 or ρ = - 1 . In either case, the process (1) is a random walk, and dealing with it is tricky business. We will discuss more about stationarity later in the course (the
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