MEI MT Problem Set 51

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MEI MT Problem Set 51
Timothee Carayol

January 15, 2010

1

Available on http://personal.lse.ac.uk/carayolt/ec402.htm

Question 1
Question
yi = β1 x1i + β2 x2i + εi ; A1, A2, A3F , A4GM, A5normal hold, with i = 1, .., 11. Further assume: x1 x1 = 2 , x2 x2 = 2, x1 x2 = 1, 4 x1 y = 1, x2 y = 1, y y = 3 . Finally assume that for two hypothetical observations we have x1,12 x2,12 5 −2 = . x1,13 x2,13 3 −7

Answer
Some preliminary algebra: x1 x1 x1 x2 XX= = x2 x1 x2 x2 Therefore (X X )−1 = β1 β2
1 3

2 1 1 2 2 −1 −1 2
1 3

. . 1 1 =
1 3 1 3

= (X X )−1 X y =

2 −1 −1 2

.

Question 1 (cont)
Answer
y MX y y y − y X (X X )−1 X y εε ˆˆ = = 11 − 2 9 9 2 −1 1 4 1 1 1 3 − 3 −1 2 1 9
4−2 3

s2 =

=

2 = = 9 27 We’ll start with (b), and then do (a), because (a) builds on (b).

Question 1 (b)
Answer
(b) Want to find 80% prediction intervals for the expected value of the dependent variable y for observation 12 and 13. Note that we talk about prediction rather than confidence intervals here, because we want to find an interval around a predicted value than around a population parameter. But the method is the same: find a pivotal quantity whose distribution is known, and which allows us to build an interval which has a known probability to include our quantity of interest.

ˆ ˆ E (y12 ) = x12 β ⇒ E (y12 ) − x12 β = x12 β − x12 β ˆ ˆ ⇒ E (y12 ) − x12 β = x12 (β − β)

Question 1 (b) (cont)
Answer
ˆ ˆ β − β ∼ N 0, σ 2 (X X )−1 ⇒ x12 (β − β) ∼ 2 x (X X )−1 x N 0, σ 12 12 ˆ E (y12 ) − x12 β ∼ N 0, σ 2 x12 (X X )−1 x12 ∼N 0, σ 2 5 −2 1 3 2 −1 −1 2 5 −2

∼ N 0, σ 2 (26) ∼ N 0, 26σ 2

Question 1 (b) (cont)

Answer
But we do not know σε , so we use instead ˆ ˆ E (y12 )−x β E (y12 )−x β T = √ 212 = √1.9312 ∼ t(N − K ) = t(9).
26s

Our prediction√ interval for E (y12 ) is, then, √ ˆ ˆ [x12 β − t9,10% 1.93, x12 β + t9,10% 1.93] = √ √ [1 − 1.383 1.93, 1 + 1.383 1.93]. Likewise, we√ could find an 80% prediction interval for E (y13 ): √ [− 4 − 1.383 3.901, − 4 + 1.383 3.901]. 3 3

Question 1 (a)
Answer
(a) Want to find 80% prediction intervals for the dependent variable y for observation 12 and 13. ˆ ˆ y12 = x12 β + ε12 ⇒ y12 − x12 β = x12 β − x12 β + ε12 ˆ ˆ ⇒ y12 − x12 β = x12 (β − β) + ε12 Note that the only difference from (b) is this extra randomness from the error term, which will increase the variance, and hence lead to a wider prediction interval compared to (b).
2 ˆ y12 − x12 β ∼ N 0, σ 2 x12 (X X )−1 x12 + σε

∼ N 0, σ 2 (26 + 1) ∼ N 0, 27σ 2

Question 1 (a) (cont)

Answer
2 Otherwise, same thing as (b): substitute s 2 for σε , etc.

Our prediction√ interval for y12 is, then, √ ˆ ˆ [x12 β − t9,10% 2, x12 β + t9,10% 2] = √ √ [1 − 1.383 2, 1 + 1.383 2]. Likewise, we√ could find an 80% prediction interval for y13 : √ 4 [− 3 − 1.383 3.975, − 4 + 1.383 3.975]. 3

Question 1 (c)

Answer
Our answers differ because our prediction for y has to take into account the added randomness of the error term, which our prediction for E (y ) ignores. Hence our prediction interval for y is wider.

Question 2
Question
Consider the model y = X β + ε, X : T × K , with (say) A1, A2, A3Rmi.

Answer
First note that you can see both of the transformations suggested as a multiplication of the matrix of regressors X on the right by some K × K non-singular matrix, say A. To see this, note that for any matrix A, each column of XA is a linear combination of the columns of X . Conversely, any transformation of X whereby each regressor is transformed into a linear combination of the columns of X can be written as Z = XA for some square matrix A; if the transformation is such that no information is lost2 , then A is non-singular.