Econometrics I. September, Part I. Department of Economics Stanford University

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1 Econometrics I Deartment of Economics Stanfor University Setember, 2008 Part I

2 Samling an Data Poulation an Samle. ineenent an ientical samling. (i.i..) Samling with relacement. aroximates samling without relacement. Large samle theory eveloe uner i.i. assumtion. Samling theory is a toic by itself.

3 Asymtotic Statistics Large Samle Theory. Concets of Convergence. Why o we nee large samle theory? How large is large samle? Consistency an Asymtotic Distribution.

4 Moes of convergence Convergence of a sequence of numbers: α n α as n, or lim n α n = α for all ɛ > 0, Sequence of ranom variables. lim α n α < ɛ. n Convergence in robability: X n X 0 if for all ɛ > 0: lim P ( X n X > ɛ) = 0 or lim P ( X n X < ɛ) = 1. n n Alternative efinition: ɛ > 0, δ > 0, N > 0, such that P ( X n X > ɛ) < δ, n N. Convergence almost surely: P (lim n X n = X ) = 1. a.s. X. X n

5 Convergence in mean square: X n X m.s. 0 if lim E (X n X ) 2 = 0. n Can be generalize to L convergence: X n X X n X aroun. L 0 imlies X n X lim E X n X = 0. n L 0 if 0, but not the other way This follows from Chebyshev inequality an Markov inequality: Proof: P ( X n > ɛ) ɛ E X n. E X n E X n 1 ( X n > ɛ) ɛ E1 ( X n > ɛ) = ɛ P ( X n > ɛ).

6 Convergence in istribution lim P (X n x) = P (X x) n at continuity oints x of the istribution of X. Both X n an X can be scalar or vector. Why only continuity oints? X n X = X n X. However, X n X X n X unless X α. Continuity maing theorem: Let X n X. Let g ( ) be a function such that its set of iscontinuity oints E is close an P (X E) = 0, then g (X n ) g (X ). Some elements of X can be constants.

7 Theorem in Amemiya is a secial case of the continuous maing theorem: Let X n α. If g ( ) is continuous at α, then g (X n ) g (α). Theorem (Slutsky) is also a secial case of continuous maing: If X n X an Y n α, then X n + Y n X n Y n X n /Y n X + α. αx X /α if α 0. Where will these theorems be use?

8 Law of Large Numbers Weak Law of Large Numbers(WLLN): X t, t = 1,.... Let X n = 1 n n t=1 X t, unerwhat conitions oes X n E X n 0. Sufficient to show E X n E X n 0, say = 2 but other > 0 also works. An easy WLLN: X t uncorrelate with mean 0 an Var (X t ) = σ 2. E X 2 = V ( X ) = 1 n Var (X t) 0. Theorem (Khinchine) Let {X i } be ineenent an ientically istribute (i.i..) with EX i = µ. Then X n µ.

9 Central Limit Theorem Uner what conitions oes Z n = ( V X n ) 1/2 ( Xn E X n ) N (0, 1). Lineberg-Levy CLT: Let {X i } be i.i.. with EX i = µ an VX i = σ 2. Then Z n N (0, 1). Note that V X n = 1 n σ2. So n ( Xn µ ) σ N (0, 1) or n ( Xn µ ) N ( 0, σ 2). This is an exact statement, the next one is aroximate X n A N ( µ, σ 2 /n ).

10 LLN is a secial case of convergence in robability. CLT is a secial case of convergence in istribution. X n X limn E [f (X n )] = E [f (X )] for every boune continuous f. However EX n may not converge to EX. Class of convergence etermining functions. Characteristic function is such a class: convergence in characteristic function imlies convergence in istribution.

11 Proerties of characteristic functions Definition: φ X (λ) = Ee iλx = E [cos (λx ) + i sin (λx )]. Examles: X N (0, 1), φ X (λ) = e λ2 /2. Cauchy f (x) = 1 π(1+x 2 ), then φ X (λ) = e λ. Proerties: φ ax +b (λ) = Ee iλ(ax +b) = e ibλ φ X (aλ). Let X t be i.i., an let S n = n t=1 X t, then φ Sn (λ) = [φ Xt (λ)] n.

12 Examles If X i i.i.. normal N (0, 1), let S n = 1 n n t=1 X t, then φ Sn (λ) = [ φ Xt ( λ/ n )] n = [ex Still N (0, 1) If X t ii Cauchy, let S n = 1 n n t=1 X t, then ( ( λ/ n ) )] ) 2 n = ex ( λ2. 2 φ Sn (λ) = [φ Xt (λ/n)] n = [ex ( λ/n )] n = ex ( λ ). Still Cauchy. So No LLN for Cauchy ranom variables, because it has no mean.

13 Intuitive roof of Lineberg-Levy CLT: For any ranom variable, Y, with zero mean an unit variance (var(y) = 1), the characteristic function of Y is, by Taylor s theorem. φ Y (t) = 1 t2 2 + o ( t 2), t 0. Letting Y i be (X i µ)/σ, the stanarise value of X i, the stanarise mean of the observations X 1, X 2,..., X n is just Z n = X n µ n σ/ n = Y i n By simle roerties of characteristic functions, the characteristic function of Z n is [ ( )] n ( )] t φ Y = [1 t2 t 2 n n 2n + o e t2 /2, n. n But, this limit is just the characteristic function of a stanar normal istribution. i=1

14 More Central Limit Theorem Liaounov CLT: allow for ineenent but nonientically istribute {X i }. {X t }, t = 1,..., ineenent an mean 0. Requires conitions on the thir moment. Can relax the assumtion to only 2 + δ moment for δ > 0. Both are secial cases of the Lineberg-Feller CLT. See, e.g. Billingsley or Avance Amemiya.

15 Lineberg Conition: σ 2 t = Var (X t ), an let C 2 n = n t=1 σ2 t : lim n 1 C 2 n n E [ Xt 2 1 ( X t > ɛc n ) ] = 0. ɛ > 0. t=1 Requires the tail of X t to be small. Liaounov Conition: imlies Lineberg conition ( [ ] n ) 1/3 lim C 2 1/2 n E X t 3 = 0. n t=1 Liaounov Conition can be weakene: for δ > 0, ( [ ] n ) 1/(2+δ) lim C 2 1/2 n E X t 2+δ = 0. n t=1

16 Stochastic Orer X n = o (1) if X n X n = O (1) if In articular, if X n Facts 0. lim M lim su P ( X n > M) = 0. n X, X n = O (1). X n = O (a n ) means a 1 n X n = O (1). O (1) o (1) = o (1). O (a n ) O (b n ) = O (a n b n ). O (a n ) + O (b n ) = O (a n + b n ) = O (max (a n, b n )).

17 The Delta metho Let the function g ( ) be continuously ifferentiable aroun µ, an suose n ( Xn µ ) N ( 0, σ 2), then n ( g ( Xn ) g (µ) ) g (µ) N ( 0, σ 2) ( = N 0, g (µ) 2 σ 2). Can be roven using the mean value theorem: n ( g ( Xn ) g (µ) ) = g (X n ) n ( Xn µ ) for X n between X n an µ. Obviously X n µ since X n µ. The results follow from continuous maing an slutsky.

18 Examle: g (x) = x 2. n ( Xn µ ) N ( 0, σ 2). Using the elta metho n ( X 2 n µ 2) This only works when µ 0. If µ = 0, then n ( ) X 2 n 0. 2µN ( 0, σ 2) = N ( 0, 4µ 2 σ 2) This is not a useful asymtotic aroximation. Using continuous maing theorem n X 2 n σ 2 χ 2 1.

19 Multivariate CLT Wal-Device: a sequence of ranom vectors X n converges in istribution to X if an only if λ X n converges to λ X for all vector λ. Multivariate Lineber-Levy: {X i } be i.i.. ranom vectors with EX i = µ an V X i = Σ. Then n ( Xn µ ) N (0, Σ) Alternative statement: for Σ 1/2 Σ 1/2 = Σ Multivariate Slutsky. nσ 1/2 ( X µ ) Multivariate Delta Metho. N (0, I)