80 X 1, X 2,, X n ( λ ) λ P(X = x) = f (x; λ) = λx e λ, x = 0, 1, 2, x! l(λ) = n f (x i ; λ) = i=1 i=1 n λ x i e λ i=1 x i! = λ n i=1 x i e nλ n i=1 x

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1 80 X 1, X 2,, X n ( λ ) λ P(X = x) = f (x; λ) = λx e λ, x = 0, 1, 2, x! l(λ) = n f (x i ; λ) = n λ x i e λ x i! = λ n x i e nλ n x i! n n log l(λ) = log(λ) x i nλ log( x i!) log l(λ) λ = 1 λ n x i n = 0

2 81 λ λ λ = 1 n n X i = X λ λ X λ E(X) = V(X) = λ E( λ) = E( 1 n n X i ) = 1 n n E(X i ) = 1 n n λ = λ

3 82 n n n V( λ) = V( 1 n X i ) = 1 n 2 V(X i ) = 1 n 2 λ = λ n 1 ( ) 2 log f (X; λ) ne λ 1 = ( ) 2 = (X log λ λ log X!) ne λ = λ 2 ne[(x λ) 2 ] = λ2 nv(x) = λ2 nλ = λ n 1 [( X ) 2] ne λ 1 V( λ) = 1 ( ) 2 log f (X; λ) ne λ V( λ) λ

4 83 n f (x i ; λ) = λ n x i e nλ n x i! = λnx e nλ (nx)! (nx)! n x i! = g(x; λ) h(x 1, x 2,, x n ) E(X) = λ, V(X) = λ n P( X λ > ɛ) < λ nɛ AR(1) y t = φy t 1 + ɛ t, ɛ t N(0, σ 2 )

5 84 1. Mean of y t given y t 1, y t 2, E(y t y t 1, y t 2, ) = φy t 1 2. Variance of y t given y t 1, y t 2, V(y t y t 1, y t 2, ) = σ 2 3. Thus, y t y t 1, y t 2, N(0, σ 2 ). = Conditional distribution of y t given y t 1, y t 2, 4. The stationarity condition is: the solution of φ(x) = 1 φx = 0, i.e., x = 1/φ, is greater than one in absolute value, or equivalently, φ < Rewriting the AR(1) model, y t = φy t 1 + ɛ t

6 85 = φ 2 y t 2 + ɛ t + φɛ t 1 = φ 3 y t 3 + ɛ t + φɛ t 1 + φ 2 ɛ t 2. = φ s y t s + ɛ t + φɛ t φ s 1 ɛ t s+1. As s is large, φ s approaches zero. = Stationarity condition 6. For stationarity, y t = φy t 1 + ɛ t is rewritten as: y t = ɛ t + φɛ t 1 + φ 2 ɛ t Mean of y t E(y t ) = E(ɛ t + φɛ t 1 + φ 2 ɛ t 2 + ) = E(ɛ t ) + φe(ɛ t 1 ) + φ 2 E(ɛ t 2 ) + = 0

7 86 8. Variance of y t 9. Thus, y t N ( 0, 10. Estimation of AR(1) model: (a) Log-likelihood function V(y t ) = V(ɛ t + φɛ t 1 + φ 2 ɛ t 2 + ) = V(ɛ t ) + V(φɛ t 1 ) + V(φ 2 ɛ t 2 ) + = σ 2 (1 + φ 2 + φ 4 + ) = σ2 1 φ 2 σ 2 1 ρ 2 ). = Unconditional distribution of yt log f (y T,, y 1 ) = log f (y 1 ) + T log f (y t y t 1,, y 1 ) t=1 = 1 2 log(2π) 1 ( ) 2 log σ φ 2 σ 2 /(1 φ 2 ) y2 1

8 87 T 1 log(2π) T 1 log(σ 2 ) σ 2 T (y t φy t 1 ) 2 t=2 = T 2 log(2π) T 2 log(σ2 ) 1 ( ) 2 log 1 1 φ 2 1 2σ 2 /(1 φ 2 ) y2 1 1 T (y 2σ 2 t φy t 1 ) 2 Note as follows: ) 1 f (y 1 ) = ( 2πσ2 /(1 φ 2 ) exp 1 2σ 2 /(1 φ 2 ) y2 1 ( 1 f (y t y t 1,, y 1 ) = exp 1 ) 2πσ 2 2σ (y 2 t φy t 1 ) 2 t=2 log f (y T,, y 1 ) = T 1 σ 2 2 σ σ 4 /(1 φ 2 ) y σ 4 T (y t φy t 1 ) 2 = 0 t=2

9 88 log f (y T,, y 1 ) φ = φ 1 φ 2 + φ σ 2 y σ 2 T (y t φy t 1 )y t 1 = 0 The MLE of φ and σ 2 satisfies the above two equation. t=2

10 y t = X t β + u t, u t = ρu t 1 + ɛ, ɛ t N(0, σ 2 ) Log of distribution function of u t log f (u T,, u 1 ) = log f (u 1 ) + T log f (u t u t 1,, y 1 ) t=1 = 1 2 log(2π) 1 ( ) 2 log σ ρ 2 σ 2 /(1 ρ 2 ) u2 1 T 1 2 log(2π) T 1 2 log(σ 2 ) 1 σ 2 T (u t ρu t 1 ) 2 t=2 = T 2 log(2π) T 2 log(σ2 ) 1 ( ) 2 log 1 1 ρ 2 1 2σ 2 /(1 ρ 2 ) u2 1 1 T (u 2σ 2 t ρu t 1 ) 2 t=2

11 90 Log of distribution function of y t log f (y T,, y 1 ) T = log f (y 1 ) + log f (y t y t 1,, y 1 ) t=1 = 1 2 log(2π) 1 ( ) 2 log σ ρ 2 σ 2 /(1 ρ 2 ) (y 1 X 1 β) 2 T T 1 2 log(2π) T 1 2 log(σ 2 ) 1 σ 2 t=2 = T 2 log(2π) T 2 log(σ2 ) 1 ( ) 2 log 1 1 ρ 2 ( (yt X t β) ρ(y t 1 X t 1 β) ) 2 1 2σ 2 T (y t Xt β) 2, t=2 where 1 ρ2 y y t, for t = 1, t = y t ρy t 1, for t = 2, 3,, T, 1 ρ2 X Xt t, for t = 1, = X t ρx t 1, for t = 2, 3,, T,

12 91 log f (y T,, y 1 ) is maximized with respect to β, ρ and σ 2. OLS, AR(1), AR(1)+X StataSE Data Data Editor Excel 123, var1, var2, var3,... command Y= + X+ Z reg Y X Z results Y, X, Z

13 92 gen t=_n tsset t t reg Y X Z dwstat scatter Y X X Y line Y X time time X Y Stata ( ) \2,940 (1995) t x y

14 93. gen t=_n. tsset t. reg y x Source SS df MS Number of obs = F( 1, 2) = 7.35 Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE = y Coef. Std. Err. t P> t [95% Conf. Interval] x _cons arima y, ar(1) nocons (setting optimization to BHHH) Iteration 0: log likelihood =

15 94 Iteration 1: log likelihood = Iteration 2: log likelihood = Iteration 3: log likelihood = Iteration 4: log likelihood = (switching optimization to BFGS) Iteration 5: log likelihood = Iteration 6: log likelihood = Iteration 7: log likelihood = Iteration 8: log likelihood = Iteration 9: log likelihood = ARIMA regression Sample: 1-4 Number of obs = 4 Wald chi2(1) = Log likelihood = Prob > chi2 = OPG y Coef. Std. Err. z P> z [95% Conf. Interval] ARMA ar L /sigma Note: The test of the variance against zero is one sided, and the two-sided confidence interval is truncated at zero.

16 95. arima y x,ar(1) (setting optimization to BHHH) Iteration 0: log likelihood = Iteration 1: log likelihood = (backed up) Iteration 2: log likelihood = (backed up) Iteration 3: log likelihood = (backed up) Iteration 4: log likelihood = (backed up) (switching optimization to BFGS) Iteration 5: log likelihood = (backed up) Iteration 6: log likelihood = Iteration 7: log likelihood = Iteration 8: log likelihood = Iteration 9: log likelihood = Iteration 10: log likelihood = Iteration 11: log likelihood = Iteration 12: log likelihood = ARIMA regression Sample: 1-4 Number of obs = 4 Wald chi2(2) = Log likelihood = Prob > chi2 = OPG y Coef. Std. Err. z P> z [95% Conf. Interval]

17 y x _cons ARMA ar L /sigma Note: The test of the variance against zero is one sided, and the two-sided confidence interval is truncated at zero. 96

18 97 7 Qualitative Dependent Variable ( ) 1. Discrete Choice Model ( ) 2. Limited Dependent Variable Model ( ) 3. Count Data Model ( ) Usually, the regression model is given by: y i = X i β + u i, u i N(0, σ 2 ), i = 1, 2,, n, where y i is a continuous type of random variable within the interval from to. When y i is discrete or truncated, what happens?

19 Discrete Choice Model ( ) Binary Choice Model ( ) Example 1: Consider the regression model: y i = X i β + u i, u i (0, σ 2 ), i = 1, 2,, n, where y i is unobserved, but y i is observed as 0 or 1, i.e., y i = Consider the probability that y i takes 1, i.e., 1, if y i > 0, 0, if y i 0. P(y i = 1) = P(y i > 0) = P(u i > X i β) = P(u i > X i β ) = 1 P(u i X i β ) = 1 F( X i β ) = F(X i β ), (if the dist. of u i is symmetric.),

AR(1) y t = φy t 1 + ɛ t, ɛ t N(0, σ 2 ) 1. Mean of y t given y t 1, y t 2, E(y t y t 1, y t 2, ) = φy t 1 2. Variance of y t given y t 1, y t

87 6.1 AR(1) y t = φy t 1 + ɛ t, ɛ t N(0, σ 2 ) 1. Mean of y t given y t 1, y t 2, E(y t y t 1, y t 2, ) = φy t 1 2. Variance of y t given y t 1, y t 2, V(y t y t 1, y t 2, ) = σ 2 3. Thus, y t y t 1,