1 Appendix A: Matrix Algebra

Transcription

1 Appendix A: Matrix Algebra. Definitions Matrix A =[ ]=[A] Symmetric matrix: = for all and Diagonal matrix: 6=0if = but =0if 6= Scalar matrix: the diagonal matrix of = Identity matrix: the scalar matrix of = Triangular matrix: =0if Idempotent matrix: A = AA = A Symmetric idempotent matrix: A 0 A = A = AA Orthogonal matrix: A = A 0 Unitary matrix: A 0 A = AA 0 = I Trace of A : (A) = P = sum of diagonal terms. (ABC) = (BCA) = (CBA) if A B C are symmetric. (A) = [ (A)] (A + B) = (A)+ (B) Matrix Addition: A + B =[ + ] (A + B)+C = A+(B + C) (A + B) 0 = A 0 +B 0

2 Matrix Multiplication AB 6= BA : = = (AB) C = A (BC) A (B + C) =AB + AC (AB) 0 = B 0 A 0 Idempotent (projection) Matrix y = x + z + u where y x z and u are vectors and are scalars. Let M = ³ 0 z (z 0 z) z then, is an idempotent matrix. M M = ³ z (z 0 z) z 0 ³ 0 z (z 0 z) z = z (z 0 z) z 0 z (z 0 z) z 0 + z (z 0 z) z 0 z (z 0 z) z 0 = z (z 0 z) z 0 z (z 0 z) z 0 + z (z 0 z) z 0 = z (z 0 z) z 0 = Further note that M z = ³ z (z 0 z) z 0 z =0 Hencewehave M y = M x + M z + M z u = M x + M z u

3 Vector Lenth of a vector: Norm is defined as e = µ P e 0 e = = Orthogonal vectors: Two nonzero vectors a and b are orthogonal, written a b iff a 0 b = b 0 a =0 Regression in a Matrix form y = Xb + u TheOLSestimateis TheOLSresidualsare ˆb = (X 0 X) X 0 y = (X 0 X) X 0 Xb +(X 0 X) X 0 u = b +(X 0 X) X 0 u û = y Xˆb = Xb Xˆb + u = u X ³ˆb b Hencewehave X 0 û = X ³u 0 X ³ˆb b = X 0 u X 0 X ³ˆb b = X 0 u X 0 X (X 0 X) X 0 u = 0 Matrix Inverse (AB) = B A A 0 = A 0 0 A 0 A 3

4 A A A A =? (see p.966 A-74) Kronecker Products Let A = then A B = B B B B A is and B is Then A B is () () A (B + C) =A B + A C (A + B) C = A C + B C (A) B = A (B) = (A B) where is a scalar (A B) C = A (B C) (A B) = A B (A B) 0 = A 0 B 0 tr(a B) =tr(a)tr(b) (A B)(C D) =AC BD. Eigen values and Eigen vectors Eigen Values Characteristic vectors = Eigen vectors, c Characteristic roots = Eigen values. Ac = c Ac Ic = 0 A I c = 0 4

5 Example: Find eigen values of A : A = 3 0 Solutions: 3 A I = 0 = =0 Eigen Vector: The characteristic vectors of a symmetric matrix are orthogonal. That is, C 0 C = I where C =[c c c ] Alternatively C is a unitary matrix. Let Λ = ( )= Then we have Ac = c or AC = CΛ C 0 AC = C 0 CΛ = Λ = ( ) Alternatively we have A = CΛC 0 which is the spectral decomposition of A 5

6 Some Facts: Prove the followings. (A) = (Λ) (A) = (CΛC 0 )= (ΛCC 0 )= (ΛI) = (Λ) The trace of a matrix equals the sum of its eigen values.. A = Λ 3. AA = A = CΛ C 0 4. A = CΛ C 0 5. Suppose that A is a nonsigular symmetric matrix. Then A = CΛ C 0 6. Consider a matrix P such that then P 0 P = A P = Λ C 0 Matrix Decomposition LU decomposition (Cholesky Decomposition) A = LU where L is lower triangular and U is upper triangular matrix. L = U 0 Example: Hence the solution is given by = 0 = + 0 = = =? 6

7 Spectral (Eigen) Decomposition A = CΛC 0 Schur Decomposition A = USU 0 where U is an orthogonal matrix and S is a upper triangular matrix. Quadratic forms Let be a symmetric matrix. Then all eigen values of are positive (negative), then is a positive (negative) definite matrix. If has both negative and positive eigen values, then is indefinite..3 Matrix Algebra (Ax) x (x 0 Ax) x (x 0 Ax) A = A = Ax = xx 0 (Ax) x 0 = A 0 7

8 .4 Sample Questions: Part I: Calculation A = B = 0 0 Q: Find eigen values of A Q: Find the lower triangular matrix of A Q3: A B Q4: (A B) Q5: tr(a) Part II: Matrix Algebra Consider the following regression = () Q6: If you wrote (9) as y = Xβ + u () Define, and Q7: Consider the following problem Show the first derivertives of function. arg min = u 0 u Q8: Show the solution satisfies β =(X 0 X) X 0 y 8

9 Probability and Distribution Theory. Probability Distributions () = Prob ( = ). 0 Prob( = ). P () =(discrete case), R () =(continuous case) Cumulative Distribution Function P () = Prob ( ) : discrete () = R () = Prob ( ) : continuous. 0 (). If then () () 3. (+ ) = 4. ( ) =0 Expectations of a Random Variable Mean or expected value of a random variable is P () discrete [] = R () continuous Median: used when the distribution is not symmetric Mode: the value of at which () take its maximum 9

10 Functional expectation Let () be a function of Then P () () discrete [ ()] = R () () continuous Variance Note that () = ( ) P = ( ) () discrete R ( ) () continuous () = ( ) = + = Now we consider third and fourth central moments Skewness : ( ) 3 Kurtosis : ( ) 4 Skewness is a measure of the asymmetry of a distribution. For symmetric distribution, we have ( ) = ( + ) and ( ) 3 =0. Some Specific Probabilty Distributions.. Normal Distribution or we note = Ã! exp ( ) 0

11 Properties:. Addition and Multiplication + +. Standard normal function ³ exp (0 ) ( 0 ) = () = exp µ y x Chi-squared, and distributions (; ) = ³ Γ () exp [ 0]. If (0 ) then chi-squared with one degress of freedom.

12 . If are independent variables, then X = 3. If are independent (0 ) variables, then X = 4. If are independent (0 ) variables, then X = 5. If and are independent and variables, then If and are independent and variables, then ( ) 7. If is a (0 ) variable and is and is independent of then the ratio = p and it has the density function given by () = Γ + Γ where = and Γ () is the gamma function 8. (0 ) as Γ () = Z 0 + µ+ 9. If then ()

13 0. Noncentral distribution: If ( ) then () has a noncentral distribution.. If and have a joint normal distribution, then w =( ) 0 has w (0 Σ) where Σ =. If w (μ Σ) where w has elements, then w 0 Σ w has a noncentral. The noncentral parameter is μ 0 Σ μ.. Other Distributions Lognormal distribution () = Ã exp! ln Note that Hence () = exp µ + ³ () = + = ln () µ ln + () () µ = ln + () () Mode () = Median () = 3

14 exp [ln ] y x Properties. If ( ) then exp () ( ). If ( ) then ln () ( ) 3. If ( ) then = + is a shifted LN of. () = () + () = ( + ) = () 4. If ( ) then = is also LN. (ln + ) 5. If ( ) then = is also LN. ( ) 6. If ( ) ( ) and they are independent, then + + 4

15 Gamma Distribution where () = Γ () exp ( ) for 000 Γ () = Note that if is a positive integer, then Z 0 Γ () =( )! When = gamma distribution becomes exponential distribution exp ( ) for 0 () = 0 for 0 When = = gamma dist. = dist. When is a positive integer, gamma dist. is called Erlang family. Beta distribution For a variable constrained between 0 and 0 the beta distribution has its density as Γ ( + ) ³ ³ () = Γ () Γ () Usually s range becomes (0 ) that is =. symmetric if =. = = becomes (0 ) 3., becomes 4. = strictly convex 5. = = straight line 6. = strictly concave 7. Mean: + 8. Variance: + for = (++)(+) (++)(+) for = 5

16 Figure : Logistic Distribution () = When =and =0 we have (+ ) () = ( ) ( + ( ) ) 0 () = + ( ) ( + ) () = + y x + 6

17 y x + 0 y x Exponential distribution () = Weibull Distribution () = () for 0 0 for 0 When = Weibell becomes the exponential distribution. 7

18 The standard Cauchy distri- Cauchy Distribution () = ( 0 ) + where 0 is the location parameter, is the scale parameter. bution is the case where 0 =0and = () = ( + ) (+ ) y x Note that the distribution with =becomes a standard Cauchy distribution. Also note that the mean and variance of the Cauchy distribution don t exist..3 Representations of A Probability Distribution Survival Function () = () =Prob [ ] where is a continuous random variable. Hazard Function (Failure rate) () = () () = () () Let () = (exponential density function), then we have () = () () = 8

19 which implies that the harzard rate is a constant with respect to time. However for Weibull distribution or log normal distribution, the harzard function is not a constant any more. Moment Generating Function (mdf) The mgf of a random variable is () = for Note that mgf is an alternate definition of probability distribution. Hence there is one for one relationship between the pdf and mgf. However mgf does not exist sometimes. For example, the mgf for the Cauchy distribution is not able to be defined. Characteristic Function (cf) Alternatively, the following characteristic function is used frequently in Finance to define probability function. Even when the mdf does not exist, cf always exist. () = = For example, the cf for the Cauchy distribution is exp ( 0 ) Cumulants The cumulants of a random variable are defined by the cumulant generating function which is the logarithm of the mgf. () =log Then, the cumulants are gievn by = = 0 (0) = = 00 (0) = () (0) 9

20 .4 Joint Distributions The joint distribution for and denoted ( ) is defined as P P ( ) Prob ( ) = R R ( ) where Consider the following bivariate normal distribution as an example. µ ( ) = p exp + = = = Suppose that = = = =0 =05 Then we have ( ) = µ 05 exp + 05 exp 05 ( + ) ( 05 ) 0.5 z y x We denote 0

21 Marginal distribution It is defined as Prob ( = 0 )= X P ( ) Prob ( = 0 = 0 ) Prob ( = 0 )= () = R ( ) Note that ( ) = () () iff and are independent Also note that if and are independent, then ( ) = () () alternatively Prob ( )=Prob ( ) Prob ( ) For a bivariate normal distribution case, the marginal distribution is given by () = () = Expectations in a joint distribution Mean: P () = () = P P ( ) R () = R R ( ) Variance: See B-50. Covariance and Correlation [ ] = ( ) = ( ) = ( + ) = ()+ ()+ ( )

22 .5 Conditioning in a bivariate distribution ( ) = ( ) () For a bivariate normal distribution case, the conditional distribution is given by ( ) = + where = = If =0 then and are independent. Regression: The Conditional Mean The conditional mean is the mean of the conditional distribution which is defined as P ( ) ( ) = R ( ) The conditional mean function ( ) is called the regression of on = ( )+( ( )) = ( )+ Example: = + + Then ( ) = + Conditional Variance ( ) = ( ( )) = ( )

23 .6 The Multivariate Normal Distribution Let x =( ) 0 and have a multivariate normal distribution. Then we have µ (x) =() Σ exp (x μ)0 Σ (x μ). If x (μ Σ) then Ax + b (Aμ + b AΣA 0 ). If x (μ Σ) then (x μ) 0 Σ (x μ) 3. If x (μ Σ) then Σ (x μ) (0 I ) 3

24 .7 Sample Questions Q: Write down the definitionsofskewnessandkurtosis.whatisthevalueofskewnessfor the symmetric distribution. Q: Let (0 ) for = Further assume that is independent each other.then.. P = 3. Q3: Write down the standard normal density Q4: Let ( ). ln (). Prove that = (ln + ) 3. Prove that = ( ) Q5: Write down the density function of the Gamma distribution. Write down the values of and when Gamma =. Write down the values of and when Gamma = exponential distribution Q6: Write down the density function of the logistic distribution. Q7: Write down the density function of Cauchy distribution. Write down the value of when Cauchy= distribution Q8: Write down the definition of Moment Generating and Characteristic function. Q9: Suppose that x =( 3 ) 0 and x ( Σ). Write down the normal density in this case.. y = Ax + b 3. z = Σ (x c) where c 6= μ 4

25 3 Estimation and Inference 3. Definitions. Random variable and constant: A random variable is believed to change over time across individual. Constant is believed not to change either dimension. It becomes an issue in the panel data. = + Here we decompose ( = ; = ) into its mean (time invariant) and time varying components. Now is random or constant. According to the definition of random variables, can be a constant since it does not change over time. However, if has a pdf, then it becomes a random variable.. IID: independent, identically distributed: Consider the following sequence x =( )=(3) Now we are asking if each number is a random variable or constant. If they are random, then we have to ask the pdf of each number. Suppose that 0 Now we have to know that is an independent event. If they are independent, then next we have to know is identical or not. Typical assumption is IID. 3. Mean: 4. Standard error (deviation) 5. Covariance = " = = X = # X ( ) = X ( )( ) = 5

26 6. Correlation = Population values and estimates = +. Estimate: it is a statistic computed from a sample ³ˆ. Samplemeanisthestatistic for the population mean ().. Standard deviation and error: is the standard deviation and is the standard error of the population. 3. regression error and residual: is the error, ˆ is the residual 4. Estimator: a rule or method for using the data to estimate the parameter. OLS estimator is consistent should read the estimation method using OLS is consistently estimated a parameter. 5. Asymptotic = Approximated. Asymptotic theory = approximation property. We are interested in how an approximation works as Estimation in the Finite Sample. Unbiased: An estimator of a parameter is unbiased if the mean of its sampling distribution is ³ˆ =0for all. Efficient: An unbiased estimator ˆ is more efficient than another unbiased estimator ˆ is the sampling variance of ˆ is less than that of ˆ ³ˆ ³ˆ 6

27 3. Mean Squared Error: ³ˆ = ³ˆ = ³ˆ ˆ + ˆ = ³ˆ + h ³ˆ i 4. Likelihood Function: rewrite = and consider the joint density of If are independent, then ( ) = ( ) ( ) ( ) Y = ( ) = ( ) The function ( u) is called the likelihood function for given the data u. = 5. Cramer-Rao Lower Bound: Under regularity condition, the variance of an unbiased estimator of a parameter will always be at least as large as µ [ ()] Ã! ln () ln () = = where the quantity () is the information number for the sample. 4 Large Sample Distribution Theory Definition and Theorem (Consistency and Convergence in Probability). Convergence in probability: The random variable converges in probability to a constant if We denote lim Prob ( )=0for any positive or plim = 7

28 Carefully look at the subscript. This means is dependent on the size of For an example, the sample mean, P = is a function of. Almost sure convergence: ³ Prob lim = = Note that almost sure convergence is stronger than convergence in probability. We denote 3. Convergence in the th mean lim ( )=0 and denote it as When = we say convergence in quardratic mean. 4. Consistent Estimator: An estimator ˆ of a parameter is a consistent estimator of iff plim ˆ = 5. Khinchine s weak law of large number: If is a random sample from a distribution with finite mean ( )= then plim = plim X = = 6. Chebychev s weak law of large number: If is a sample of observations such that ( )= ( )=, = P = 0 as then where = P = plim ( )=0 8

29 7. Kolmogorov s Strong LLN: If is a sequence of independently distributed random variables such that ( )= and ( )= such that P = as then 0 8. (Corollary of 7) If is a sequence of iid variables such that ( )= and then 0 9. Markov s Strong LLN: If is a sequence of independent random variables with ( )= and if for some 0, P = h +i + then 0 Properties of Probability Limits. If and are random variables with plim = and plim = then plim ( + ) = + plim = plim = if 6= 0. W is a matrix whose elements are random variables and if plimw = Ω then plimw = Ω 3. If X and Y are random matrices with plimx = B and plimy = C then plimx Y = BC 9

30 Convergence in Distribution. converges in distribution to a random variable with cdf () if lim ( ) () =0at all continuity points of () In this case, () is the limiting distribution of andthisiswritten. Cramer-Wold Device: If x x then c 0 x c 0 x where c 3. Lindeberg-Levy CLT (Central limit theorem): If arearandomsamplefroma probability distribution with finite mean and finite variance, then it sample mean, = P = have the following limiting distribution ( ) 0 4. Lindegerg-Feller CLT: Suppose that are a random sample from a probability distribution with finite mean and finite variance.let = X = = X = where lim max ( ) ( )=0 Further assume that lim = then it sample mean, = P = have the following limiting distribution ( ) 0 or ( ) (0 ) 30

31 5. Liapounov CLT: Suppose that { } is a sequnece of independent random variables with finite mean and finite positive variance such that ³ + for some 0 If is positive and finite for all sufficiently large, then ( ) (0 ) 6. Multivariate Lindeberg-Feller CLT: ( x μ ) (0 Q) where (x )=Q and we assume that lim Q = Q 7. Asymptotic Covariance Matrix: Suppose that ³ˆθ θ (0 V) thenitsasymptoticcovariance matrixisdefined as Asy. Var ³ˆθ = V Order of A Sequence. A sequence is of order denoted iff (a) == () (b) = = ( ):plim = plim = (c) = ( +0)= ( ): plim( +0) =. A sequence is of order less than iff (a) =0= () : plim0 =0 (b) = then = () plim =0 3

32 Order in Probability. A sequence random variable is ( ()) ifthereexistssome such that 0 and all Prob µ () where is a finite constant (a) If (0 ) then = () Since given there is always some such that Prob ( ) (b) ( ) = + (c) If ( ) (0 ) then ( )= but = (). The notation = ( ) means 0 (a) If (0 ) then = and = () (b) ( ) = + 3

33 Sample Qestions Part I: Consider the following model = + (0 ) and Q: Show that = () =0 for all and Q: Consider = P = and = P = Suppose that 6=0but 6= 0 then is = = ()? Q3: Suppose that =0but 6= 0 then is = and =? Q4: Suppose that =0but =0 then is = and =? P Q5: Consider = P = = Suppose that = 0 but 6= 0 then is =? Part II: Consider the following model y = Xb + u Q6: Obtain the limiting distribution of θ = x 0 u Q7: Obtain the limiting distribution of ˆb Q8: Suppose that (0 ) Find the asymptotic variance of ˆb 33

34 5 Chapters through 4: The Classical Assumptions. Linear y = Xb + u. X is a nonstochastic and finite matrix 3. X 0 X is nonsingular for all 4. (u) =0 5. u (0 I) When X is stochastic 4. Exogeneity of the independent variables: (u X) =0 5-. Homoscedasticity and no-autocorrelation. 5-. Normal distribution. Properties: A. Existence: Given,,3, ˆβ exists for all and it unique B. Unbiasedness: Given -4, ³ˆβ = β C. Normality: Given -5, ˆβ ³β (X 0 X) D. Gauss-Markov Theorem: OLS estimator is the minimum variance linear unbiased estimator (whether X is stochastic or nonstochastic). Linear Model Consider two models. = + + =

35 Consider log, semilog, level: = + + ln = + ln + ln =

36 5. Brainstorm I: Sample Mean Notation: = earning at time Classification: Male, Female, Skilled Worker, Non-skilled worker. Question : How to test the difference between male and female earning. = sample mean of male earning = sample mean of female earning Question : How to explain the earning difference between male and female. By using skill data. 3 = sample mean of skilled worker 4 = sample mean of nonskilled worker If so how? Question 3: Deriving the limiting distributions for Q and Q as Question 4: Form a null hypothesis to test if male and female earning difference. 36

37 5. Brainstorm II: Trend Regression Notation: The true model is given by = + 0 Now consider two regressions = + = + Question : Deriving the limiting distribution of ˆ Question : Write down as a function of, and Question 3: Deriving the limiting distribution of ˆ 37

38 5.3 Brainstorm I Continue: Dummy Regression Notation: = earning. = Decision for Ph.D. program. = Decision for taking Econometric class Consider the following decision tree. If =0 then the values for does not matter. Question : Construct a dummy regression for the first example of Brainstorm I Question : Construct a dummy regression for the current example Question 3: Derive the limiting distribution for Q. 38

39 5.4 Sample Questions Basic 0: Sample Mean Suppose that is i.i.d. ( ). Let ˆ P = = ( ) Show that ˆ is biased but consistent. Obtain the unbiased estimator.. Let = for = Obtain the unbiased variance for 3. Let = + for =3 Obtain the unbiased variance for 4. Let ˆ ˆ ˆ 3 be the unbiased variance for Q,,3. Find the smallest variance. Basic I: Single Regressor Consider the following model = + We assume that all classical assumptions hold.. Show the OLS estimator ˆ is unbiased.. Show the OLS estimator is minimizing the following quadratic loss X = 3. Show that Basic II: Multiple Regressors X ˆ =0 = Consider the following model = X β + = + + or equivalently y = Xβ + u = X +X +u We assume the classical assumptions hold. 39

40 . Show the OLS estimator ˆ is unbiased.. Show the OLS estimator is minimizing the following quadratic loss u 0 u 3. Show that 4. Show that X 0 û =0 ˆ =(X 0 X) X 0 y 5. Define M and M and show the followings ˆ = (X 0 M X ) X 0 M y ˆ = (X 0 M X ) X 0 M y 6. Suppose that =0 Consider the following two regressions y = X + e (3) y = X + X + u (4) (a) Show that in (3) is smaller than that in (4). (b) Write down the relationship between and in general. (c) Suppose that 6=0and ratio for ˆ is greater than. Show that in (4) is greater than in (3). 7. Consider (3) as the true model. (0 ) Let = û0 û (a) Show that ˆ and ˆ be independent. (b) Show that ˆ q (X 0X ) 40

41 8. Consider (4) as the true model. Let = + is independent on Also is independent on. (a) Find plimˆ (b) Let (0 ) for 0 Find plimˆ and plimˆ 9. Consider (4) as the true model. 6=0 and is independent on Suppose that you are interested in testing Derive the limiting distribution ˆ = ˆ ˆ 0 : = =0 4

42 5.5 Answer and Additional Notes True Model (when is nonstochastic) = + = + = + Regression = + + = + ( + )+ = + + (5). How to obtain ˆ and ˆ (OLS estimators) Takethesampleaverage () - () yields X = + X + X (6) = + or equivalently = + since = X = ˆ = + P P ˆ = P P = P P Sinceweassume is nonstochastic, we have P P ³ˆ = P = P =0 P ³ˆ = P 4 = ( P ) P (7)

43 Note that ³X =( + + ) = ( + + ) Hence ³X = ( + + ) = + + +( + + ) We will assume ( ) = 0 for 6= : independent = :identical Then we have = µ X µ = X + ³X = ³X +( + + ) = ³X + ( + + ) = + µ = = (8) since =0if 6= and + =0for all Also note that + = µ = X µ + X µ + X + X + + ³X = 0 + = = 43

44 Now, we have ³X = + + +( + + ) = X ( + + ) = " X X # = ( + + ) = = à X X! = = = = à X = X = =! X = = = à X! X = since X = = X = X Ã! X X = X X = = = = X = = = + = Now from (3), we have ³ˆ P = P = P = P = P = ( P ) P = P = P Note that lim P =0if X = In order to have a finite variance, we need to have ³ˆ = 44 P P

45 Nowitiseasytoshowthat(byusingLLCLT) ³ˆ or ˆ p P µ 0 (0 ) P To Students: Q. Now you do more simple model = + (9) and get the limiting distribution of ˆ Here we assume = =0 Q. (Example of nonstochastic ) Consider = + (0) This is a regression of on That is, we let = and =for all in (9) then we have (0). Find the limiting distribution of ˆ When is stochastic We don t work with expectation term here. Instead of this, we take probability limit (since we are obtaining the limiting distribution, so we are caring about consistency only. The previous case, both two unbiaseness and consistency becomes identical problem since was nonstochastic.) Note that Similarly, we have plim ³ˆ = plim P P = plim P P =0 plim plim = + = = + = 45

46 Hence ³X = ( + + ) = + + +( + + ) µ µ µ µ = ( ) " # = ( ) ( ) = ( ) ( ) = 3 + = 3 = + Also note that X µ plim X X µ = plim X plim = = + Then we have X µ 0 + X 0 From Cramer-Wold Device, we have P P µ 0 Usually in textbooks, we don t follow the above derivation. Simply others use the conditional expectation. Let x =( ) 0 Then we consider ³ˆ x and ³ˆ x 46

47 which is equivalent to treat like a nonstochastic variable. Asymptotically we note that ³ˆ x 0 Q where µ x 0 x Q = lim 47

48 Dummy Variable Regression: Let s go back to our orignial example = + + where =0for = for = Suppose that = total number of female = total number of male. Then = + = or = Treat as if is nonrandom. Then we have ˆ = + P P X = X =0 Next Note so that X = X Ã! X = X = 0 if female = = if male ³X X = X = : total number of female or male Hencewehave X = X Next, find the limiting distribution of ³X = = 4 = = 4 4 () X We know plim X =0 48

49 also we know since X = X Ã = X = X =! X = X X Ã = Ã! X X = X! X = = 0 if is female. = if is male Hence X = X = Ã = X = = X = X = X =! Ã X = X = = X = X = +! where = if is male if is female Note that the stochastic properties of isthesameas as long as is independent and identically distributed. Next, we know already the value of ( ) from (8) Ã! Ã X = X X! ( ) = = = = µ 4 4 = 4 4 =

50 Hence X µ 0 4 and ³ˆ = P P = P 4 since P = 4 from (). Therefore finally we have ³ˆ 0 4 = 4 X or ˆ p 4 (0 ) Note that the asymptotic variance of ˆ is Asy Var ³ˆ = 4 Now consider the quantity of ³X = 4 ³ˆ = Asy Var Two Dependent Dummies Consider two models = = + + Now = + where Further note that 0 if is non-skilled = if is skilled X 6=0 50

51 Consider the following pay-off matrix where indiciates the total number of observations. Unskilled Skilled Total Female + Male + Total + + Assume the total number of female = that of male. Further consider the following assumptions. + = + = = Then we have Unskilled Skilled Total Female + Male + Total 3 3 and the probability matrix becomes Unskilled Skilled Total Female + Male + Total 3 3 Note that + so that finally we have = + = = = 0 = 6 Unskilled Skilled Total Female 3 Male 6 Total 6 3 5

52 Consider the following expectation Expected earning Female & unskilled; = =0 Female & skilled; =0 = Male & unskilled; = =0 Male & skilled; = = ( )= ( )= + ( )= + ( )= + + Unskilled Skilled Total Female + + ( + ) 3 3 Male ( + )+ ( + + ) 3 3 Total + ( + ) ( + )+ ( + + ) Now let =0but 0Then we have Unskilled Skilled Total Female + + ( + ) = Male + + ( + ) = Total + Hence unskilled worker s earning is lower than skilled workers, and female earning is lower than male earning because of 0 not of 0 We can do the above analysis (very tedious) but rather also can do the following regression analysis to test if =0but 6= 0 Construct the following null hypothesis Further note that when we run = () 0 : =0 0 : =0 3 0 : = =0 = + + (3) the OLS estimator ˆ may not be zero. But the OLS estimator ˆ in () could be zero. Why? 5

53 Omitted Variable If ( ) 6= 0, or in other words, ( ) 6= 0 then the OLS estimator ˆ in (3) becomes inconsistent. (To students: Prove it) Cross Dummies Suppose that among unskilled workers, there is no gender earning difference. However among skilled workers, there is gender earning difference. How to test? Expected earning Female & unskilled; = =0 Female & skilled; =0 = Male & unskilled; = =0 Male & skilled; = = ( )= ( )= + ( )= + ( )= In this case, we will have = Then test =0but 6= 0 What is the meaning of = =0but 6= 0? What is the meaning of =0and =0? What is the meaning of =0but 6= 0and 6= 0? Sequential Dummies Consider the following learning choice: Expected earning Ph.D & taking Econometrics III Ph.D & not taking Econometrics III No Ph.D ( )= + + ( )= + ( )= Construct dummy variable regression: 53

54 Another Example of Nonstochastic Regressor = + 0 Q: Find the limiting distribution of ˆ = P P Consider X =0 ³X = ( ) = ( + ) = = X Note that Hencewehave X = = ( +)( +) 6 X µ ( +)( +) 0 6 and P P P = ( +)( +) 6 Now fine which makes The answer is P P 0 µ 0 6 ( +)( +) r ( +)( +) = 6 = r r + ( )= ( ) Hencewehave 3 ³ˆ µ0 3 54

55 Linear and Nonlinear Restrictions (Chapter 5) Consider the following regression y = Xb + u = x + x + u Then in general we have ³ˆb b (0 Σ b ) or ˆ 0 ˆ 0 Next consider the following linear restriction 0ˆ + ˆ = Alternatively we may let ³ˆ ˆ = 0ˆ + ˆ + := ˆ (4) Taking Taylor expansion around their true values yields ³ˆ ˆ = ( )+ ( ) ³ˆ + ( ) ³ˆ + ( ) + Note that from (4), we have ³ˆ + ( ) ( ) ³ˆ ³ˆ + ³ˆ ( ) = 0 ( ) = ( ) = ( ) = ( ) =0 (5) so that ˆ = ³ˆ + ³ˆ ³ˆ = ( 0 ) ³ˆ 55

56 or Finally we have (ˆ ) = h 0 i h 0 = 0 0 ³ˆ ³ˆ i (ˆ ) p (0 ) 0 Now compare this limiting result with Greene. (page 84) Let h i R = and q = Then we have 0 =(Rb q) 0 h R (X 0 X) R 0 i (Rb q) Note that (X 0 X) = Σ b in our notation. Next we consider a nonlinear restriction. Usually for a nonlinear restriction case, the second and cross terms in (4) are not equal to zero but become small terms. To see this, consider ³ˆ ˆ ( )= ( ) ³ˆ + ( ) ³ˆ + or ˆ = ( ) ³ˆ + ( ) ³ˆ + where is the remainder term. Note that ³ˆ is (), sothat ³ˆ ³ˆ is ( ) and ³ˆ ³ˆ = ( ) Hence we have = is Therefore, ( ) (ˆ ) = ³ˆ + ( µ ) ³ˆ + 56

57 and (ˆ ) = h ( ) ( ) h 0 i ( ) ( ) ³ˆ ³ˆ i µ + ( ) ( ) Example: Suppose that you want to test if ˆ ˆ =0 Then we have ( ) = ( ) = 57