Matematické Metody v Ekonometrii 5.

Transcription

1 Matematické Metody v Ekonometrii 5. Multiple Regression Analysis - goodness of fit Blanka Šedivá KMA zimní semestr 2016/2017 Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

2 Illustration of the least square method for simple linear regression line Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

3 Goodness of fit and coefficient of determination for simple linear regression model Y = X β + ε (y i y) = (ŷ i y) + (y i ŷ i ) i (y i y) 2 = i (ŷ i y) 2 + i (y i ŷ i ) 2 (hold only for model with absolute part) Total sum of squares TSS = S y = i (y i y) 2 Explained sum of squares ESS = Sŷ = i (ŷ i y) 2 Residual sum of squares RSS = i (y i ŷ i ) 2 TSS = ESS + RSS coefficient of determination R 2 = ESS TSS = 1 RSS TSS = 1 RSS i (y i y) 2 R 2 measure the proportion or percentage of total variation in Y explained by the regression model for sample line linear regression R 2 = ĉor (X, Y ) Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

4 Adjusted coefficient of determination Mean Total sum of squares MTSS = TSS/(n 1) = s y = i (y i y) 2 /(n 1) Mean Explained sum of squares MESS = ESS/(p 1) = sŷ = i (ŷ i y) 2 /(p 1) Mean Residual sum of squares MRSS = RSS/(n p) = s R = i (y i ŷ i ) 2 /(n p) R 2 = 1 RSS TSS = 1 (n p)s2 R (n 1)s 2 y R 2 adj = 1 s2 R s 2 y R 2 adj = 1 ( 1 R 2) n 1 n p Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

5 Regression through the Origin (RTO) Y i = βx i + ε i the RTO residuals will usually have a nonzero mean, because forcing the regression line through the origin is generally inconsistent with the best fit equation i (y i y) 2 = i (ŷ i y) 2 + i (y i ŷ i ) 2 is not valid if the RTO model provides a sufficiently poor fit, the data may exhibit more variation around the regression line than around y, in which case i (y i ŷ i ) 2 > i (y i y) 2 (y i 0) = (ŷ i 0) + (y i ŷ i ) i (y i) 2 = i (ŷ i) 2 + i (y i ŷ i ) 2 RRTO 2 = i ŷ 2 i i y i 2 Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

6 Wald statistics for Testing Multiple Hypothesis Normal linear regression model Y N ( X β, s 2 ) I Let R be a q p matrix with q p and restriction of vektor can by describe be equation Rβ = r W = (Rb r) (R T ( X T ) ) 1 1 T X R (Rb r)/s 2 assymp. If s 2 = RSS/(n p) than W /q F q,n p We can test multiple hypotheses test or a joint hypotheses test H 0 : β 3 = 0, β 4 = 0, β 5 = 0 or H 0 : β 2 = β 4, β 1 = β 5 or H 0 : β 2 + β 3 = 1 or χ 2 q Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

7 Testing the overall significance of regression x 1, x 2,..., x n do not help to explain y H 0 : β 1 = 0; β 2 = 0;... ; β k = 0, all slope coefficients are simultaneously zero H 1 = nonh 0 exist β j 0 for normally distributed disturbances the variable F = ESS/(p 1) RSS/(n p) is distributed as F distribution with ν 1 = p 1 a ν 2 = n p F = R2 1 R 2 n p p 1 if the F value exceeds the critical F value from F table at the α percent level of significance, we reject H 0 otherwise we do not reject it joint testing of hypotheses Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

8 Testing model with restriction RSS r sum of square residuals from the restricted model RSS ur sum of square residuals from the unrestricted model F = (RSS r RSS ur )/q RSS ur /(n p) using the fact RSS = TSS(1 R 2 ) we can obtain R square form of statistics F F = (R2 ur Rr 2 )/q (1 Rur 2 )/(n p) Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

9 Testing linear equality restrictions - example H 0 : β 1 + β 2 = 1 One restrictionq = 1 Testing statistic F = b 1 + b 2 (β 1 + β 2 ) var b1 + var b 2 + 2cov (b 1, b 2 ) If the F value exceeds the critical F value with degrees of freedom with ν 1 = q a ν 2 = n p from F table at the α percent level of significance, we reject H 0 otherwise we do not reject it. For this restriction F ν1 =1,ν 2 =n 2 = t ν=n 2 Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

10 Example I y = (1, 2, 3, 4) T a vysvětlující proměnné x = ( 2, 1, 1, 2) T maticově Y = X β + ɛ X = 1 1 Y 1 1 = = [ β1 β 1 ] b = ( X T ) 1 T X X y [ X T 4 0 X = 0 10 [ ] X T 10 y = 7 ] ( ) [ 1 X T 1/4 0 X = 0 1/10 ] Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

11 Example II ŷ = b = [ 5/2 ] 7/ residua e = [ 5/2 7/10 1/10 2/10 2/10 1/10 ] = RSE = e T e = 10/100 = 1/10 s 2 R sb = 11/10 18/10 32/10 39/10 = RSS/2 = 1/20 = 0.05 ( ) , = (0.1118, ) Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

12 Example III confidence interval β for α = 5% and df = 2 is according tables t0.975 = P ( < β 1 < ) = 0.95 P ( < β2 < ) = 0.95 H 0 : β 1 = 0 statistics t = 2.5 1/80 = the critical region for α = 5% is W = (, ) (4.3027, ) t W reject H 0 H 0 : β 2 = 0 t = 0.7 1/200 = the critical region for α = 5% is W = (, ) (4.3027, ) t W reject H 0 total sum of squares TSS = (1 2.5) 2 + (2 2.5) 2 + (3 2.5) 2 + (4 2.5) 2 = 5 Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13

13 Example IV R 2 = = 0.98 H 0 : β 2 = 0 F = = 98 the critical region for α = 5% is W = (18.513, ) t W reject H 0 Blanka Šedivá (KMA) Matematické Metody v Ekonometrii 5. zimní semestr 2016/ / 13