Smoothing of sign test and approximation of its p -value Mengxin LU - PowerPoint PPT Presentation

Smoothing of sign test and approximation of its p -value Mengxin LU Yoshihiko MAESONO Kyushu Univ., Grad. Sch. Math. Kyushu Univ., Fac. Math. 1

1. Introduction X 1 , X 2 , · · · , X n : i.i.d. F ( x − θ ) F is symmetric distr., i.e. F ( − x ) = 1 − F ( x ) ( f ( − x ) = f ( x )) One sample testing problem: Null-hypothesis: H 0 : θ = 0 vs. Alternative: H 1 : θ > 0 [Test statistics] t -test √ n ( X ) n n 1 ( X i − X ) 2 � � √ T = , X = X i , V = n − 1 V i =1 i =1 2

Sign test n � 1 , x ≥ 0 � S = ψ ( X i ) , ψ ( x ) = 0 , x < 0 i =1 Mann-Whitney test （ Wilcoxon’s signed rank test ） � W = ψ ( X i + X j ) 0 ≤ i ≤ j ≤ n 3

Under null hypothesis H 0 , distributions of S, W do not depend on F (distribution-free) For observed values s, w of S, W , we evaluate significance probabilities P 0 ( S ≥ s ) , P 0 ( W ≥ w ) If p -value is small enough, we reject H 0 For large n it is quite difficult to obtain exact distribution Normal approximation S − E 0 ( S ) L − → N (0 , 1) , � V ar 0 ( S ) 4

W − E 0 ( W ) L − → N (0 , 1) � V ar 0 ( W ) Using （ continuity correction ） p -vales are given by P 0 ( S ≥ s ) = P 0 (2( S − n ≥ 2( s − n 2 − 0 . 5) 2 − 0 . 5) √ n √ n ) ≈ 1 − Φ(2( s − n 2 − 0 . 5) √ n ) , P 0 ( W ≥ w ) √ √ 24( W − n ( n +1) 24( w − n ( n +1) − 0 . 5) − 0 . 5) 4 4 = P 0 ( ≥ ) � � n ( n + 1)(2 n + 1) n ( n + 1)(2 n + 1) √ 24( w − n ( n +1) − 0 . 5) 4 ≈ 1 − Φ( ) � n ( n + 1)(2 n + 1) 5

2. Smoothing sign based on kernel estimation Purpose of this talk ： constructing smoothing sign test Brown, Hall, Young (2001, B.K.) ： Smoothed sign test Median ˜ θ n ˜ � θ = argmin | x i − θ | i =1 Smoothed median: { ( x i − θ ) 2 + ( x j − θ ) 2 } 1 / 2 ˆ � θ = argmin i<j Smoothed sign test : based on the smoothed median estimator 6

Proposed test has different Pitman A.R.E. The proposed test statistic is not distribution-free The asymptotic distribution depend on the underlying distribution We have to make estimators They also obtained an Edgeworth expansion, but it contains unknown parameters 7

Let us consider empirical distribution n F n ( x 0 ) = 1 � I ( X i ≤ x 0 ) n i =1 If F ( x ) is symmetric around 0, F (0) = 1 2 If alternative H 1 is true, F n (0) is an unbiased estimator of F ( − θ ) Under H 1 , F n (0) takes smaller value than 1 2 , stochastically Under H 0 , nF n (0) is a binomial B ( n, 1 2 ) Normal approximation is valid, but the Edgeworth expansion is invalid 8

[Test statistic based on kernel estimator] Let us assume a kernel function k ( u ) is 2nd order � ∞ � ∞ k ( u ) du = 1 , uk ( u ) du = 0 −∞ −∞ Let us define � t W ( t ) = k ( u ) du −∞ Kernel estmator of F ( x 0 ) is given by n F n ( x 0 ) = 1 W ( x 0 − X i ˜ � ) n h n i =1 where h n is bandwidth h n → 0 , nh n → ∞ ( n → ∞ ) For testing H 0 , we can use S = ˜ ˜ F n (0) 9

This test is regarded as smoothed test of S （ F n (0) ） For observed value ˜ s , significance probability is P 0 ( ˜ S ≤ ˜ s ) ˜ S is not distribution-free. In the sequel, we use the bandwidth h n = n − 1 4 , or h n = n − 1 3 ˜ S is a sum of i.i.d. random variables. If n →∞ V ar [ W ( − X i 0 < lim )] < ∞ , h n ˜ S is asymptotically normal S − E ( ˜ ˜ S ) L − → N (0 , 1) � V ar ( ˜ S ) 10

Its asymptotic variance is given by n →∞ nV ar θ ( ˜ lim S ) = F ( − θ )(1 − F ( − θ )) = (1 − F ( θ )) F ( θ ) Further we have S n ( θ ) = E θ ( ˜ µ ˜ S ) = F ( − θ ) + O ( h n ) , S ) = F (0)(1 − F (0)) + O ( h n ) = 1 nV ar 0 ( ˜ 4 + O ( h n ) , µ ′ S n (0) ˜ = e ( ˜ lim S ) = − 2 f (0) � n →∞ nV ar 0 ( ˜ S n ) This efficacy coinsides to one of the sign test S . Thus S and ˜ S have same Pitman A.R.E. 11

3. Approximation of p -value based on Edgeworth Edgeworth expansion for kernel type estimator ˜ F n ( x 0 ) of F ( x 0 ) Garc´ ıa-Soid´ an et al. (1997): They showed the validity of the expansion, but not obtained an explicit form of the expansion Huang and Maesono (2013) Assuming h n = n − 1 4 or h n = n − 1 3 , obtained the explicit form of the expansion 12

[When h n = n − 1 4 ] Under some regularity conditions ˜ F n ( x 0 ) − F ( x 0 ) ≤ y ) = G ( y ) + o ( n − 1 P ( 2 ) (1) √ n � n W ( x 0 − X 1 V ar ( 1 )) h n where G ( y ) = Φ( y ) + D 1 + n − 1 4 D 2 + n − 1 2 D 3 . (2) D 1 = − φ ( y ) C 2 + 1 2 − 1 2 + 1 2 φ ′ ( y ) C 2 6 φ (2) ( y ) C 3 24 φ (3) ( y ) C 4 2 − 1 2 + 1 120 φ (4) ( y ) C 5 720 φ (5) ( y ) C 6 2 , D 2 = − φ ( y ) C 2 + φ ′ ( y ) C 2 C 3 − 1 2 C 3 + 1 6 φ (2) ( y ) C 2 24 φ (3) ( y ) C 3 2 C 3 , 13

D 3 = − φ ( y ) C 4 + 1 3 + C 2 C 4 ) − 1 6 B 3 , 0 ( φ ( y )( y 2 − 1) 2 φ ′ ( y )( C 2 − C 2 [ φ ′ ( y )( y 2 − 1) + 2 yφ ( y )] + C 2 2 [ φ ( y ) + 2 yφ ′ ( y ) +1 2 [ φ ′ ( y ) + yφ (2) ( y ) + 1 2( y 2 − 1) φ (2) ( y )] − C 3 6( y 2 − 1) φ (3) ( y )]) Further � ∞ W i ( u ) k ( u ) u j du, A i,j = −∞ 1 − 2 F ( x 0 ) B 3 , 0 = , � F ( x 0 )(1 − F ( x 0 )) 14

B 3 , 1 = 3 f ( x 0 )( A 1 , 1 − A 1 , 2 ) , 3 [ F ( x 0 )(1 − F ( x 0 ))] 2 f ′ ( x 0 ) A 0 , 2 C 2 = , � F ( x 0 )(1 − F ( x 0 )) f ′′ ( x 0 ) A 0 , 3 + f ( x 0 ) f ′ ( x 0 ) A 0 , 2 A 1 , 1 C 3 = − , 3 � 6 F ( x 0 )(1 − F ( x 0 )) 2[ F ( x 0 )(1 − F ( x 0 )] 2 f (3) ( x 0 ) A 0 , 4 − f ( x 0 ) f ′′ ( x 0 ) A 0 , 3 A 1 , 1 C 4 = 3 � F ( x 0 )(1 − F ( x 0 )) 6[ F ( x 0 )(1 − F ( x 0 ))] 2 f 2 ( x 0 ) f ′ ( x 0 ) A 0 , 2 A 2 − ( f ′ ( x 0 )) 2 A 0 , 2 ( A 1 , 2 − F ( x 0 ) A 0 , 2 ) 1 , 1 + 3 5 4[ F ( x 0 )(1 − F ( x 0 ))] 4[ F ( x 0 )(1 − F ( x 0 )] 2 2 15

When we use ˜ S , note that x 0 = 0 If k ( − u ) = k ( u ) is symmetric, we have W ( u ) = 1 − W ( − u ), and so � ∞ � ∞ W ( u ) k ( u ) u 2 du = { 1 − W ( − u ) } k ( u ) u 2 du A 1 , 2 = −∞ −∞ � ∞ � ∞ � ∞ { 1 − W ( t ) } k ( − t )( − t ) 2 dt k ( t ) t 2 dt − W ( t ) k ( t ) t 2 dt = −∞ −∞ −∞ � ∞ W ( t ) k ( t ) t 2 dt = − A 1 , 2 = − −∞ Thus A 1 , 2 = 0. If the following conditions on the kernel � ∞ � ∞ k ( z ) z 2 dz = k ( z ) z 4 dz = 0 , k ( − u ) = k ( u ) , (3) −∞ −∞ � ∞ W ( z ) k ( z ) zdz = 0 (4) −∞ 16

are satisfied, we have C 2 = C 3 = C 4 = 0. Thus we have the equation (1) 2 B 3 , 0 G ( y ) = Φ( y ) − n − 1 6 φ ( y )( y 2 − 1) Further, since F (0) = 1 2 , we have B 3 , 0 = 0 Also S ) = 1 2 + o ( n − 3 E 0 ( ˜ 2 ) , S ) = 1 4 n + o ( n − 3 V 0 ( ˜ 2 ) Therefore the conditions (3) and (4) are satisfied, we have an improve- 17

ment of the normal approximation P (2 √ n ( ˜ S − 1 2) ≤ y ) = Φ( y ) + o ( n − 1 2 ) [Remark 1] In addition to (3), (4), if the condition � ∞ W ( z ) k ( z ) z 3 dz = 0 A 1 , 3 = −∞ is satisfied, we have the Edgeworth expansion with remainder term o ( n − 3 4 ), and the expansion does not include unknown parameters ． 18

[When h n = n − 1 3 ] • If (3), (4) are satisfied, the residual term is o ( n − 3 4 ) • If (3), (4) and A 1 , 3 = A 2 , 2 = 0 are satisfied, the residual term is o ( n − 1 ). And the expansion does not have unknown parameters. The n − 1 term is 1 − 3 F (0) + 3 F 3 (0) F (0)(1 − F (0)) [When h n = n − δ ( 1 3 < δ < 1 2 ) のとき ] • (3), (4) are satisfied, the residual term is o ( n − 1 ), and the expansion does not have unknown parameters. 19

[Remark 2] The kernel k ( u ) which satisfies (3) and (4) is given by k ( u ) = ( a 0 + a 1 | u | + a 2 u 2 + a 3 | u | 3 ) I ( | u | ≤ 1) where √ √ a 0 = 45 65) , a 1 = 9 64(1 + 4(5 − 3 65) , √ √ a 2 = 105 64 ( − 23 + 9 65) , a 3 = 9(3 − 65) It is possible to make a kernel which satisfies A 1 , 3 = A 2 , 2 = 0. The kernel is quite complicate. 20

[Sketch of the proofs ] Let us define W 1 = W ( x 0 − X 1 ) h n Evaluation of bias Using Taylor expansion, we have W ( x 0 − y � E ( W 1 ) = ) f ( y ) dy h n � = h n W ( z ) f ( x 0 − h n Z ) dz = h n ( − 1 ) W ( z ) F ( x 0 − h n z ) | + ∞ −∞ h n 21

− h n ( − 1 � ) k ( z ) F ( x 0 − h n z ) dz h n � = k ( z ) F ( x 0 − h n z ) dz k ( z )( F ( x 0 ) − h n zf ( x 0 ) + 1 � ′ ( x 0 ) + · · · ) dz 2( h n z ) 2 f = � � = F ( x 0 ) k ( z ) dz − f ( x 0 ) h n zk ( z ) dz +1 � ′ ( x 0 ) h 2 z 2 k ( z ) dz + · · · 2 f n = F ( x 0 ) + 1 ′ ( x 0 ) A 0 , 2 + 1 2 h 2 6 h 3 n f ′′ ( x 0 ) A 0 , 3 n f + 1 24 h 4 n f (3) ( x 0 ) A 0 , 4 + O ( h 5 n ) 22

If k ( u ) is 4-th order kernel, bias is O ( h 4 n ) and A 0 , 1 = A 0 , 2 = A 0 , 3 = 0 If k ( u ) is 6-th order kernel, bias is O ( h 5 n ) and A 0 , 1 = A 0 , 2 = A 0 , 3 = A 0 , 4 = A 0 , 5 = 0 If k ( u ) is ℓ -th order kernel � ∞ � ∞ u m − 1 k ( u ) du = 0 (2 ≤ m ≤ ℓ − 1) , u ℓ k ( u ) du � = 0 −∞ −∞ 23