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Kolmogorov goodness-of-fit test ! for -symmetric distributions S in climate and weather modeling Authors: Z.N. Zenkova L.A. Lanshakova Reporter: Dr. Zhanna Zenkova PhD, MBA, TSU Associate Professor Agenda ! Title: Kolmogorov


  1. Kolmogorov goodness-of-fit test ! for -symmetric distributions S in climate and weather modeling Authors: Z.N. Zenkova L.A. Lanshakova Reporter: Dr. Zhanna Zenkova PhD, MBA, TSU Associate Professor

  2. Agenda ! Title: Kolmogorov goodness-of-fit test for -symmetric distributions S in climate and weather modeling 1. Mathematical statistics in climate and weather modeling; 2. Kolmogorov goodness-of-fit test; ! 3. -symmetric distributions; S 4. Example .

  3. Mathematical statistics in climate and weather modeling Water lifting 3

  4. Mathematical statistics in climate and weather modeling Уровень подъема воды 4

  5. Kolmogorov goodness-of-fit test ( ) = X X , X ,..., X 1 2 N = H : F ( x ) G ( x ) 0 N = d sup F ( x ) G ( x ) (1) N x R , N 1 ) ( ) = F ( x ) I X (2) ( N ; x i N = i 1 ( ) + 2 2 ( ) k 2 k z < = = lim P N d z K z ( 1 ) e (3) N N = k 5

  6. Kolmogorov goodness-of-fit test G ( x ) F N ( x ) d N 6

  7. ! -symmetric distribution S = = + < < < Definition . If for , , k >1, p F ( c ) F ( c 0 ) c c ... c = i 1 , k i i i 1 2 k = = for and = = = x c p F ( c ) 1 , j 1 , k p F ( 0 c ) 0 , + + + j 1 k 1 k 1 0 c.d.f. F ( x ) satisfies the conditions: p p ( { } ) ( { } ) , + j 1 j (4) + = + F max x , S ( x ) 0 p F min x , S ( x ) 0 + j j 1 j p , j S j ( x ) where are continuous monotonically decreasing functions, ( ) 1 = S ( x ) S ( x ) j j ( ) ( ) , , , = = x c S c c S c c + j 1 + j i i j j 1 0 then c.d.f. F ( x ) is - symmetric c.d.f. 7

  8. ! -symmetric distribution S = = = For k = 1, p F ( 1 c ) 0 . 5 , S ( x ) 2 c x 1 1 1 c we can obtain classical symmetry of c.d.f. around : 1 ( ) = F ( x ) 1 F 2 c x (5) 1 8

  9. Test modification S = S F ( x ) G ( x ), H : x R F , G , 0 against ( ) , S = S H : F ( x ) V G ( x ) V , F , G , x R 1 S S * = d p sup F ( x ) G ( x ) (6) 1 N N x R * G ( x ) = (7) G ( x ) p 1 9

  10. Test modification F S is e.d.f. based on k times symmetrized sample ( x ) N ( ) * ,..., * * = X X X 1 N { ( ) } ( j ) ( j 1 ) j 1 = (8) j = X min X , S X 1 k , , + k j 1 i i i ( k ) * = X X i i ( 0 ) = = X X , i 1 , N i i 10

  11. Test modification ( ) , For = > F ( x ) V G ( x ) y 0 , ( ) + j i 1 ( ) V p a V p b 1 j 1 i 0 1 1 ( ) p p 1 1 S = N < N ! det P d y (9) + ( j i 1 ) ! = i,j 1 , N , i y i 1 z 1 = 1 = = + a i 1 0 , i 1 , N b i 0 1 , N p N p 1 1 p p j j = = z min z , y , y max y , z , = j 1 , k + + + + j j 1 j 1 j j 1 j 1 p p p p + j 1 j + j 1 j = = y z y 11 + + k 1 k 1

  12. Test modification > For z 0 ( ) z S < = lim P N d z K , (10) N p N 1 12

  13. Example The uniform c.d.f. U ( x ) is satisfied by (4), if for arbitrary c = < p 1 < p 0 1 , 1 1 1 p 1 1 x , 0 x c , 1 p 1 (11) = S ( x ) 1 1 x < p , c x 1 . 1 1 1 p 1 13

  14. Example [ ] , S = = H 0 : F ( x ) U ( x ) x , x 0 , 1 against [ ] , S = H 1 : F ( x ) U ( x ), x 0 , 1 1 m x [ ] p , x 0 , p , 1 1 p 1 = U ( x ) (12) 1 m 1 x ] ( ) ( 1 1 p , x p , 1 . 1 1 1 p 1 14

  15. Example m = 3, = p = c 0 . 8 1 1 1 0,9 = p = Classical symmetric cdf, c 0 . 5 1 1 0,8 0,7 0,6 U ( x ) U 1 x ( ) 0,5 0,4 0,3 0,2 0,1 0 0 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 15

  16. Example 1 = m = 3, p 0 , 7 16

  17. Thank you for your attention!

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