International Scholarly Research Notices

Research Article

On the Order Statistics of Standard Normal-Based Power Method Distributions

Table 1

General expressions for the expected values of the order statistics for 𝑝 𝑡 = 1 , 2 , 3 ( 𝑍 ) in (1.7) and sample sizes of 𝑛 = 1 , … , 9 . 𝐼 2 𝑟 − 1 denotes an integral in (2.1) where 𝑟 = 1 , … , 4 .
Sample size ( 𝑛 )Expected value
1 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 1 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 1 ] = 0
2 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 2 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 2 ] = 𝐼 1 √ = 1 / 𝜋
3 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 3 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 3 ] = 0
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 3 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 3 ] = ( 3 / 2 ) 𝐼 1 √ = 3 / ( 2 𝜋 )
4 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 4 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 4 ] = ( 3 / 2 ) ( 𝐼 1 − 𝐼 3 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 4 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 4 ] = ( 1 / 2 ) ( 3 𝐼 1 + 𝐼 3 )
5 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 5 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 5 ] = 0
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 5 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 5 ] = ( 5 / 2 ) ( 𝐼 1 − 𝐼 3 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 5 ∶ 5 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 5 ] = ( 5 / 4 ) ( 𝐼 1 + 𝐼 3 )
6 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 6 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 6 ] = ( 1 5 / 8 ) ( 𝐼 1 − 2 𝐼 3 + 𝐼 5 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 5 ∶ 6 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 6 ] = ( 1 5 / 1 6 ) ( 3 𝐼 1 − 2 𝐼 3 − 𝐼 5 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 6 ∶ 6 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 6 ] = ( 3 / 1 6 ) ( 5 𝐼 1 + 1 0 𝐼 3 + 𝐼 5 )
7 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 7 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 7 ] = 0
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 5 ∶ 7 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 7 ] = ( 1 0 5 / 3 2 ) ( 𝐼 1 − 2 𝐼 3 + 𝐼 5 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 6 ∶ 7 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 7 ] = ( 2 1 / 8 ) ( 𝐼 1 − 𝐼 5 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 7 ∶ 7 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 7 ] = ( 7 / 3 2 ) ( 3 𝐼 1 + 1 0 𝐼 3 + 3 𝐼 5 )
8 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 5 ∶ 8 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 8 ] = ( 3 5 / 1 6 ) ( 𝐼 1 − 3 𝐼 3 + 3 𝐼 5 − 𝐼 7 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 6 ∶ 8 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 8 ] = ( 2 1 / 1 6 ) ( 3 𝐼 1 − 5 𝐼 3 + 𝐼 5 + 𝐼 7 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 7 ∶ 8 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 8 ] = ( 7 / 1 6 ) ( 5 𝐼 1 + 5 𝐼 3 − 9 𝐼 5 − 𝐼 7 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 8 ∶ 8 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 8 ] = ( 1 / 1 6 ) ( 7 𝐼 1 + 3 5 𝐼 3 + 2 1 𝐼 5 + 𝐼 7 )
9 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 5 ∶ 9 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 5 ∶ 9 ] = 0
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 6 ∶ 9 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 4 ∶ 9 ] = ( 6 3 / 1 6 ) ( 𝐼 1 − 3 𝐼 3 + 3 𝐼 5 − 𝐼 7 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 7 ∶ 9 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 3 ∶ 9 ] = ( 6 3 / 1 6 ) ( 𝐼 1 − 𝐼 3 − 𝐼 5 + 𝐼 7 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 8 ∶ 9 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 2 ∶ 9 ] = ( 9 / 1 6 ) ( 3 𝐼 1 + 7 𝐼 3 − 7 𝐼 5 + 3 𝐼 7 )
𝐸 [ 𝑝 𝑡 ( 𝑍 ) 9 ∶ 9 ] = − 𝐸 [ 𝑝 𝑡 ( 𝑍 ) 1 ∶ 9 ] = ( 9 / 3 2 ) ( 𝐼 1 + 7 𝐼 3 + 7 𝐼 5 + 𝐼 7 )