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International Scholarly Research Notices
/
2012
/
Article
/
Tab 6
/
Research Article
On the Order Statistics of Standard Normal-Based Power Method Distributions
Table 6
Expected values of order statistics for
𝑝
1
(
𝑍
)
=
𝑍
for
𝑛
=
6
,
7
.
𝐸
[
𝑝
1
(
𝑍
)
4
∶
6
]
=
1
5
0
𝜀
2
𝜋
5
/
2
−
3
0
t
a
n
−
1
√
(
1
/
2
)
𝜋
3
/
2
+
1
5
0
t
a
n
−
1
√
(
1
/
2
)
2
𝜋
5
/
2
=
0
.
2
0
1
5
4
6
8
3
…
𝐸
[
𝑝
1
(
𝑍
)
5
∶
6
]
=
−
1
5
2
√
𝜋
−
7
5
𝜀
2
𝜋
5
/
2
+
6
0
t
a
n
−
1
√
(
1
/
2
)
𝜋
3
/
2
−
7
5
t
a
n
−
1
√
(
1
/
2
)
2
𝜋
5
/
2
=
0
.
6
4
1
7
7
5
0
3
…
𝐸
[
𝑝
1
(
𝑍
)
6
∶
6
]
=
1
5
2
√
𝜋
+
1
5
𝜀
2
𝜋
5
/
2
−
3
0
t
a
n
−
1
√
(
1
/
2
)
𝜋
3
/
2
+
1
5
t
a
n
−
1
√
(
1
/
2
)
2
𝜋
5
/
2
=
1
.
2
6
7
2
0
6
3
6
…
𝐸
[
𝑝
1
(
𝑍
)
4
∶
7
]
=
0
𝐸
[
𝑝
1
(
𝑍
)
5
∶
7
]
=
5
2
5
𝜀
2
2
𝜋
5
/
2
−
1
0
5
t
a
n
−
1
√
(
1
/
2
)
2
𝜋
3
/
2
+
5
2
5
t
a
n
−
1
√
(
1
/
2
)
2
2
𝜋
5
/
2
=
0
.
3
5
2
7
0
6
9
5
…
𝐸
[
𝑝
1
(
𝑍
)
6
∶
7
]
=
−
2
1
2
√
𝜋
−
2
1
0
𝜀
2
𝜋
5
/
2
+
1
0
5
t
a
n
−
1
√
(
1
/
2
)
𝜋
3
/
2
−
2
1
0
t
a
n
−
1
√
(
1
/
2
)
2
𝜋
5
/
2
=
0
.
7
5
7
3
7
4
2
7
…
𝐸
[
𝑝
1
(
𝑍
)
7
∶
7
]
=
2
1
2
√
𝜋
+
1
0
5
𝜀
2
2
𝜋
5
/
2
−
1
0
5
t
a
n
−
1
√
(
1
/
2
)
2
𝜋
3
/
2
+
1
0
5
t
a
n
−
1
√
(
1
/
2
)
2
2
𝜋
5
/
2
=
1
.
3
5
2
1
7
8
3
7
…
