6
The Probable Error of a Mean
By continuing this process we find
I
n
−
2
=
(
σ
2
n
)
n
−
2
2
(
n
−
3
)
(
n
−
5
)
.
.
.3
.1
I
0
{\displaystyle I_{n-2}=\left({\frac {\sigma ^{2}}{n}}\right)^{\frac {n-2}{2}}(n-3)(n-5)...3.1I_{0}}
or
=
(
σ
2
n
)
n
−
3
2
(
n
−
3
)
(
n
−
5
)
.
.
.4
.2
I
1
{\displaystyle =\left({\frac {\sigma ^{2}}{n}}\right)^{\frac {n-3}{2}}(n-3)(n-5)...4.2I_{1}}
according as
n
{\displaystyle n}
But
I
0
{\displaystyle I_{0}}
∫
0
∞
e
−
n
x
2
2
σ
2
d
x
=
π
2
n
σ
{\displaystyle \int _{0}^{\infty }e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}dx={\sqrt {\frac {\pi }{2n}}}\sigma }
and
I
1
{\displaystyle I_{1}}
∫
0
∞
x
e
−
n
x
2
2
σ
2
d
x
=
[
−
σ
2
n
e
−
n
x
2
2
σ
2
]
x
=
0
x
=
∞
=
σ
2
n
{\displaystyle \int _{0}^{\infty }xe^{-{\frac {nx^{2}}{2\sigma ^{2}}}}dx=\left[-{\frac {\sigma ^{2}}{n}}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}\right]_{x=0}^{x=\infty }={\frac {\sigma ^{2}}{n}}}
Hence if
n
{\displaystyle n}
A
=
A
r
e
a
(
n
−
3
)
(
n
−
5
)
.
.
.3
.1
π
2
(
σ
2
n
)
n
−
1
2
{\displaystyle A={\frac {Area}{(n-3)(n-5)...3.1{\sqrt {\frac {\pi }{2}}}\left({\frac {\sigma ^{2}}{n}}\right)^{\frac {n-1}{2}}}}}
and if
n
{\displaystyle n}
A
=
A
r
e
a
(
n
−
3
)
(
n
−
5
)
.
.
.4
.2
(
σ
2
n
)
n
−
1
2
{\displaystyle A={\frac {Area}{(n-3)(n-5)...4.2\left({\frac {\sigma ^{2}}{n}}\right)^{\frac {n-1}{2}}}}}
Hence the equation may be written
y
=
N
(
n
−
3
)
(
n
−
5
)
.
.
.3
.1
2
π
(
n
σ
2
)
n
−
1
2
x
n
−
2
e
−
n
x
2
2
σ
2
{\displaystyle y={\frac {N}{(n-3)(n-5)...3.1}}{\sqrt {\frac {2}{\pi }}}\left({\frac {n}{\sigma ^{2}}}\right)^{\frac {n-1}{2}}x^{n-2}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}}
n
{\displaystyle n}
or
y
=
N
(
n
−
3
)
(
n
−
5
)
.
.
.4
.2
(
n
σ
2
)
n
−
1
2
x
n
−
2
e
−
n
x
2
2
σ
2
{\displaystyle y={\frac {N}{(n-3)(n-5)...4.2}}\left({\frac {n}{\sigma ^{2}}}\right)^{\frac {n-1}{2}}x^{n-2}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}}
n
{\displaystyle n}
where
N
{\displaystyle N}
To show that there is no correlation between (a ) the distance of the mean of a sample from the mean of the population and (b ) the standard deviation of a sample with normal distribution.
(1) Clearly positive and negative positions of the mean of the sample are equally likely, and hence there cannot be correlation between the absolute value of the distance of the mean from the mean of the population and the standard deviation, but (2) there might be correlation between the square of the distance and the square of the standard deviation.
Let
u
2
=
(
S
(
x
1
)
n
)
2
{\displaystyle u^{2}=\left({\frac {S(x_{1})}{n}}\right)^{2}}
s
2
=
S
(
x
1
2
)
n
−
(
S
(
x
1
)
n
)
2
{\displaystyle s^{2}={\frac {S(x_{1}^{2})}{n}}-\left({\frac {S(x_{1})}{n}}\right)^{2}}
Then if
m
1
′
{\displaystyle m'_{1}}
M
1
′
{\displaystyle M'_{1}}
u
2
{\displaystyle u^{2}}
s
2
{\displaystyle s^{2}}
M
1
′
=
μ
2
(
n
−
1
)
n
{\displaystyle M'_{1}=\mu _{2}{\frac {(n-1)}{n}}}
m
1
′
=
μ
2
n
{\displaystyle m'_{1}={\frac {\mu _{2}}{n}}}
![{\displaystyle \int _{0}^{\infty }xe^{-{\frac {nx^{2}}{2\sigma ^{2}}}}dx=\left[-{\frac {\sigma ^{2}}{n}}e^{-{\frac {nx^{2}}{2\sigma ^{2}}}}\right]_{x=0}^{x=\infty }={\frac {\sigma ^{2}}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/501380516d015b614acebf892fa6614d71249517)
