Otoshajonta eli otoskeskihajonta
s
{\displaystyle s}
otosvarianssi n
s
2
{\displaystyle s^{2}}
s
2
=
∑
i
=
1
n
(
x
i
−
x
¯
)
2
n
−
1
{\displaystyle s^{2}=\sum _{i=1}^{n}{\frac {(x_{i}-{\overline {x}})^{2}}{n-1}}\,\,\,\,{\ }}
ja
x
¯
=
∑
i
=
1
n
x
i
n
{\displaystyle {\overline {x}}=\sum _{i=1}^{n}{\frac {x_{i}}{n}}\,\,\,\,\,{\ }}
on tutkittavan muuttujan
x
{\displaystyle x}
Kun luvut
x
1
,
x
2
,
…
,
x
n
{\displaystyle x_{1},x_{2},\ldots ,x_{n}}
X ,
s
{\displaystyle s}
harhaton estimaatti joukon X keskihajonnasta. Intuitiivisesti tämä selittyy sillä, että otoskeskiarvo
x
¯
{\displaystyle {\overline {x}}}
X todellisesta keskiarvosta otoksen suuntaan, mikä tuottaisi keskihajonnan (
s
2
{\displaystyle s^{2}}
X todellinen keskiarvo, nimittäjässä pitäisi olla n kuten yleensäkin keskihajonnan kaavassa.
Otosvarianssin harhattomuus
muokkaa
Tiedetään, että satunnaismuuttujan varianssi
Var
(
x
i
)
=
σ
2
=
E
[
x
i
2
]
−
(
E
[
x
i
]
)
2
=
{\displaystyle \operatorname {Var} (x_{i})=\sigma ^{2}=\mathbb {E} [x_{i}^{2}]-(\mathbb {E} [x_{i}])^{2}=}
E
[
x
i
2
]
−
μ
2
⟺
E
[
x
i
2
]
=
σ
2
+
μ
2
{\displaystyle \mathbb {E} [x_{i}^{2}]-\mu ^{2}\iff \mathbb {E} [x_{i}^{2}]=\sigma ^{2}+\mu ^{2}}
Var
(
x
¯
)
=
σ
2
n
=
E
[
x
¯
2
]
−
(
E
[
x
¯
]
)
2
=
{\displaystyle \operatorname {Var} ({\bar {x}})={\frac {\sigma ^{2}}{n}}=\mathbb {E} [{\bar {x}}^{2}]-(\mathbb {E} [{\bar {x}}])^{2}=}
E
[
x
¯
2
]
−
μ
2
⟺
E
[
x
¯
2
]
=
σ
2
n
+
μ
2
{\displaystyle \mathbb {E} [{\bar {x}}^{2}]-\mu ^{2}\iff \mathbb {E} [{\bar {x}}^{2}]={\frac {\sigma ^{2}}{n}}+\mu ^{2}}
n
{\displaystyle n}
E
[
s
2
]
=
E
[
∑
i
=
i
n
(
x
i
−
x
¯
)
2
n
−
1
]
=
1
n
−
1
E
[
∑
i
=
i
n
(
x
i
2
−
2
x
i
x
¯
+
x
¯
2
)
]
=
{\displaystyle \mathbb {E} [s^{2}]=\mathbb {E} \left[\sum _{i=i}^{n}{\frac {(x_{i}-{\bar {x}})^{2}}{n-1}}\right]={\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2}-2x_{i}{\bar {x}}+{\bar {x}}^{2})\right]=}
1
n
−
1
E
[
∑
i
=
i
n
(
x
i
2
)
−
∑
i
=
i
n
(
2
x
i
x
¯
)
+
∑
i
=
i
n
(
x
¯
2
)
]
=
{\displaystyle {\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2})-\sum _{i=i}^{n}(2x_{i}{\bar {x}})+\sum _{i=i}^{n}({\bar {x}}^{2})\right]=}
1
n
−
1
E
[
∑
i
=
i
n
(
x
i
2
)
−
2
x
¯
∑
i
=
i
n
(
x
i
)
+
∑
i
=
i
n
(
x
¯
2
)
]
=
{\displaystyle {\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2})-2{\bar {x}}\sum _{i=i}^{n}(x_{i})+\sum _{i=i}^{n}({\bar {x}}^{2})\right]=}
1
n
−
1
E
[
∑
i
=
i
n
(
x
i
2
)
−
2
n
x
¯
2
+
∑
i
=
i
n
(
x
¯
2
)
]
=
{\displaystyle {\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2})-2n{\bar {x}}^{2}+\sum _{i=i}^{n}({\bar {x}}^{2})\right]=}
1
n
−
1
(
n
E
[
x
i
2
]
−
2
n
E
[
x
¯
2
]
+
n
E
[
x
¯
2
]
)
=
{\displaystyle {\frac {1}{n-1}}\left(n\mathbb {E} [x_{i}^{2}]-2n\mathbb {E} [{\bar {x}}^{2}]+n\mathbb {E} [{\bar {x}}^{2}]\right)=}
1
n
−
1
(
n
E
[
x
i
2
]
−
n
E
[
x
¯
2
]
)
=
{\displaystyle {\frac {1}{n-1}}\left(n\mathbb {E} [x_{i}^{2}]-n\mathbb {E} [{\bar {x}}^{2}]\right)=}
1
n
−
1
(
n
(
σ
2
+
μ
2
)
−
n
(
σ
2
n
+
μ
2
)
)
=
{\displaystyle {\frac {1}{n-1}}\left(n(\sigma ^{2}+\mu ^{2})-n({\frac {\sigma ^{2}}{n}}+\mu ^{2})\right)=}
1
n
−
1
(
n
σ
2
+
n
μ
2
−
σ
2
−
n
μ
2
)
)
=
{\displaystyle {\frac {1}{n-1}}\left(n\sigma ^{2}+n\mu ^{2}-\sigma ^{2}-n\mu ^{2})\right)=}
1
n
−
1
(
n
−
1
)
σ
2
=
σ
2
{\displaystyle {\frac {1}{n-1}}(n-1)\sigma ^{2}=\sigma ^{2}}
Siis otosvarianssin odotusarvo on sama kuin satunnaismuuttujan varianssi, joten otosvarianssi on harhaton estimaatti.
![{\displaystyle \operatorname {Var} (x_{i})=\sigma ^{2}=\mathbb {E} [x_{i}^{2}]-(\mathbb {E} [x_{i}])^{2}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e6144e25ca15aba823f6b5c2a6d8a6d8aff298e)
![{\displaystyle \mathbb {E} [x_{i}^{2}]-\mu ^{2}\iff \mathbb {E} [x_{i}^{2}]=\sigma ^{2}+\mu ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf2743cc3c5c3e4c2a7baca0117dc00521f702e2)
![{\displaystyle \operatorname {Var} ({\bar {x}})={\frac {\sigma ^{2}}{n}}=\mathbb {E} [{\bar {x}}^{2}]-(\mathbb {E} [{\bar {x}}])^{2}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0719ea5bdb44f301808bb777fea89c4b40e6b6a0)
![{\displaystyle \mathbb {E} [{\bar {x}}^{2}]-\mu ^{2}\iff \mathbb {E} [{\bar {x}}^{2}]={\frac {\sigma ^{2}}{n}}+\mu ^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aadfb1f37bd554da7327ebea68c5183dc1160af9)
![{\displaystyle \mathbb {E} [s^{2}]=\mathbb {E} \left[\sum _{i=i}^{n}{\frac {(x_{i}-{\bar {x}})^{2}}{n-1}}\right]={\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2}-2x_{i}{\bar {x}}+{\bar {x}}^{2})\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a6674f1988aaedd335a0582afc3160c44e06c1e)
![{\displaystyle {\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2})-\sum _{i=i}^{n}(2x_{i}{\bar {x}})+\sum _{i=i}^{n}({\bar {x}}^{2})\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fba5552f16311f031ce03c72a29512fc9767382)
![{\displaystyle {\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2})-2{\bar {x}}\sum _{i=i}^{n}(x_{i})+\sum _{i=i}^{n}({\bar {x}}^{2})\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b18192815b0e94961a32a0ca88354f0ce0d382a)
![{\displaystyle {\frac {1}{n-1}}\mathbb {E} \left[\sum _{i=i}^{n}(x_{i}^{2})-2n{\bar {x}}^{2}+\sum _{i=i}^{n}({\bar {x}}^{2})\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae838eff8ad4c1bfc47c84e58d26d323c3a9dc95)
![{\displaystyle {\frac {1}{n-1}}\left(n\mathbb {E} [x_{i}^{2}]-2n\mathbb {E} [{\bar {x}}^{2}]+n\mathbb {E} [{\bar {x}}^{2}]\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b51515ca41ba638f221acac2b8c5dd527909e5c)
![{\displaystyle {\frac {1}{n-1}}\left(n\mathbb {E} [x_{i}^{2}]-n\mathbb {E} [{\bar {x}}^{2}]\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0003e5ed6c8bb9fb0018e0600454542ad6c2194b)
