Ohne Beschränkung der Allgemeinheit gelte
a
=
0
,
b
=
1
{\displaystyle a=0,b=1}
Sei
K
{\displaystyle K}
n
{\displaystyle n}
p
{\displaystyle p}
E
[
K
n
]
=
p
{\displaystyle \mathbb {E} \left[{\frac {K}{n}}\right]=p}
Mit dem schwachen Gesetz der großen Zahlen (
E
[
K
/
n
]
=
p
{\displaystyle \mathbb {E} [K/n]=p}
lim
n
→
∞
P
(
|
K
n
−
p
|
>
δ
)
=
0
{\displaystyle \lim _{n\to \infty }{P\left(\left|{\frac {K}{n}}-p\right|>\delta \right)}=0}
für alle
δ
>
0
{\displaystyle \delta >0}
K
/
n
{\displaystyle K/n}
p
{\displaystyle p}
Diese Konvergenz bzgl.
n
{\displaystyle n}
p
{\displaystyle p}
K
/
n
{\displaystyle K/n}
p
{\displaystyle p}
Die Varianz von
K
/
n
{\displaystyle K/n}
1
4
n
{\displaystyle {\frac {1}{4n}}}
Da
K
{\displaystyle K}
V
a
r
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K
/
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)
=
V
a
r
(
K
)
n
2
=
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p
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1
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n
2
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p
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1
−
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n
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+
p
n
{\displaystyle {\begin{aligned}Var(K/n)&={\frac {Var(K)}{n^{2}}}\\&={\frac {np(1-p)}{n^{2}}}\\&={\frac {p(1-p)}{n}}\\&={\frac {-p^{2}+p}{n}}\end{aligned}}}
. Wir suchen das globale Maximum bezüglich
p
{\displaystyle p}
[
0
,
1
]
{\displaystyle [0,1]}
0
=
∂
V
a
r
(
K
/
n
)
∂
p
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p
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0
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p
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1
−
1
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p
p
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2
{\displaystyle {\begin{aligned}0&={\frac {\partial Var(K/n)}{\partial p}}={\frac {-2p+1}{n}}\\0&=-2p+1\\-1&=-2p\\p&={\frac {1}{2}}\end{aligned}}}
. Bei
p
^
:=
1
2
{\displaystyle {\hat {p}}:={\frac {1}{2}}}
∂
2
V
a
r
(
K
/
n
)
∂
2
p
=
−
2
>
0
{\displaystyle {\frac {\partial ^{2}Var(K/n)}{\partial ^{2}p}}=-2>0}
an der Stelle
p
^
{\displaystyle {\hat {p}}}
p
=
0
{\displaystyle p=0}
p
=
1
{\displaystyle p=1}
p
^
{\displaystyle {\hat {p}}}
V
a
r
(
K
/
n
)
(
1
2
)
=
1
4
n
{\displaystyle Var(K/n)\left({\frac {1}{2}}\right)={\frac {1}{4n}}}
Das Intervall
[
0
,
1
]
{\displaystyle [0,1]}
f
{\displaystyle f}
p
{\displaystyle p}
f
{\displaystyle f}
f
{\displaystyle f}
Also ist
f
{\displaystyle f}
p
{\displaystyle p}
Daraus folgt für alle
ε
>
0
{\displaystyle \varepsilon >0}
p
{\displaystyle p}
gleichmäßigen Gesetz der großen Zahl , ), also
{\displaystyle }
lim
n
→
∞
P
(
|
f
(
K
n
)
−
f
(
x
)
|
>
ε
)
=
0
{\displaystyle \lim _{n\to \infty }{P\left(\left|f\left({\frac {K}{n}}\right)-f\left(x\right)\right|>\varepsilon \right)}=0}
(siehe auch hier ????????) Aus der Beschränktheit von
f
{\displaystyle f}
x
{\displaystyle x}
lim
n
→
∞
E
[
|
f
(
K
n
)
−
f
(
x
)
|
]
=
0
{\displaystyle \lim _{n\to \infty }{E\left[\left|f\left({\frac {K}{n}}\right)-f\left(x\right)\right|\right]}=0}
Für alle Funktionen
f
{\displaystyle f}
n
{\displaystyle n}
f
(
x
)
=
∑
k
=
0
n
f
(
x
)
(
n
k
)
p
k
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1
−
p
n
−
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)
{\displaystyle f(x)=\sum _{k=0}^{n}f(x){n \choose k}p^{k}(1-p^{n-k})}
f
(
x
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=
f
(
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×
1
n
=
f
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x
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×
(
p
+
1
−
p
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n
=
f
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x
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×
∑
k
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0
n
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n
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p
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1
−
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n
−
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=
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k
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f
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{\displaystyle {\begin{aligned}f(x)&=f(x)\times 1^{n}\\&=f(x)\times (p+1-p)^{n}\\&=f(x)\times \sum _{k=0}^{n}{n \choose k}p^{k}(1-p)^{n-k}\\&=\sum _{k=0}^{n}f(x){n \choose k}p^{k}(1-p)^{n-k}\end{aligned}}}
aufgrund des Binomischen Lehrsatzes.
Gemäß dem Lemma 2 gilt
|
f
(
K
/
n
)
−
f
(
p
)
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∑
k
=
0
n
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f
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−
f
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p
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n
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{\displaystyle |f(K/n)-f(p)|=\sum _{k=0}^{n}|f(K/n)-f(p)|{n \choose k}p^{k}(1-p)^{n-k}}
ε
>
0
{\displaystyle \varepsilon >0}
f
{\displaystyle f}
δ
>
0
{\displaystyle \delta >0}
∀
x
,
y
∈
[
a
,
b
]
:
|
x
−
y
|
<
δ
⟹
|
f
(
x
)
−
f
(
y
)
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<
ε
/
2.
{\displaystyle \forall x,y\in [a,b]:|x-y|<\delta \implies |f(x)-f(y)|<\varepsilon /2.}
Zerlege die Summe in zwei Teile:
einen Teil
A
{\displaystyle A}
k
{\displaystyle k}
|
k
/
n
−
x
|
<
δ
{\displaystyle |k/n-x|<\delta }
einen Teil
B
{\displaystyle B}
k
{\displaystyle k}
Wegen der Stetigkeit von
f
{\displaystyle f}
A
{\displaystyle A}
|
f
(
K
(
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/
n
)
−
f
(
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)
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<
ε
/
2
{\displaystyle |f(K(x)/n)-f(x)|<\varepsilon /2}
B
{\displaystyle B}
|
f
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K
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x
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/
n
)
−
f
(
x
)
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<
M
+
M
=
2
M
{\displaystyle |f(K(x)/n)-f(x)|<M+M=2M}
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
E
[
|
f
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K
/
n
)
−
f
(
x
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]
=
E
[
∑
k
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0
n
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f
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/
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−
f
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|
(
n
k
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p
k
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n
−
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≤
E
[
(
1
k
wie in
A
)
×
ε
/
2
]
+
E
[
(
1
k
wie in
B
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×
2
M
]
=
P
(
k
wie in
A
)
×
ε
/
2
+
P
(
k
wie in
B
)
×
2
M
≤
1
×
ε
/
2
+
2
M
ε
4
n
=
ε
{\displaystyle {\begin{aligned}\mathbb {E} \left[|f(K/n)-f(x)|\right]&=\mathbb {E} \left[\sum _{k=0}^{n}|f(K/n)-f(x)|{n \choose k}p^{k}(1-p)^{n-k}\right]\\&\leq \mathbb {E} \left[(\mathbf {1} _{k{\text{ wie in }}A})\times \varepsilon /2\right]+\mathbb {E} \left[(\mathbf {1} _{k{\text{ wie in }}B})\times 2M\right]\\&=P(k{\text{ wie in }}A)\times \varepsilon /2+P(k{\text{ wie in }}B)\times 2M\\&\leq 1\times \varepsilon /2+2M{\frac {\varepsilon }{4n}}\\&=\varepsilon \end{aligned}}}
. Mit der Dreiecksgleichung des Erwartungswertes und seiner Linearität folgt für ein beliebiges, fixes
x
{\displaystyle x}
E
[
|
f
(
K
/
n
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−
f
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x
)
|
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≥
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E
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f
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{\displaystyle {\begin{aligned}\mathbb {E} \left[|f(K/n)-f(x)|\right]&\geq |\mathbb {E} \left[f(K/n)-f(x)\right]|\\&=|\mathbb {E} \left[f(K/n)\right]-\mathbb {E} \left[f(x)\right]|\\&=|\mathbb {E} \left[f(K/n)\right]-f(x)|\end{aligned}}}
. Definiere die Bernstein-Polynome durch
B
n
(
f
)
(
x
)
:=
∑
ν
=
0
n
f
(
ν
n
)
b
ν
,
n
(
x
)
{\displaystyle {\begin{aligned}B_{n}(f)(x):=\sum _{\nu =0}^{n}f\left({\frac {\nu }{n}}\right)b_{\nu ,n}(x)\end{aligned}}}
mit
b
ν
,
n
(
x
)
=
(
n
ν
)
x
ν
(
1
−
x
)
n
−
ν
,
ν
=
0
,
…
,
n
.
{\displaystyle b_{\nu ,n}(x)={n \choose \nu }x^{\nu }\left(1-x\right)^{n-\nu },\quad \nu =0,\ldots ,n.}
Dann genügt es, Lemma 3 zu zeigen, denn dann ist zusammengefasst (mit
P
:=
B
n
(
f
)
{\displaystyle \mathbf {P} :=B_{n}(f)}
|
B
n
(
f
)
(
x
)
−
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)
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=
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E
[
f
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/
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−
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≤
E
[
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f
(
K
/
n
)
−
f
(
x
)
|
]
≤
ε
.
{\displaystyle {\begin{aligned}|B_{n}(f)(x)-f(x)|&=|\mathbb {E} [f(K/n)]-f(x)|\\&\leq \mathbb {E} [|f(K/n)-f(x)|]\\&\leq \varepsilon .\end{aligned}}}
E
[
f
(
K
/
n
)
]
=
B
n
(
f
)
(
x
)
{\displaystyle \mathbb {E} [f(K/n)]=B_{n}(f)(x)}
Es folgt schrittweise aus dem Gesetz des bewusstlosen Statistikers (»law of unconscious statistician«), der Berechnung der Wahrscheinlichkeitsfunktion und dem Einsetzen der Wahrscheinlichkeitsfunktion der Binomialverteilung das Ergebnis.
E
[
f
(
K
/
n
)
]
=
∑
ν
=
0
n
f
(
ν
/
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)
f
K
/
n
(
ν
/
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=
∑
ν
=
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f
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f
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−
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{\displaystyle {\begin{aligned}\mathbb {E} [f(K/n)]&=\sum _{\nu =0}^{n}f(\nu /n)f_{K/n}(\nu /n)\\&=\sum _{\nu =0}^{n}f(\nu /n)f_{K}(n\nu /n)\left|{\frac {dn\nu /n}{d\nu }}\right|\\&=\sum _{\nu =0}^{n}f(\nu /n)f_{K}(\nu )\left|{\frac {d\nu }{d\nu }}\right|\\&=\sum _{\nu =0}^{n}f(\nu )f_{K}(\nu )\left|1\right|\\&=\sum _{\nu =0}^{n}f(\nu )f_{K}(\nu )\\&=\sum _{\nu =0}^{n}f\left({\frac {\nu }{n}}\right){n \choose \nu }x^{\nu }\left(1-x\right)^{n-\nu }\end{aligned}}}
![{\displaystyle f:[a,b]\mapsto \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4932ba702175b9e0bc40b79efbc051c7ff44949)
![{\displaystyle \forall x\in [a,b]:|f(x)-\mathbf {P} (x)|<\epsilon }](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb6df93254626ebe5297c7fa8f5db64ab5e5c470)
![{\displaystyle \mathbb {E} \left[{\frac {K}{n}}\right]=p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5a43631c2ed2752c20db5da762bfa76bd81e2ce)
![{\displaystyle \mathbb {E} [K/n]=p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f350ade53540b38f5177e6083cd1ed31d106984f)
![{\displaystyle [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
![{\displaystyle \lim _{n\to \infty }{E\left[\left|f\left({\frac {K}{n}}\right)-f\left(x\right)\right|\right]}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/084e2a08de6799586bfe82d126b7af5ed1f06017)
![{\displaystyle \forall x,y\in [a,b]:|x-y|<\delta \implies |f(x)-f(y)|<\varepsilon /2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6c28b572675473d6537efc31625987741243df7)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle {\begin{aligned}\mathbb {E} \left[|f(K/n)-f(x)|\right]&=\mathbb {E} \left[\sum _{k=0}^{n}|f(K/n)-f(x)|{n \choose k}p^{k}(1-p)^{n-k}\right]\\&\leq \mathbb {E} \left[(\mathbf {1} _{k{\text{ wie in }}A})\times \varepsilon /2\right]+\mathbb {E} \left[(\mathbf {1} _{k{\text{ wie in }}B})\times 2M\right]\\&=P(k{\text{ wie in }}A)\times \varepsilon /2+P(k{\text{ wie in }}B)\times 2M\\&\leq 1\times \varepsilon /2+2M{\frac {\varepsilon }{4n}}\\&=\varepsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bf77c8ad23ff63baa8efe26a890279fe87b20a9)
![{\displaystyle {\begin{aligned}\mathbb {E} \left[|f(K/n)-f(x)|\right]&\geq |\mathbb {E} \left[f(K/n)-f(x)\right]|\\&=|\mathbb {E} \left[f(K/n)\right]-\mathbb {E} \left[f(x)\right]|\\&=|\mathbb {E} \left[f(K/n)\right]-f(x)|\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/774875646507c2bd13e56f610f0094db65918577)
![{\displaystyle {\begin{aligned}|B_{n}(f)(x)-f(x)|&=|\mathbb {E} [f(K/n)]-f(x)|\\&\leq \mathbb {E} [|f(K/n)-f(x)|]\\&\leq \varepsilon .\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/efec34e94ac254a2d1b6012abd5531a01182483e)
![{\displaystyle \mathbb {E} [f(K/n)]=B_{n}(f)(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7727b697ef4e1583b8dde9b6760c06d1ad9bf9b5)
![{\displaystyle {\begin{aligned}\mathbb {E} [f(K/n)]&=\sum _{\nu =0}^{n}f(\nu /n)f_{K/n}(\nu /n)\\&=\sum _{\nu =0}^{n}f(\nu /n)f_{K}(n\nu /n)\left|{\frac {dn\nu /n}{d\nu }}\right|\\&=\sum _{\nu =0}^{n}f(\nu /n)f_{K}(\nu )\left|{\frac {d\nu }{d\nu }}\right|\\&=\sum _{\nu =0}^{n}f(\nu )f_{K}(\nu )\left|1\right|\\&=\sum _{\nu =0}^{n}f(\nu )f_{K}(\nu )\\&=\sum _{\nu =0}^{n}f\left({\frac {\nu }{n}}\right){n \choose \nu }x^{\nu }\left(1-x\right)^{n-\nu }\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ba3b894c1f8b56b743a60ac84c7c852fb635534)