Satz von Moivre-Laplace
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Sei X 1 , X 2 , X 3 , ⋯ {\displaystyle X_{1},X_{2},X_{3},\cdots } bernoulli-verteilter Zufallsvariablen und deren Summe S n {\displaystyle S_{n}} binomialverteilt mit Parametern n ∈ N {\displaystyle n\in \mathbb {N} } p ∈ ] 0 , 1 [ {\displaystyle p\in ]0,1[} σ 2 = p ( 1 − p ) {\displaystyle \sigma ^{2}=p(1-p)}
(1) P ( S n = k ) = B ( k ∣ p , n ) ≈ 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) {\displaystyle \quad \operatorname {P} \left(S_{n}=k\right)\;=\;B(k\mid p,n)\;\approx \;{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)}
lim n → ∞ P ( x 1 ≤ S n − n p n ≤ x 2 ) = 1 2 π σ 2 ∫ x 1 x 2 exp ( − x 2 2 σ 2 ) d x {\displaystyle \quad \lim _{n\to \infty }\operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x} x 1 , x 2 ∈ R {\displaystyle x_{1},x_{2}\in \mathbb {R} } x 1 < x 2 {\displaystyle x_{1}<x_{2}}
Nach Gauß lässt sich die Gammafunktion Γ ( x ) {\displaystyle \Gamma (x)} x ∈ C ∖ { 0 , − 1 , − 2 , … } {\displaystyle x\in \mathbb {C} \setminus \{0,-1,-2,\dotsc \}} n ∈ N 0 {\displaystyle n\in \mathbb {N} _{0}}
Γ ( x ) = lim n → ∞ n ! n x x ( x + 1 ) ( x + 2 ) ⋯ ( x + n ) {\displaystyle \Gamma (x)=\lim _{n\to \infty }{\frac {n!\,n^{x}}{x(x+1)(x+2)\dotsm (x+n)}}} Bemerkungen:
Es gilt n ! = Γ ( n + 1 ) = n Γ ( n ) {\displaystyle n!=\Gamma (n+1)=n\Gamma (n)}
Die Stirlingformel lautet n ! = lim n → ∞ ( 2 π n ( n e ) n + ϵ n ) {\displaystyle n!=\lim _{n\to \infty }\left({\sqrt {2\pi n}}\;\left({\tfrac {n}{\mathrm {e} }}\right)^{n}+\epsilon _{n}\right)}
Nachfolgend wird das Näherungszeichen "≈ {\displaystyle \approx } = {\displaystyle =} Satz (Stirling-Formel und Gammafunktion)
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In der Halbebene Re ( x ) > 0 {\displaystyle \operatorname {Re} (x)>0}
log Γ ( x ) = ( x − 1 2 ) log x − x + log 2 π + O ( 1 | x | ) {\displaystyle \log \Gamma (x)=(x-{\tfrac {1}{2}})\log x-x+\log {\sqrt {2\pi }}+{\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}} Dabei ist log x {\displaystyle \log x} x {\displaystyle x} log Γ ( x ) {\displaystyle \log \Gamma (x)} x > 0 {\displaystyle x>0}
Nach Gauß ist
Γ ( x ) = lim n → ∞ n ! n x x ( x + 1 ) ( x + 2 ) ⋯ ( x + n ) {\displaystyle \Gamma (x)=\lim _{n\to \infty }{\frac {n!\,n^{x}}{x(x+1)(x+2)\dotsm (x+n)}}} also
log Γ ( x ) = lim n → ∞ ( log n ! + x log n − ∑ k = 0 n log ( x + k ) ) {\displaystyle \log \Gamma (x)=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-\sum _{k=0}^{n}\log(x+k){\bigg )}} x {\displaystyle x} d 2 f ( t ) d t 2 | t = k = d 2 log ( x + t ) d t 2 | t = k = 1 ( x + t ) 2 {\displaystyle {\bigg .}{\tfrac {\mathrm {d} ^{2}f(t)}{\mathrm {d} t^{2}}}{\bigg |}_{t=k}={\bigg .}{\tfrac {\mathrm {d} ^{2}\log(x+t)}{\mathrm {d} t^{2}}}{\bigg |}_{t=k}={\tfrac {1}{(x+t)^{2}}}} ∑ k = 0 n log ( x + k ) = ∫ 0 n log ( x + t ) d t + 1 2 ( log x + log ( x + n ) ) + ∫ 0 n φ ( t ) 1 ( x + t ) 2 d t {\displaystyle \sum _{k=0}^{n}\log(x+k)=\int _{0}^{n}\log(x+t)\mathrm {d} t+{\frac {1}{2}}{\bigg (}\log x+\log(x+n){\bigg )}\;+\;\int _{0}^{n}\varphi (t){\tfrac {1}{(x+t)^{2}}}\mathrm {d} t}
log Γ ( x ) = lim n → ∞ ( log n ! + x log n − ∫ 0 n log ( x + t ) d t − 1 2 ( log x + log ( x + n ) ) − ∫ 0 n φ ( t ) 1 ( x + t ) 2 d t ) {\displaystyle \log \Gamma (x)=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-\int \limits _{0}^{n}\log(x+t)\mathrm {d} t-{\frac {1}{2}}{\Big (}\log x+\log(x+n){\Big )}-\int \limits _{0}^{n}\varphi (t){\frac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg )}} | φ ( t ) | ≤ 1 8 {\displaystyle {\big |}\varphi (t){\big |}\leq {\tfrac {1}{8}}} | ∫ 0 n φ ( t ) 1 ( x + t ) 2 d t | ≤ 1 8 | ∫ 0 n 1 ( x + t ) 2 d t | {\displaystyle {\bigg |}\int _{0}^{n}\varphi (t){\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}\leq {\tfrac {1}{8}}{\bigg |}\int _{0}^{n}{\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}} 1 8 | ∫ 0 n 1 ( x + t ) 2 d t | = 1 8 | [ − 1 x + t ] 0 n | = 1 8 | − 1 x + n − − 1 x | = 1 8 | − x − n + x − x ( x + n ) | = 1 8 | 1 x ( x n + 1 ) | ≤ 1 8 1 | x | ≈ O ( 1 | x | ) {\displaystyle {\tfrac {1}{8}}{\bigg |}\int _{0}^{n}{\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}={\tfrac {1}{8}}{\bigg |}{\bigg [}{\tfrac {-1}{x+t}}{\bigg ]}_{0}^{n}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-1}{x+n}}-{\tfrac {-1}{x}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-x-n+x}{-x(x+n)}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {1}{x({\tfrac {x}{n}}+1)}}{\bigg |}\leq {\tfrac {1}{8}}{\tfrac {1}{|x|}}\approx {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}} | ∫ 0 n φ ( t ) 1 ( x + t ) 2 d t | ≈ O ( 1 | x | ) {\displaystyle {\bigg |}\int _{0}^{n}\varphi (t){\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}\approx {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}} O ( 1 | x | ) {\displaystyle {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}
log Γ ( x ) = lim n → ∞ ( log n ! + x log n − ∫ 0 n log ( x + t ) d t − 1 2 ( log x + log ( x + n ) ) ) {\displaystyle \log \Gamma (x)=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-\int \limits _{0}^{n}\log(x+t)\mathrm {d} t-{\frac {1}{2}}{\Big (}\log x+\log(x+n){\Big )}{\bigg )}} ∫ 0 n log ( x + t ) d t = [ ( x + t ) log ( x + t ) − 1 ] 0 n = ( x + n ) log ( x + n ) − x log x + C {\displaystyle \int _{0}^{n}\log(x+t)\mathrm {d} t={\Big [}(x+t)\log(x+t)-1{\Big ]}_{0}^{n}=(x+n)\log(x+n)-x\log x+C} o.B.d.A. gewählt werden kann, sei C = − n {\displaystyle C=-n}
log Γ ( x ) = lim n → ∞ ( log n ! + x log n − ( x + n ) log ( x + n ) + x log x + n − 1 2 ( log x + log ( x + n ) ) ) = lim n → ∞ ( log n ! + x log n − ( x + n + 1 2 ) log ( x + n ) + ( x − 1 2 ) log x + n ) {\displaystyle {\begin{aligned}\log \Gamma (x)&=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-{\Big (}x+n{\Big )}\log(x+n)+x\log x+n-{\frac {1}{2}}{\Big (}\log x+\log(x+n){\Big )}{\bigg )}\\&=\lim _{n\to \infty }{\bigg (}\log n!+x\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}\log(x+n)+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+n{\bigg )}\end{aligned}}} n ! = lim n → ∞ ( 2 π n ( n e ) n + ϵ n ) {\displaystyle n!=\lim _{n\to \infty }\left({\sqrt {2\pi n}}\;\left({\tfrac {n}{\mathrm {e} }}\right)^{n}+\epsilon _{n}\right)} log e = 1 {\displaystyle \log e=1} log ϵ n = ϵ n ′ {\displaystyle \log \epsilon _{n}=\epsilon _{n}'} log n ! = lim n → ∞ ( ( n + 1 2 ) log n − n + log 2 π + ϵ n ′ ) {\displaystyle \log n!=\lim _{n\to \infty }\left((n+{\tfrac {1}{2}})\log n-n+\log {\sqrt {2\pi }}+\epsilon _{n}'\right)}
log Γ ( x ) = lim n → ∞ ( ( n + 1 2 ) log n − n + x log n − ( x + n + 1 2 ) log ( x + n ) + ( x − 1 2 ) log x + n + ϵ n ′ ) + log 2 π = lim n → ∞ ( ( x + n + 1 2 ) log n − ( x + n + 1 2 ) log ( x + n ) + ϵ n ′ ) + ( x − 1 2 ) log x + log 2 π {\displaystyle {\begin{aligned}\log \Gamma (x)&=\lim _{n\to \infty }{\bigg (}{\Big (}n+{\frac {1}{2}}{\Big )}\log n-n+x\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}\log(x+n)+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+n+\epsilon _{n}'{\bigg )}+\log {\sqrt {2\pi }}\\&=\lim _{n\to \infty }{\bigg (}{\Big (}x+n+{\frac {1}{2}}{\Big )}\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}\log(x+n)+\epsilon _{n}'{\bigg )}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+\log {\sqrt {2\pi }}\end{aligned}}}
Nun ist log ( x + n ) = log ( n ( 1 + x n ) ) = log n + log ( 1 + x n ) {\displaystyle \log(x+n)=\log(n(1+{\tfrac {x}{n}}))=\log n+\log(1+{\tfrac {x}{n}})} x {\displaystyle x} n {\displaystyle n} log ( 1 + x n ) ≈ x n + O ( 1 n 2 ) {\displaystyle \log(1+{\tfrac {x}{n}})\approx {\tfrac {x}{n}}+{\mathcal {O}}{\Big (}{\tfrac {1}{n^{2}}}{\Big )}} O ( 1 n 2 ) {\displaystyle {\mathcal {O}}{\Big (}{\tfrac {1}{n^{2}}}{\Big )}} log ( x + n ) = log n + x n {\displaystyle \log(x+n)=\log n+{\tfrac {x}{n}}}
log Γ ( x ) = log 2 π + ( x − 1 2 ) log x + lim n → ∞ ( ( x + n + 1 2 ) log n − ( x + n + 1 2 ) ( log n + x n ) + ϵ n ′ ) = log 2 π + ( x − 1 2 ) log x + lim n → ∞ ( − x n ( x + n + 1 2 ) + ϵ n ′ ) = log 2 π + ( x − 1 2 ) log x − x {\displaystyle {\begin{aligned}\log \Gamma (x)&=\log {\sqrt {2\pi }}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+\lim _{n\to \infty }{\bigg (}{\Big (}x+n+{\frac {1}{2}}{\Big )}\log n-{\Big (}x+n+{\frac {1}{2}}{\Big )}{\Big (}\log n+{\frac {x}{n}}{\Big )}+\epsilon _{n}'{\bigg )}\\&=\log {\sqrt {2\pi }}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x+\lim _{n\to \infty }{\bigg (}-{\frac {x}{n}}(x+n+{\frac {1}{2}})+\epsilon _{n}'{\bigg )}\\&=\log {\sqrt {2\pi }}+{\Big (}x-{\frac {1}{2}}{\Big )}\log x-x\end{aligned}}} ◻ {\displaystyle \Box } An dieser Stelle sei erwähnt, dass diese Näherung für x ≥ 9 {\displaystyle x\geq 9} 1 % {\displaystyle 1\%}
Für die weitere Beweisführung seien folgende Umformungen angegeben:
Γ ( x ) ≈ 2 π x ( x − 1 2 ) e − x {\displaystyle \Gamma (x)\approx {\sqrt {2\pi }}x^{{\Big (}x-{\frac {1}{2}}{\Big )}}e^{-x}} n ! = Γ ( n + 1 ) = n Γ ( n ) {\displaystyle n!=\Gamma (n+1)=n\Gamma (n)}
n ! = n Γ ( n ) ≈ n 2 π n ( n − 1 2 ) e − n = 2 π n ( n e ) n = n n e − n 2 π n {\displaystyle n!=n\Gamma (n)\approx n{\sqrt {2\pi }}n^{{\Big (}n-{\tfrac {1}{2}}{\Big )}}e^{-n}={\sqrt {2\pi n}}{\bigg (}{\frac {n}{e}}{\bigg )}^{n}=n^{n}e^{-n}{\sqrt {2\pi n}}}
Der Beweis wird in zwei Schritten durchgeführt.
Zunächst wird gezeigt:
B ( k ∣ p , n ) ≈ n n e − n 2 π n k k e − k 2 π k ( n − k ) n − k e − ( n − k ) 2 π ( n − k ) p k ( 1 − p ) n − k {\displaystyle B(k\mid p,n)\approx {\frac {n^{n}e^{-n}{\sqrt {2\pi n}}}{k^{k}e^{-k}{\sqrt {2\pi k}}(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}}}p^{k}(1-p)^{n-k}} Dazu werden mit Hilfe der Stirlingformel die Fakultäten ersetzt, also folgende Näherungen vorgenommen:
n ! ≈ 2 π n ( n e ) n = n n e − n 2 π n {\displaystyle n!\approx {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}=n^{n}e^{-n}{\sqrt {2\pi n}}} k ! ≈ 2 π k ( k e ) k = k k e − k 2 π k {\displaystyle k!\approx {\sqrt {2\pi k}}\left({\frac {k}{e}}\right)^{k}=k^{k}e^{-k}{\sqrt {2\pi k}}} ( n − k ) ! ≈ 2 π ( n − k ) ( n − k e ) n − k = ( n − k ) n − k e − ( n − k ) 2 π ( n − k ) {\displaystyle (n-k)!\approx {\sqrt {2\pi (n-k)}}\left({\frac {n-k}{e}}\right)^{n-k}=(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}} Damit lässt sich die Binomialverteilung wie folgt ausdrücken:
B ( k ∣ p , n ) = ( n k ) p k ( 1 − p ) n − k = n ! k ! ( n − k ) ! p k ( 1 − p ) n − k ≈ n n e − n 2 π n k k e − k 2 π k ( n − k ) n − k e − ( n − k ) 2 π ( n − k ) p k ( 1 − p ) n − k = n 2 π k ( n − k ) n n k k ( n − k ) n − k p k ( 1 − p ) n − k = n 2 π k ( n − k ) ( n p k ) k ( n ( 1 − p ) n − k ) n − k = 1 2 π n k n ( 1 − k n ) ( n p k ) k ( n ( 1 − p ) n − k ) n − k {\displaystyle {\begin{aligned}B(k\mid p,n)&={n \choose k}p^{k}(1-p)^{n-k}\\&={\frac {n!}{k!(n-k)!}}p^{k}(1-p)^{n-k}\\&\approx {\frac {n^{n}e^{-n}{\sqrt {2\pi n}}}{k^{k}e^{-k}{\sqrt {2\pi k}}(n-k)^{n-k}e^{-(n-k)}{\sqrt {2\pi (n-k)}}}}p^{k}(1-p)^{n-k}\\&={\sqrt {\frac {n}{2\pi k\left(n-k\right)}}}{\frac {n^{n}}{k^{k}\left(n-k\right)^{n-k}}}p^{k}(1-p)^{n-k}\\&={\sqrt {\frac {n}{2\pi k\left(n-k\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\\&={\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\end{aligned}}} Mit der Näherung k n ≈ p {\displaystyle {\frac {k}{n}}\approx p} Taylorapproximation wird gezeigt:
B ( k ∣ p , n ) ≈ 1 2 π n k n ( 1 − k n ) ( n p k ) k ( n ( 1 − p ) n − k ) n − k ≈ 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) {\displaystyle B(k\mid p,n)\approx {\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\approx {\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)} n ≫ k {\displaystyle n\gg k} k n ≈ p {\displaystyle {\frac {k}{n}}\approx p} 1 2 π n k n ( 1 − k n ) ≈ 1 2 π n p ( 1 − p ) = 1 2 π n σ 2 {\displaystyle {\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\approx {\frac {1}{\sqrt {2\pi np(1-p)}}}={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}}
B ( k ∣ p , n ) ≈ 1 2 π n σ 2 ( n p k ) k ( n ( 1 − p ) n − k ) n − k = 1 2 π n σ 2 ( p k / n ) k ( ( 1 − p ) 1 − k / n ) n − k = 1 2 π n σ 2 exp ( n n log ( ( p k / n ) k ( 1 − p 1 − k / n ) n − k ) ) = 1 2 π n σ 2 exp ( n n log ( ( k / n p ) − k ( 1 − k / n 1 − p ) − ( n − k ) ) ) = 1 2 π n σ 2 exp ( n ( − k n log ( k / n p ) ⏟ f ( x ) − ( 1 − k n ) log ( 1 − k / n 1 − p ) ⏟ g ( x ) ) ) {\displaystyle {\begin{aligned}B(k\mid p,n)&\approx {\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\left({\frac {p}{k/n}}\right)^{k}\left({\frac {(1-p)}{1-k/n}}\right)^{n-k}\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left({\frac {n}{n}}\log \left(\left({\frac {p}{k/n}}\right)^{k}\left({\frac {1-p}{1-k/n}}\right)^{n-k}\right)\right)\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left({\frac {n}{n}}\log \left(\left({\frac {k/n}{p}}\right)^{-k}\left({\frac {1-k/n}{1-p}}\right)^{-(n-k)}\right)\right)\\&={\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp {\Bigg (}n{\bigg (}-\underbrace {{\frac {k}{n}}\log \left({\frac {k/n}{p}}\right)} _{f(x)}-\underbrace {\left(1-{\frac {k}{n}}\right)\log \left({\frac {1-k/n}{1-p}}\right)} _{g(x)}{\bigg )}{\Bigg )}\end{aligned}}} Taylorapproximation in der Annäherung durch die Schmiegparabel . Wir erhalten mit x = k n {\displaystyle x={\frac {k}{n}}} f ( x ) = x ln x p {\displaystyle f(x)=x\ln {\frac {x}{p}}} g ( x ) = ( 1 − x ) ln 1 − x 1 − p {\displaystyle g(x)=(1-x)\ln {\frac {1-x}{1-p}}} x = p {\displaystyle x=p}
T 2 f ( x , p ) = ( x ln x p ) | x = p + d d x ( x ln x p ) | x = p ⋅ ( x − p ) + 1 2 d 2 d x 2 ( x ln x p ) | x = p ⋅ ( x − p ) 2 = ( ln x p + x ⋅ p x ⋅ 1 p ) | x = p ⋅ ( x − p ) + 1 2 d d x ( ln x p + 1 ) | x = p ⋅ ( x − p ) 2 = x − p + 1 2 ( p x ⋅ 1 p ) | x = p ⋅ ( x − p ) 2 = x − p + 1 2 p ⋅ ( x − p ) 2 {\displaystyle {\begin{aligned}T_{2}f(x,p)&=\left.\left(x\ln {\frac {x}{p}}\right)\right|_{x=p}+\left.{\frac {\mathrm {d} }{\mathrm {d} x}}\left(x\ln {\frac {x}{p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left(x\ln {\frac {x}{p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=\left.\left(\ln {\frac {x}{p}}+x\cdot {\frac {p}{x}}\cdot {\frac {1}{p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(\ln {\frac {x}{p}}+1\right)\right|_{x=p}\cdot (x-p)^{2}\\&=x-p+\left.{\frac {1}{2}}\left({\frac {p}{x}}\cdot {\frac {1}{p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=x-p+{\frac {1}{2p}}\cdot (x-p)^{2}\end{aligned}}}
und
T 2 g ( x , p ) = ( ( 1 − x ) ln 1 − x 1 − p ) | x = p + d d x ( ( 1 − x ) ln 1 − x 1 − p ) | x = p ⋅ ( x − p ) + 1 2 d 2 d x 2 ( ( 1 − x ) ln 1 − x 1 − p ) | x = p ⋅ ( x − p ) 2 = ( − ln 1 − x 1 − p + ( 1 − x ) ⋅ 1 − p 1 − x ⋅ − 1 1 − p ) | x = p ⋅ ( x − p ) + 1 2 d d x ( − ln 1 − x 1 − p − 1 ) | x = p ⋅ ( x − p ) 2 = − ( x − p ) − 1 2 ( 1 − p 1 − x ⋅ − 1 1 − p ) | x = p ⋅ ( x − p ) 2 = − ( x − p ) + 1 2 ( 1 − p ) ⋅ ( x − p ) 2 {\displaystyle {\begin{aligned}T_{2}g(x,p)&=\left.\left((1-x)\ln {\frac {1-x}{1-p}}\right)\right|_{x=p}+\left.{\frac {\mathrm {d} }{\mathrm {d} x}}\left((1-x)\ln {\frac {1-x}{1-p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} ^{2}}{\mathrm {d} x^{2}}}\left((1-x)\ln {\frac {1-x}{1-p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=\left.\left(-\ln {\frac {1-x}{1-p}}+(1-x)\cdot {\frac {1-p}{1-x}}\cdot {\frac {-1}{1-p}}\right)\right|_{x=p}\cdot (x-p)+\left.{\frac {1}{2}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left(-\ln {\frac {1-x}{1-p}}-1\right)\right|_{x=p}\cdot (x-p)^{2}\\&=-(x-p)-\left.{\frac {1}{2}}\left({\frac {1-p}{1-x}}\cdot {\frac {-1}{1-p}}\right)\right|_{x=p}\cdot (x-p)^{2}\\&=-(x-p)+{\frac {1}{2(1-p)}}\cdot (x-p)^{2}\end{aligned}}}
Bemerkungen: zu beachten ist d d x ln h ( x ) = h ′ ( x ) h ( x ) {\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\ln h(x)={\frac {h'(x)}{h(x)}}} Integralrestglied R 2 f ( x , p ) {\displaystyle R_{2}f(x,p)} R 2 g ( x , p ) {\displaystyle R_{2}g(x,p)}
Die Zusammenfassung beider Taylorapproximationen liefert dann mit σ 2 = p ( 1 − p ) {\displaystyle \sigma ^{2}=p(1-p)} T 2 f ( x , p ) + T 2 g ( x , p ) = x − p + 1 2 p ⋅ ( x − p ) 2 − ( x − p ) + 1 2 ( 1 − p ) ⋅ ( x − p ) 2 = 1 2 p ⋅ ( x − p ) 2 + 1 2 ( 1 − p ) ⋅ ( x − p ) 2 = 1 − p + p 2 p ( 1 − p ) ⋅ ( x − p ) 2 = 1 2 p ( 1 − p ) ⋅ ( x − p ) 2 = 1 2 σ 2 ⋅ ( x − p ) 2 {\displaystyle {\begin{aligned}T_{2}f(x,p)+T_{2}g(x,p)&=x-p+{\frac {1}{2p}}\cdot (x-p)^{2}-(x-p)+{\frac {1}{2(1-p)}}\cdot (x-p)^{2}\\&={\frac {1}{2p}}\cdot (x-p)^{2}+{\frac {1}{2(1-p)}}\cdot (x-p)^{2}\\&={\frac {1-p+p}{2p(1-p)}}\cdot (x-p)^{2}\\&={\frac {1}{2p(1-p)}}\cdot (x-p)^{2}\\&={\frac {1}{2\sigma ^{2}}}\cdot (x-p)^{2}\end{aligned}}}
B ( k ∣ p , n ) ≈ 1 2 π n k n ( 1 − k n ) ( n p k ) k ( n ( 1 − p ) n − k ) n − k ≈ 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) {\displaystyle B(k\mid p,n)\approx {\frac {1}{\sqrt {2\pi n{\frac {k}{n}}\left(1-{\frac {k}{n}}\right)}}}\left({\frac {np}{k}}\right)^{k}\left({\frac {n(1-p)}{n-k}}\right)^{n-k}\;\approx \;{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)} ◻ {\displaystyle \Box }
Gezeigt wird, dass aus dem lokalen Grenzwertsatz im Limes n → ∞ {\displaystyle n\rightarrow \infty } Riemann-Summe folgt:
P ( x 1 ≤ S n − n p n ≤ x 2 ) = 1 2 π σ 2 ∫ x 1 x 2 exp ( − x 2 2 σ 2 ) d x {\displaystyle \operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;={\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x} x 1 , x 2 ∈ R {\displaystyle x_{1},x_{2}\in \mathbb {R} } x 1 < x 2 {\displaystyle x_{1}<x_{2}} Zunächst folgt mit dem lokalen Grenzwertsatz:
P ( x 1 ≤ S n − n p n ≤ x 2 ) = ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 P ( S n = k ) = ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) ⋅ ( 1 + ϵ n , p ( k ) ) {\displaystyle \operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;=\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}\operatorname {P} \left(S_{n}=k\right)\;=\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\cdot (1+\epsilon _{n,p}(k))} wobei lim n → ∞ ϵ n , p ( k ) = 0 {\displaystyle \lim _{n\to \infty }\epsilon _{n,p}(k)=\;0}
Setzen wir nun ϵ ¯ n , p := sup k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 | ϵ n , p ( k ) | {\displaystyle {\bar {\epsilon }}_{n,p}:=\sup _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}|\epsilon _{n,p}(k)|}
∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) ⋅ ϵ n , p ( k ) ) ≤ ϵ ¯ n , p ⋅ ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) {\displaystyle \sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\cdot \epsilon _{n,p}(k))\;\leq \;{\bar {\epsilon }}_{n,p}\cdot \sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)} und im Limes n → ∞ {\displaystyle n\to \infty }
lim n → ∞ ϵ n , p ⏟ → 0 ⋅ ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) ⏟ → ∫ a b … < ∞ = 0 {\displaystyle \lim _{n\to \infty }\;\underbrace {\quad \epsilon _{n,p}\quad } _{\to 0}\cdot \underbrace {\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)} _{\to \int _{a}^{b}\ldots <\infty }=0} Und daher gilt im Limes n → ∞ {\displaystyle n\to \infty }
P ( x 1 ≤ S n − n p n ≤ x 2 ) = ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) {\displaystyle \operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;=\;\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)} Im Folgenden wird für die Riemann-Summe die Integraldarstellung gezeigt:
lim n → ∞ ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) = 1 2 π σ 2 ∫ x 1 x 2 exp ( − x 2 2 σ 2 ) d x {\displaystyle \lim _{n\to \infty }\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\;=\;{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x} Wir bilden ein äquidistantes Gitter Γ n := { k − n p n | k = 0 , 1 , … , n } ⊆ R {\displaystyle \Gamma _{n}:=\left\{\left.{\tfrac {k-np}{\sqrt {n}}}\right|\,k=0,1,\ldots ,n\right\}\subseteq \mathrm {R} } Maschenweite 1 n {\displaystyle {\frac {1}{\sqrt {n}}}} Riemannintegral überführen:
lim n → ∞ ∑ k ∈ { 0 , 1 … , n } x 1 ≤ k − n p n ≤ x 2 1 2 π n σ 2 exp ( − n 2 σ 2 ( k n − p ) 2 ) = lim n → ∞ ∑ x ∈ Γ n x 1 ≤ x ≤ x 2 1 2 π σ 2 exp ( − n 2 σ 2 ( k − n p n ) 2 ) ⋅ 1 n = 1 2 π σ 2 ∫ x 1 x 2 exp ( − x 2 2 σ 2 ) d x {\displaystyle {\begin{aligned}\lim _{n\to \infty }\sum _{k\in \{0,1\ldots ,n\} \atop x_{1}\leq {\frac {k-np}{\sqrt {n}}}\leq x_{2}}{\frac {1}{\sqrt {2\pi n\sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k}{n}}-p\right)^{2}\right)\;&=\;\lim _{n\to \infty }\sum _{x\in \Gamma _{n} \atop x_{1}\leq x\leq x_{2}}{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\,\exp \left(-{\frac {n}{2\sigma ^{2}}}\left({\frac {k-np}{n}}\right)^{2}\right)\cdot {\frac {1}{\sqrt {n}}}\;\\&=\;{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x\end{aligned}}} Unter Verwendung der Ergebnisse aus Schritt 1 und Schritt 2 folgt die eingangs formulierte Behauptung:
lim n → ∞ P ( x 1 ≤ S n − n p n ≤ x 2 ) = 1 2 π σ 2 ∫ x 1 x 2 exp ( − x 2 2 σ 2 ) d x {\displaystyle \quad \lim _{n\to \infty }\operatorname {P} \left(x_{1}\leq {\frac {S_{n}-np}{\sqrt {n}}}\leq x_{2}\right)\;=\;{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\int _{x_{1}}^{x_{2}}\exp \left(-{x^{2} \over 2\sigma ^{2}}\right)\mathrm {d} x} x 1 , x 2 ∈ R {\displaystyle x_{1},x_{2}\in \mathbb {R} } x 1 < x 2 {\displaystyle x_{1}<x_{2}} ◻ {\displaystyle \Box } 
![{\displaystyle p\in ]0,1[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cd07a5e96ded88e85782453bdae4353f369a648)
![{\displaystyle f:[n_{0},N]\to \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eddf7cfe6165a006c3b09737e53998a912ba8e63)
![{\displaystyle {\begin{aligned}\int \limits _{0}^{1}f(x)\mathrm {d} x&=-\int \limits _{0}^{1}\varphi ''(x)f(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg [}\varphi '(x)f(x){\bigg ]}_{0}^{1-\epsilon }+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg (}{\bigg .}\varphi '(x){\bigg |}_{x=1-\epsilon }f(1-\epsilon )-\varphi '(0)f(0){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg (}{\frac {1}{2}}{\Big (}1-2(1-\epsilon )+0{\Big )}f(1-\epsilon )-{\frac {1}{2}}f(0){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}{\bigg (}{\frac {1}{2}}(2\epsilon -1)f(1-\epsilon )-{\frac {1}{2}}f(0){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;+\;\lim _{\epsilon \to 0}\int \limits _{0}^{1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;+\;\lim _{\epsilon \to 0}{\Bigg (}{\bigg [}\varphi (x)f'(x){\bigg ]}_{0}^{1-\epsilon }\;-\;\int \limits _{0}^{1-\epsilon }\varphi (x)f''(x)\mathrm {d} x{\Bigg )}\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;-\;\lim _{\epsilon \to 0}\;\int \limits _{0}^{1-\epsilon }\varphi (x)f''(x)\mathrm {d} x\\&={\frac {1}{2}}{\bigg (}f(0)+f(1){\bigg )}\;-\;\int \limits _{0}^{1}\varphi (x)f''(x)\mathrm {d} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5526fe44fb84afadd6d91789781dda89b1c94cc4)
![{\displaystyle {\begin{aligned}\int \limits _{n_{0}}^{N}f(x)\mathrm {d} x&=-\int \limits _{n_{0}}^{N}\varphi ''(x)f(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg [}\varphi '(x)f(x){\bigg ]}_{k}^{k+1-\epsilon }+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg (}{\bigg .}\varphi '(x){\bigg |}_{x=k+1-\epsilon }\cdot f(k+1-\epsilon )-\varphi '(k)f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg (}{\frac {1}{2}}{\Big (}1-2(k+1-\epsilon )+2k{\Big )}f(k+1-\epsilon )-{\frac {1}{2}}f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=-\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\bigg (}{\frac {1}{2}}(2\epsilon -1)f(k+1-\epsilon )-{\frac {1}{2}}f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N-1}{\bigg (}{\frac {1}{2}}f(k+1)+{\frac {1}{2}}f(k){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}-1}^{N-1}{\frac {1}{2}}f(k+1)-{\frac {1}{2}}f(n_{0})+\sum _{k=n_{0}}^{N}{\frac {1}{2}}f(k)-{\frac {1}{2}}f(N)\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}{\frac {1}{2}}f(k)-{\frac {1}{2}}f(n_{0})+\sum _{k=n_{0}}^{N}{\frac {1}{2}}f(k)-{\frac {1}{2}}f(N)\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi '(x)f'(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;+\;\lim _{\epsilon \to 0}\sum _{k=n_{0}}^{N-1}{\Bigg (}{\bigg [}\varphi (x)f'(x){\bigg ]}_{0}^{1-\epsilon }\;-\;\int \limits _{k}^{k+1-\epsilon }\varphi (x)f''(x)\mathrm {d} x{\Bigg )}\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;-\;\lim _{\epsilon \to 0}\;\sum _{k=n_{0}}^{N-1}\int \limits _{k}^{k+1-\epsilon }\varphi (x)f''(x)\mathrm {d} x\\&=\sum _{k=n_{0}}^{N}f(k)-{\frac {1}{2}}{\bigg (}f(n_{0})+f(N){\bigg )}\;-\;\int \limits _{n_{0}}^{N}\varphi (x)f''(x)\mathrm {d} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1553de6000283dee5cc1aaf6d3ad2d86ef71d6a7)
![{\displaystyle {\tfrac {1}{8}}{\bigg |}\int _{0}^{n}{\tfrac {1}{(x+t)^{2}}}\mathrm {d} t{\bigg |}={\tfrac {1}{8}}{\bigg |}{\bigg [}{\tfrac {-1}{x+t}}{\bigg ]}_{0}^{n}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-1}{x+n}}-{\tfrac {-1}{x}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {-x-n+x}{-x(x+n)}}{\bigg |}={\tfrac {1}{8}}{\bigg |}{\tfrac {1}{x({\tfrac {x}{n}}+1)}}{\bigg |}\leq {\tfrac {1}{8}}{\tfrac {1}{|x|}}\approx {\mathcal {O}}{\Big (}{\tfrac {1}{|x|}}{\Big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b450c65c4798ff0daa9e89245dbcef2a2534b04a)
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