Statistical Proof of Stirling's Formula
📂Mathematical Statistics Statistical Proof of Stirling's Formula 정리 lim n → ∞ n ! e n ln n − n 2 π n = 1
\lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi n} }} = 1
n → ∞ lim e n l n n − n 2 πn n ! = 1
설명 The Stirling approximation, or Stirling’s formula , is usefully applied in various fields, including statistics and physics. If one is well-versed in probability theory, the mathematical statistical proof can be more intuitive and easier to understand compared to other proofs.
같이보기 증명 Let’s denote it as X 1 , ⋯ , X n ∼ iid exp ( 1 ) X_{1} , \cdots , X_{n} \overset{\text{iid}}{\sim} \exp (1) X 1 , ⋯ , X n ∼ iid exp ( 1 ) .
Relationship between Exponential Distribution and Gamma Distribution :
Γ ( 1 , 1 λ ) ⟺ exp ( λ )
\displaystyle \Gamma \left(1, { 1 \over \lambda } \right) \iff \text{exp} (\lambda)
Γ ( 1 , λ 1 ) ⟺ exp ( λ )
Sum of Gamma Distributions : If it is X i ∼ Γ ( k i , θ ) X_i \sim \Gamma ( k_{i}, \theta) X i ∼ Γ ( k i , θ ) , then
∑ i = 1 n X i ∼ Γ ( ∑ i = 1 n k i , θ )
\displaystyle \sum_{i=1}^{n} X_{i} \sim \Gamma \left( \sum_{i=1}^{n} k_{i} , \theta \right)
i = 1 ∑ n X i ∼ Γ ( i = 1 ∑ n k i , θ )
Since X 1 , ⋯ , X n X_{1} , \cdots , X_{n} X 1 , ⋯ , X n is iid , Y = ∑ k = 1 n X k Y = \sum_{k=1}^{n} X_{k} Y = ∑ k = 1 n X k follows a Gamma Distribution with a degree of freedom of n , 1 n,1 n , 1 , denoted as Γ ( n , 1 ) \Gamma (n,1) Γ ( n , 1 ) . Meanwhile, due to E X k = Var X k = 1 E X_{k} = \operatorname{Var} X_{k} = 1 E X k = Var X k = 1 , according to the Central Limit Theorem when n → ∞ n \to \infty n → ∞
∑ k = 1 n X k / n − 1 1 / n → D N ( 0 , 1 )
{{ \sum_{k=1}^{n} X_{k}/n - 1 } \over { 1 / \sqrt{n} }} \overset{D}{\to} N (0,1)
1/ n ∑ k = 1 n X k / n − 1 → D N ( 0 , 1 )
In other words, when Z Z Z is a random variable following the standard normal distribution , for all x ∈ R x \in \mathbb{R} x ∈ R
P ( Y / n − 1 1 / n ≤ x ) → P ( Z ≤ x ) ⟹ P ( Y / n ≤ x n + 1 ) → P ( Z ≤ x ) ⟹ P ( Y ≤ n x + n ) → P ( Z ≤ x )
\begin{align*}
& P \left( {{ Y/n - 1 } \over { 1 / \sqrt{n} }} \le x \right) \to P \left( Z \le x \right)
\\ \implies & P \left( Y/n \le {{ x } \over { \sqrt{n} }} + 1 \right) \to P \left( Z \le x \right)
\\ \implies & P \left( Y \le \sqrt{n} x + n \right) \to P \left( Z \le x \right)
\end{align*}
⟹ ⟹ P ( 1/ n Y / n − 1 ≤ x ) → P ( Z ≤ x ) P ( Y / n ≤ n x + 1 ) → P ( Z ≤ x ) P ( Y ≤ n x + n ) → P ( Z ≤ x )
For sufficiently large n ∈ N n \in \mathbb{N} n ∈ N , using an approximation formula for the cumulative distribution function F □ F_{\square} F □
F Y ( n x + n ) ≈ F Z ( x )
F_{Y} \left( \sqrt{n} x + n \right) \approx F_{Z} (x)
F Y ( n x + n ) ≈ F Z ( x )
can be expressed as follows.
Probability Density Function of Gamma Distribution
f ( x ) = 1 Γ ( k ) θ k x k − 1 e − x / θ , x > 0
f(x) = {{ 1 } \over { \Gamma ( k ) \theta^{k} }} x^{k - 1} e^{ - x / \theta} \qquad , x > 0
f ( x ) = Γ ( k ) θ k 1 x k − 1 e − x / θ , x > 0
Probability Density Function of Standard Normal Distribution
f ( z ) = 1 2 π exp [ − z 2 2 ]
f(z) = {{ 1 } \over { \sqrt{2 \pi} }} \exp \left[ - {{ z^{2} } \over { 2 }} \right]
f ( z ) = 2 π 1 exp [ − 2 z 2 ]
Differentiating both sides by x x x , according to the Fundamental Theorem of Calculus
n ⋅ 1 Γ ( n ) 1 n ( n x + n ) n − 1 e − ( n x + n ) / 1 ≈ 1 2 π exp [ − x 2 2 ]
\sqrt{n} \cdot {{ 1 } \over { \Gamma ( n ) 1^{n} }} \left( \sqrt{n} x + n \right)^{n - 1} e^{ - \left( \sqrt{n} x + n \right) / 1} \approx {{ 1 } \over { \sqrt{2 \pi} }} \exp \left[ - {{ x^{2} } \over { 2 }} \right]
n ⋅ Γ ( n ) 1 n 1 ( n x + n ) n − 1 e − ( n x + n ) /1 ≈ 2 π 1 exp [ − 2 x 2 ]
Here, if it’s x = 0 x = 0 x = 0 , we obtain the following.
n ⋅ 1 Γ ( n ) ( n ) n − 1 e − n ≈ 1 2 π
\sqrt{n} \cdot {{ 1 } \over { \Gamma ( n ) }} \left( n \right)^{n - 1} e^{ - n } \approx {{ 1 } \over { \sqrt{2 \pi} }}
n ⋅ Γ ( n ) 1 ( n ) n − 1 e − n ≈ 2 π 1
Since the Gamma function is a generalization of the factorial , if Γ ( n ) = ( n − 1 ) ! \Gamma (n) = (n-1)! Γ ( n ) = ( n − 1 )! , and multiplying both sides by ( n − 1 ) ! 2 π (n-1)!\sqrt{2\pi} ( n − 1 )! 2 π ,
2 π n e ( n − 1 ) log n e − n ≈ ( n − 1 ) !
\sqrt{2 \pi n} e^{(n-1) \log n} e^{-n} \approx (n-1)!
2 πn e ( n − 1 ) l o g n e − n ≈ ( n − 1 )!
Summarizing the above,
e n log n − n e − log n 2 π n ≈ ( n − 1 ) ! ⟹ e n log n − n 1 n 2 π n ≈ ( n − 1 ) ! ⟹ e n log n − n 2 π n ≈ n !
\begin{align*}
& e^{n\log n - n} e^{-\log n} \sqrt{2 \pi n} \approx (n-1)!
\\ \implies & e^{n\log n - n} {{ 1 } \over { n }} \sqrt{2 \pi n} \approx (n-1)!
\\ \implies & e^{n\log n - n} \sqrt{2 \pi n} \approx n!
\end{align*}
⟹ ⟹ e n l o g n − n e − l o g n 2 πn ≈ ( n − 1 )! e n l o g n − n n 1 2 πn ≈ ( n − 1 )! e n l o g n − n 2 πn ≈ n !
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