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Statistical Proof of Stirling's Formula 📂Mathematical Statistics

Statistical Proof of Stirling's Formula

정리

lim⁡n→∞n!enln⁡n−n2πn=1 \lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi 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.

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증명

Let’s denote it as X1,⋯ ,Xn∼iidexp⁡(1)X_{1} , \cdots , X_{n} \overset{\text{iid}}{\sim} \exp (1).

Relationship between Exponential Distribution and Gamma Distribution: Γ(1,1λ)  ⟺  exp(λ) \displaystyle \Gamma \left(1, { 1 \over \lambda } \right) \iff \text{exp} (\lambda)

Sum of Gamma Distributions: If it is Xi∼Γ(ki,θ)X_i \sim \Gamma ( k_{i}, \theta), then ∑i=1nXi∼Γ(∑i=1nki,θ) \displaystyle \sum_{i=1}^{n} X_{i} \sim \Gamma \left( \sum_{i=1}^{n} k_{i} , \theta \right)

Since X1,⋯ ,XnX_{1} , \cdots , X_{n} is iid, Y=∑k=1nXkY = \sum_{k=1}^{n} X_{k} follows a Gamma Distribution with a degree of freedom of n,1n,1, denoted as Γ(n,1)\Gamma (n,1). Meanwhile, due to EXk=Var⁡Xk=1E X_{k} = \operatorname{Var} X_{k} = 1, according to the Central Limit Theorem when n→∞n \to \infty ∑k=1nXk/n−11/n→DN(0,1) {{ \sum_{k=1}^{n} X_{k}/n - 1 } \over { 1 / \sqrt{n} }} \overset{D}{\to} N (0,1) In other words, when ZZ is a random variable following the standard normal distribution, for all x∈Rx \in \mathbb{R} P(Y/n−11/n≤x)→P(Z≤x)  ⟹  P(Y/n≤xn+1)→P(Z≤x)  ⟹  P(Y≤nx+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*} For sufficiently large n∈Nn \in \mathbb{N}, using an approximation formula for the cumulative distribution function F□F_{\square} FY(nx+n)≈FZ(x) F_{Y} \left( \sqrt{n} x + n \right) \approx F_{Z} (x) can be expressed as follows.

Probability Density Function of Gamma Distribution f(x)=1Γ(k)θkxk−1e−x/θ,x>0 f(x) = {{ 1 } \over { \Gamma ( k ) \theta^{k} }} x^{k - 1} e^{ - x / \theta} \qquad , x > 0

Probability Density Function of Standard Normal Distribution f(z)=12πexp⁡[−z22] f(z) = {{ 1 } \over { \sqrt{2 \pi} }} \exp \left[ - {{ z^{2} } \over { 2 }} \right]

Differentiating both sides by xx, according to the Fundamental Theorem of Calculus n⋅1Γ(n)1n(nx+n)n−1e−(nx+n)/1≈12πexp⁡[−x22] \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] Here, if it’s x=0x = 0, we obtain the following. n⋅1Γ(n)(n)n−1e−n≈12π \sqrt{n} \cdot {{ 1 } \over { \Gamma ( n ) }} \left( n \right)^{n - 1} e^{ - n } \approx {{ 1 } \over { \sqrt{2 \pi} }} Since the Gamma function is a generalization of the factorial, if Γ(n)=(n−1)!\Gamma (n) = (n-1)!, and multiplying both sides by (n−1)!2π(n-1)!\sqrt{2\pi}, 2πne(n−1)log⁡ne−n≈(n−1)! \sqrt{2 \pi n} e^{(n-1) \log n} e^{-n} \approx (n-1)! Summarizing the above, enlog⁡n−ne−log⁡n2πn≈(n−1)!  ⟹  enlog⁡n−n1n2πn≈(n−1)!  ⟹  enlog⁡n−n2πn≈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*}

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