Question

4. Stirling's Formula is the claim that n! n-o0 >1. V2nn(n/e)" In this exercise, we will show how this can be obtained from the Central Limit Theorem Recall that Exponential () if fx(x)= Ae ^x, x > 0; the corresponding mgf is Mx (t) ,t<^ X = and GA)".,1 ва xa-le-px, x> 0; the corresponding mgf is Mx(t) = X~T(a,B) if fx(x)= T , t <B (a) Argue that, if Xi ~Exponential(1), i = 1,2,..., all independent, then for every x, Х, — 1 P п—00 P(Z<x) <x 1/Vn N (0,1) where Z (b) Show that L}| X;~ Г(п, 1). (c) Show that differentiating both sides of the approximation in part (a) and using part (b) suggests Vn (xVп+п)"-1.е- (xVi+n) (п—1)! 1 V27t and that settingx= 0 gives Stirling's Formula.

Answer 1

Lindeberg Ley T, Safre ^x, , . } . (49 seguene ofid randem variable wih Ei) (S-)Ne) Here xiExponential (1) E(xi) Var(xi) So by Lindeberg- Levy CLT 0)N whre Z N lim P ha Gamma (ki, e) clifubutim for i 10) N Gammuiki, Exponeutiak )Gamma () of (b) N bovideed all Xi are an Hore So i Gamma (n,)

To pnuve -(n) e iGamm() Let Gamma lr We hane to Let 7-) Z = n-1 / e , 9>0 oturu n-1 e. n-t)! dy d2 z> - n -n2+n) n-)! - e S Z 2)-n matun in por (a We 2n both tranges becomy As

griues Now selting 1-и e / i(i- i(1-u Ил UE e n-/ 27 n (eraved) V2nn /e)