lectures3 - MATH 235A Probability Theory Lecture Notes,...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 235A Probability Theory Lecture Notes, Fall 2009 Part III: The central limit theorem Dan Romik Department of Mathematics, UC Davis Draft (December 4, 2009) Lecture 10: Stirlings formula and the de Moivre-Laplace theorem We start with a motivating example that was also historically the first instance in which the phenomenon that came to be known as the Central Limit Theorem was observed. Let X 1 ,X 2 ,... be an i.i.d. of Binom(1 ,p ) random variables, and let S n = n k =1 X k , a r.v. with distribution Binom( n,p ). Theorem 1 (The de Moivre-Laplace theorem) . For any t R , P S n- np p np (1- p ) t !--- n ( t ) = 1 2 Z t- e- x 2 / 2 dx. Since this is such a concrete example, the proof will simply require us to estimate a sum of the form k t ( n k ) p k (1- p ) n- k . Knowing how to estimate such sums is a useful skill in its own right. Since the binomial coefficients are involved, we also need some preparation related to Stirlings formula. Lemma 2. The limit C = lim n n ! n ( n/e ) n exists. 1 Proof. log n ! = n X k =1 log k = n X =1 Z k 1 dx x = Z n 1 n- b x c x dx = Z n 1 n + 1 2 + ( { x } - 1 2 )- x x dx = ( n + 1 / 2) log n- n + 1 + Z n 1 { x } - 1 2 x dx = ( n + 1 / 2) log n- n + 1 + Z 1 { x } - 1 2 x dx + o (1) , where the last integral converges because R t 1 ( { x } - 1 2 ) dx is bounded and 1 /x goes to 0 as x . Note that an easy consequence of Lemma 2 is that ( 2 n n ) = (1 + o (1))2 2 n /C p n/ 2. We shall now use this to find the value of C . Lemma 3. Let f : R R be an n + 1 times continuously-differentiable function. Then for all x R , we have f ( x ) = f (0) + f (0) x + f 00 (0) 2 x 2 + ... + f ( n ) (0) n ! x n + R n ( x ) , where R n ( x ) = 1 n ! Z x f ( n +1) ( t )( x- t ) n dt. Proof. This follows by induction on n , using integration by parts. Lemma 4. C = 2 . Proof. Apply Lemma 3 with f ( x ) = (1 + x ) 2 n +1 to compute R n (1): 1 2 2 n +1 R n (1) = 1 2 2 n +1 1 n ! Z 1 (2 n + 1)(2 n ) ( n + 1)(1 + t ) n (1- t ) n dt = 2 ( 2 n n ) 2 2 n +1 ( n + 1 2 ) Z 1 (1- t 2 ) n dt = ( 2 n n ) n 2 2 n (1 + 1 2 n ) Z n 1- u 2 n n du--- n 2 C Z e- u 2 du = 2 C 2 . 2 The convergence of the integrals is justified by the fact that (1- u 2 /n ) n e- u 2 for all u n , and (1- u 2 /n ) n e- u 2 as n , uniformly on compact intervals. To finish the proof, note that 1 2 2 n +1 R n (1) = X n<k 2 n +1 ( 2 n +1 k ) 2 2 n +1 = 1 2 (this is the probability that a Binom(2 n + 1 , 1 / 2) random variable takes a value > n ). Therefore C = 2 , as claimed. Corollary 5 (Stirlings formula) . lim n n !...
View Full Document

Page1 / 23

lectures3 - MATH 235A Probability Theory Lecture Notes,...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online