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# lectures3 - MATH 235A Probability Theory Lecture Notes Fall...

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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: Stirling’s 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 →∞ Φ( t ) = 1 2 π 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 0 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 Stirling’s formula. Lemma 2. The limit C = lim n →∞ n ! n ( n/e ) n exists. 1

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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 (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 0 f ( n +1) ( t )( x - t ) n dt. Proof. This follows by induction on n , using integration by parts. Lemma 4. C = 2 π . 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 0 (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 0 (1 - t 2 ) n dt = ( 2 n n ) n 2 2 n (1 + 1 2 n ) Z n 0 1 - u 2 n n du ---→ n →∞ 2 C Z 0 e - u 2 du = 2 C · π 2 . 2
The convergence of the integrals is justified by the fact that (1 - u 2 /n ) n e - u 2 for all 0 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 π , as claimed. Corollary 5 (Stirling’s formula) . lim n →∞ n ! 2 πn ( n/e ) n = 1 . Note that the proof is based on computing P ( S 2 n +1 > n ) in two different ways, when S 2 n +1 Binom(2 n + 1 , 1 / 2). This is just the special case p = 1 / 2 , t = 0 of Theorem 1! In this very special case, by symmetry the probability is equal to 1 / 2; on the other hand, Lemma 3 enables us to relate this to the asymptotic behavior of n ! and to (half of) the gaussian integral R -∞ e - x 2 dx . The evaluation of the constant C in Stirling’s formula is the part that is attributed to James Stirling. The form that appears in Lemma 2 is due to Abraham de Moivre (1733).

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