AN UNDERGRADUATE LECTURE ON THE CENTRAL LIMIT THEOREM

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1 AN UNDERGRADUATE LECTURE ON THE CENTRAL LIMIT THEOREM N.V. KRYLOV In the first section we explain why the central limit theorem for the binomial 1/2 distributions is natural. The second section contains the proof and the third section gives a refinement of our basic estimate. 1. Setting and motivation Let S n be the number of successes in n independent Bernoulli trials with success probability p = 1/2. First, one realizes that P (S n = k) = 2 n ( n k ) = 2 n n! k!(n k)!. (1) Then comes the observation that ES n = n/2, Var S n = n/4. This shows that the random variables (S n n/2)/ n are centered and have constant variance. One may hope that their distributions converge to something as n. In other words, one may hope that for any a < b P (a < (S n n/2)/ n b) (2) has a limit as n. Let us try to guess what the limit could be. Set Observe that y kn = n 1/2 (k n/2), k =, 1,..., n. y k+1,n = y kn + n 1/2, k = y kn n + n/2. P (S n = k) = P (n 1/2 (S n n/2) = y kn ) and, for each n, introduce a function so that f n (y kn ) = P (n 1/2 (S n n/2) = y kn ), f n (y kn ) = P (S n = k). 1

2 2 N.V. KRYLOV For k n 1 simple manipulations using (1) show that P (S n = k) = 2 n n! k + 1 (k + 1)!(n k 1)! n k, implying that f n (y kn ) = f n (y kn + n 1/2 ) y kn n + n/2 + 1, n/2 y kn n f n (y kn + n 1/2 n/2 y kn n ) = f n (y kn ), (3) f n (y kn + n 1/2 ) f n (y kn ) = f n (y kn ) ( n/2 y kn n 1 ) Next, the probability in (2) is 2y kn n + 1 = f n (y kn ). (4) k:a<y kn b f n (y kn ) and this sum contains roughly speaking (b a) n terms. Therefore, each term should be of order 1/ n. One way or another, let us believe that, for a function φ, as n and y kn y. For φ n equation (4) becomes φ n (y kn ) := nf n (y kn ) φ(y) φ n (y kn + n 1/2 2y kn n + 1 ) φ n (y kn ) = φ n (y kn ). (5) Let us pretend that (5) holds for φ rather than φ n and for all y in place of y kn rather than for y kn of a special form, that is φ(y + n 1/2 2y n + 1 ) φ(y ) = φ(y) y n + n/ Devide both parts by n 1/2 and let n to see that, naturally, φ n (y) should be close to a φ that is a solution of φ (y) = 4yφ(y). All solutions of the latter equation are known to be ce 2y2, and this is a way to explain why the normal distribution arises in this particular instance of the central limit theorem.

3 AN UNDERGRADUATE LECTURE ON THE CENTRAL LIMIT THEOREM 3 2. First rigorous resulat As above we assume that p = 1/2 and add the assumption that n is an even number. Theorem 1. Let β [1/2, 3/4] and n be even. Then for k n/2 n β (that is y kn n β 1/2 ) n e 2y2 kn fn (y kn ) = e O(n4β 3) = 1 + O(n 4β 3 ), (6) where c n = P (S n = n/2), y kn = (k n/2)n 1/2. Furthermore, lim c n n = 2/π, (7) and if β [1/2, 3/4), then lim max (nπ/2) 1/2 e 2y2 knp (Sn = k) 1 =. (8) k n/2 n β Finally, for any a, b (, ) such that a < b, lim P (a n S n n/2 b n) = 2/π b a e 2y2 dy. (9) Proof. To prove (6) it suffices to concentrate on k n/2 when y kn. Set g n (y kn ) = ln f n (y kn ), x kn = 2y kn n 1/2 = 2(k n/2)n 1 = 2k/n 1 and use (3) to get g n (y k+1,n ) g n (y kn ) = ln(1 x kn ) ln(1 + x k+1,n ) = x kn x k+1,n + (1/2)[x 2 k+1,n x 2 kn] (1/3)[ x 3 kn + x 3 k+1,n], (1) where x kn x kn, x k+1,n x k+1,n. For n/2 k n/2 + n β we have y kn n β 1/2, x kn 2n β 1 2, x 3 kn + x 3 k+1,n = O(n 3β 3 ), [x 2 i+1,n x 2 in] = x 2 k+1,n = O(n 2β 2 ) = O(n 4β 3 ), where we used that β 1/2. We also observe that (x in + x i+1,n ) = 4n 1 (i n/2) + 2n 1 1 = 2n 1 (k n/2)(k + 1 n/2) + 2n 1 (k + 1 n/2) = 2n 1 (k + 1 n/2) 2 = 2y 2 k+1,n.

4 4 N.V. KRYLOV Then we find that for n/2 k n/2 + n β g n (y k+1,n ) g n () = 2y 2 k+1,n + O(n 4β 3 ), n e 2y2 kn fn (y kn ) = e O(n4β 3) = 1 + O(n 4β 3 ), (11) which is (6). After this we focus on determining the behavior of c n as n. Rewrite (6) as n 1/2 n P (S n = k) = n 1/2 e 2y2 kn + O(n 4β 3.5 ) (12) and conclude that where lim n 1/2 n P ( S n n/2 n 3/5 ) = lim J n, (13) J n := n 1/2 Here by Chebyshev s inequality e 2y2 kn k: y kn n 1/1 1 P ( S n n/2 n 3/5 ) = 1 P ( S n n/2 n 3/5 ) 1 n 1/5, implying that lim P ( S n n/2 n 3/5 ) = 1, which along with (13) yields lim n 1/2 n = lim J n. (14) To find lim J n observe that for k n/2 we have e 2y2 kn e 2y 2 for y y kn, e 2y2 kn e 2y 2 for y k,n y. Therefore, for k > n/2 ykn y k 1,n e 2y2 dy n 1/2 e 2y2 kn yk+1,n Since we infer that ȳn J n = n 1/2 + 2 e 2y2 dy (J n n 1/2 )/2 y kn e 2y2 e 2y2 kn, k>n/2:y kn n 1/1 ȳn e 2y2 dy dy. n 1/2 e 2y2 dy, where ȳ n is the largest y kn such that y kn n 1/1. Obviously, ȳ n as n. This yields lim J n = 2 e 2y2 dy = π/2.

5 AN UNDERGRADUATE LECTURE ON THE CENTRAL LIMIT THEOREM 5 Upon combining this with (14) we come to (7). Now we can replace n 1/2 n in (13) with π/2 and by the definition of integral obtain that for any a, b (, ) such that a < b π/2 lim P (a n S n n/2 b n) = lim n 1/2 k:a y kn b This proves (9). Finally, we can rewrite (6) as b e 2y2 kn = e 2y2 dy. (nπ/2) 1/2 e 2y2 kn fn (y kn ) = e O(n4β 3) = d n + O(n 4β 3 ), where d n = c n (nπ/2) 1/2 1. Then (8) follows. The theorem is proved. a 3. A refinement of Theorem 1 One can improve (6) and (8). For β = 3/4 and y kn n 1/4 equation (6) only says that n e 2y2 kn fn (y kn ) is bounded without asserting that it converges to 1 as n. The following theorem implies, in particular, that n e 2y2 kn fn (y kn ) e 4c4 /3 if n, k = k n, and y kn n 1/4 c. The refinement comes from taking two more terms in Taylor s expansions in (16). Theorem 2. For β [1/2, 1) and k n/2 n β we have where e αn6β 5 n e 2y2 kn (1+2y2 kn /(3n)) P (S n = k) e αn2β 2, (15) α = 16/3 + ( )/5. In particular, if β [1/2, 5/6), then lim max k n/2 n β n e 2y2 kn (1+2y2 kn /(3n)) P (S n = k) 1 =. Furthermore, if β [3/4, 5/7), then lim cn 1 e 2y2 kn (1+2y2 kn /(3n)) P (S n = k) 1 =. k n/2 n β

6 6 N.V. KRYLOV Proof. It follows from (3) that g n (y k+1,n ) g n (y kn ) = ln(1 x kn ) ln(1 + x k+1,n ) = x kn x k+1,n + (1/2)[x 2 k+1,n x 2 kn] (1/3)[x 3 kn + x 3 k+1,n] +(1/4)[x 4 k+1,n x 4 k,n] (1/5)[ x 5 kn + x 5 k+1,n], (16) where x kn x kn, x k+1,n x k+1,n. For n/2 k n/2 + n β we have y kn n β 1/2, x kn 2n β 1, y k+1,n n β 1/2 + n 1/2 2n β 1/2, x k+1,n 4n β 1, x 5 kn + x 5 k+1,n ( )n 5β 5, [x 2 i+1,n x 2 in] = x 2 k+1,n 16n 2β 2. We also observe that [x 4 i+1,n x 4 in] = x 4 k+1,n 2 8 n 4β n 2β 2, k n/2 [x 3 kn + x 3 k+1,n] = 8n 3 i= k+1 n/2 i 3 + 8n 3 = 2n 3 (k n/2) 2 (k + 1 n/2) 2 + 2n 3 (k + 1 n/2) 2 (k + 2 n/2) 2 = 4n 1 yk+1,n[y 2 k+1,n 2 + n 1 ] = 4n 1 yk+1,n 4 + 4n 2 yk+1,n. 2 As a consequence 4n 1 yk+1,n 4 [x 3 kn + x 3 k+1,n] 4n 1 yk+1,n n 2β 3. Now upon summing up (16) with respect to k, we find that for n/2 k n/2 + n β g n (y k+1,n ) g n () 2y 2 k+1,n (4/3)n 1 y 4 k+1,n + 72n 2(β 1), g n (y k+1,n ) g n () 2yk+1,n 2 (4/3)n 1 yk+1,n 4 n 2β 3 16/3 n 6β 5 ( )/5 2yk+1,n 2 (4/3)n 1 yk+1,n 4 αn 6β 5, where we used that β 1/2. The above inequalities yield (15), that easily implies all other assertions of the theorem. The theorem is proved. i=1 i 3