Grading: 1, 2, 3, 4, 5 (each 8 pts) e ita R. e itx dµ(x)dt = 1 2T. cos(t(x a))dt dµ(x) T (x a) 1 2T. cos(t(x a))dt =
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1 Grading:, 2, 3, 4, 5 (each 8 pts Problem : Since e ita =, by Fubini s theorem, I T = e ita ϕ(tdt = e ita e itx dµ(xdt = e it(x a dtdµ(x. Since e it(x a = cos(t(x a + i sin(t(x a and sin(t(x a is an odd function of t, We have In particular, I T = cos(t(x adtdµ(x., x = a cos(t(x adt = sin(t (x a, x a. T (x a By the dominated convergence theorem, and, since sin(t (x a, It follows that ( I T = cos(t(x adt. cos(t(x adt dµ(x, x = a cos(t(x adt = 0, x a. I T = {x = a}dµ(x = µ({a}. Problem 2: Since X n X and Y n Y, we have ϕ Xn (t ϕ X (t and ϕ Yn (t ϕ Y (t for each t. Because X n and Y n are independent, ϕ Xn+Y n (t = ϕ Xn (t ϕ Yn (t ϕ X (t ϕ Y (t = ϕ X +Y (t. Since ϕ X (t and ϕ Y (t are continuous at 0, so is ϕ X +Y (t. Hence, the continuity theorem implies that X n + Y n X + Y. Contributed by Binh Tang and Ilya Amburg
2 Problem 3: We proceed by induction. Base case n = 0: ϕ (0 (t = ϕ(t = (ix0 e itx µ(dx, and it is continuous by Theorem 3.3. in Durrett (or see the proof of continuity below which also works for n = 0. Inductive step: Suppose x n µ(dx <. Then the same is true for the (n st power, so by the inductive hypothesis, ϕ (n (t = (ixn e itx µ(dx. Thus ϕ (n (t = ϕ (n (t + h ϕ (n (t h = (ix n e itx eihx h To apply dominated convergence we must find an upper bound for that does not depend on h. Let (ixn e itx eihx h = e ihx x n hx µ(dx. g(y = eiy y = i + i2 y 2! + i3 y 2 3! +. Since g is continuous (in particular, g(0 = i we know that max{ g(y : y [, ]} <. For y we use that the numerator e iy 2, so g(y 2. Hence there is C such that g(y C < for all y, and we can take (ixn e itx eihx h C x n which is integrable with respect to µ(dx by assumption. Dominated convergence now gives To prove continuity, ϕ (n (t = (ix n e itx (ixµ(dx = (ix n e itx µ(dx. ϕ(n (t + h = (ix n e i(t+hx µ(dx. The integrand is dominated by x n, which is integrable, so dominated convergence gives ϕ(n (t + h = (ix n e itx µ(dx = ϕ (n (t. Problem 4: Suppose E[X 2 i ] =. Let X, X 2,... be an independent copy of the original sequence, and Y i = X i X i. Fix K > 0. Given A > 0 (whose value will be determined later, let U i = Y i Yi A and V i = Y i Yi >A. Since Y i = U i + V i, if U i K n and V i 0, 2
3 then Y i K n. Hence, ( P Y i K ( n P U i K n, V i 0 ( = P U i K ( n P V i 0 U i K Because of the symmetry between X i and X i, ( P V i 0 U i K ( n = P V i 0 U i K Since the sum of these conditional probabilities is at least, both of them have to be at least /2. Hence, ( P U i K n, ( V i 0 2 P U i K The central it theorem implies that n n U i Var(U i Z where Z N (0,. Hence, ( P U i K ( n P Z K Var(Ui as n. By monotone convergence, Var(U i = E[U 2 i ] E[Y 2 i ] = as A, so by setting A sufficiently large (depending on K we can ensure that P (Z K/ Var(U i.45 and therefore P ( n U i K n 2/5 for sufficiently large n. Combining all the inequalities, we have shown that for large enough n depending on K, ( P Y i K n 5. On the other hand, we assumed that S n / n converges weakly to some it W. By Problem 2, n n Y i W W where W is an independent copy of W, so ( P Y i K n P (W W K (as long as P (W W = K = 0 for continuity purposes. Setting K large enough that P (W W K < /5 gives a contradiction. 3
4 Problem 5: The characteristic function for X k is ϕ Xk (t = E[e itx k ] = k e itx dx = k cos(txdx = 2k k 2k k kt sin(kt 6 (kt2. Consequently, the characteristic function for n α S n is given by Let ϕ n (t = ϕ Xk (n α t ( (kt2. c k,n = ϕ Xk (n α t = sin(kn α t kn α t We aim to apply Fact.2 from the notes. First, { max c sin(x k,n max k n x 6n 2α (kt2 6n 2α. } : x n α t which, as long as α >, converges to 0 as n because sin(x/x as x 0. Indeed, sin(x x = x2 6 + x4 5! which implies that there is a constant C such that g(x := sin(x x ( x2 6 satisfies g(x Cx 4 for all x. Thus, if n is large enough that n α t, and c k,n = ( (kt 2 6n + 2α g(kn α t g(kn α t If we choose α = 3/2 then Ck 4 n 4α t 4 = t2 n(n + (2n + 36n 2α + g(kn α t Cn 4 4α t 4 = Cn 5 4α t 4. g(kn α t 0 as n and so c k,n = t2 8. 4
5 Finally, sup c k,n = sup (kt 2 + g(kn 3/2 t 6n 3 sup Hence, Fact.2 implies that as n, ϕ n (t = (kt 2 6n 3 + sup ( + c k,n e t2 /8 = ϕ Z (t where Z N We conclude that α = 3/2, µ = 0, and σ 2 = /9. g(kn 3/2 t = t <. ( 0,. 9 Note: Some students used the Lindeberg-Feller CLT to solve this problem. That was not the intended solution (because it was supposed to be solvable without having seen Lindeberg-Feller yet but it also should work. 5
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