But we know xnk has some as convergent subsequence

Info icon This preview shows pages 188–192. Sign up to view the full content.

But we know {Xnk} has some a.s. convergent subsequence {Xnk 1 iJ} such that Xn k!iJ --+X almost surely. Thus g(Xnk!i>) g(X) which finishes the proof. 0 Thus we see that if Xn X, it is also true that x; X 2 , and arctanXn arctanX
Image of page 188

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

6.3 Connections Between a.s. and i.p. Convergence 175 and so on. Now for the promised connection with Dominated Convergence: The statement of the Dominated Convergence Theorem holds without change when almost sure convergence is replaced by convergence i.p. Corollary 6.3.2 (Lebesgue Dominated Convergence) If Xn 2:. X and if there exists a dominating random variable ; e L 1 such that then E(Xn)-+ E(X). Proof. It suffices to show every convergent subsequence of E (X n) converges to E(X). Suppose E (X nk) converges . Then since convergence in probability is assumed, {Xnk} contains an a.s . convergent subsequence {Xnk(i)} such that Xnw> The Lebesgue Dominated Convergence Theorem implies So E(Xnk)-+ E(X) . We now list several easy results related to convergence in probability. p p (1) If Xn -+X and Yn -+ Y then p Xn + Yn -+ X + Y. To see this , just note E E [I(Xn + Yn)- (X+ Y)l > E] C [IXn- Xi > 2] U [IYn- Yl > 2]. Take probabilities, use subadditivity and let n -+ oo. (2) A multiplicative counterpart to (1) : If Xn 2:. X and Yn 2:. Y, then p XnYn-+ XY. 0 To see this, observe that given a subsequence {nk}, it suffices to find a fur- ther subsequence {nk(i)} C {nk} such that Since Xnk 2:. X, there exists a subsequence c {nk} such that X a.s. X nA. -+ .
Image of page 189
176 6. Convergence Concepts Since Yn Y, given the subsequence {nA:}, there exists a further subse- quence {nk(i)} C {nA:} such that and hence, since the product of two convergent sequences is convergent, we have x, y, nk(i) nk<i> Thus every subsequence of {XnYn} has an a.s. convergent subsequence. (3) This item is a reminder that Chebychev's inequality implies the Weak Law of Large Numbers (WLLN): If {Xn. n 2: 1} are iid with EXn = ll and Var(Xn) = cr 2 , then n LX;/n ll· i=l (4) Bernstein's version of the Weierstrass Approximation Theorem. Let f [0, 1] JR. be continuous and define the Bernstein polynomial of degree n by Then Bn(X)--+ f(x) uniformly for x E [0, 1]. The proof of pointwise convergence is easy using the WLLN: Let OJ, 8z, ... , On be iid Bernoulli random variables with P[o; = 1] =X = 1- P[o; = 0]. Define Sn = L:?=t 8; so that Sn has a binomial distribution with success probability p = x and E(Sn) = nx, Var(Sn) = nx(1- x) n. Since f is continuous on [0, 1], f is bounded. Thus, since Sn P from the WLLN, we get ---+X, n Sn P f(-) --+ f(x) n by continuity of f and by dominated convergence, we get Sn Ef(-)--+ f(x). n
Image of page 190

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

6.3 Connections Between a.s . and i.p. Convergence 177 But Sn k (n) k n k X (1-x)- =Bn(X). n k=O nk We now show convergence is uniform. Since f is continuous on [0, 1 ], f is uniformly continuous, so define the modulus of continuity as w(8) = sup lf(x)- f(y)l, lx-yl::;8 and uniform continuity means Define Now we write lim w(8) = 0 . .stO 11/11 = sup{lf(x)l: 0 x 1}. sup IBn(X)- f(x)l =sup IE(f(Sn )) - f(x)l x n supE(If(Sn)- f(x)l) x n + Sn ]+211/llsupP[I- -xl > E] x n Var(§a.) 2 n X E 211/11 nx(1- x) w(E) + - 2 - sup 2 E x n 211/111 1 =w(E) + --- ·- E 2 4 n ( by Chebychev ) where we have used So we conclude 1 sup x(1-x) = -.
Image of page 191
Image of page 192
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern