A Probability Path.pdf

# But we know xnk has some as convergent subsequence

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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

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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. -+ .
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

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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) = -.
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