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Example text

The solution of (8) is easily accomplished recursively, starting with j = 0. Substituting j = Sn and taking expectation, we obtain P (Sn ∈ A) − Pλ (A) = E{λg(Sn + 1) − Sn g(Sn )}, (9) from which the total variation distance between LSn and Pλ can be found, provided that the RHS of (9) can be uniformly estimated for all the gλ,A . To see how this works, write E(ej g(Sn )) = E(ej g(Snj + 1)) = pj Eg(Snj + 1), where Snj = ei , because of the independence of ej and Snj . Thus i=j n E{λg(Sn + 1) − Sn g(Sn )} = pj E{g(Sn + 1) − g(Snj + 1)}, j=1 and since Sn and it follows that Snj are equal unless ej = 1, an event of probability pj , n |P (Sn ∈ A) − Pλ (A)| p2j .

Of X. Remark. d, we have the following expectation inequality. d. ’s. Then E|X − Y | E|X + Y |. Proof. E|X + Y | − E|X − Y | T (P (|X + Y | > x) − P (|X − Y | > x))dx = lim T →∞ 0 T = lim T →∞ 0 (P (|X − Y | x) − P (|X + Y | x))dx 0. An alternative proof is to use the formula E|X +Y |−E|X − Y | = ∞ 0 (1 − F (u) − F (−u))(1 − G(u) − G(−u))du, where F and G are the distribution functions of X and Y respectively. ’s such that either 1 or (ii) Xn and (i) Xn and (Y1 , · · · , Yn ) are independent for all n 1.

F. F (x). If |x|dF (x) < ∞, we say that the (mathematical) expectation, or mean, of X exists. The expectation is deﬁned by xdF (x) and denoted by EX. Let k be a positive number. If EX k and E|X|k exist, then we call them the moment and the absolute moment of order k (about the origin) respectively. If E(X − EX)k and E|X − EX|k exist, they are called the central moment and the absolute central moment of order k, respectively. In particular, E(X − EX)2 is called the variance of X, usually denoted by VarX.