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Unformatted text preview: 512 Solutions Manual for Statistical Inference 2 2 c. Sn 2 in probability. By a), Sn = Sn 2 = in probability. By b), /Sn 1 in probability. 5.33 For all > 0 there exist N such that if n > N , then P (Xn + Yn > c) > 1  . Choose N1 such that P (Xn > m) > 1  /2 and N2 such that P (Yn > c + m) > 1  /2. Then P (Xn + Yn > c) P (Xn > m, +Yn > c + m) P (Xn > m) + P (Yn > c + m)  1 = 1  . 5.34 Using EXn = and VarXn = 2 /n, we obtain n(Xn ) n n E = E(Xn  ) = (  ) = 0. n(Xn ) n n n 2 Var = 2 Var(Xn  ) = 2 VarX = 2 = 1. n 5.35 a. Xi exponential(1). X = 1, VarX = 1. From the CLT, Xn is approximately n(1, 1/n). So Xn 1 Z n(0, 1) 1/ n b.
2 d d 1 P (Z x) = FZ (x) = fZ (x) = ex /2 . dx dx 2 and P Xn 1 x 1/ n P (Z x). d P dx = = Xn 1 x 1/ n d dx
n Xi x n + n n W =
i=1 Xi gamma(n, 1) i=1 d FW (x n + n) = fW (x n + n) n = dx 1 (x n + n)n1 e(x n+n) n. (n) 2 Therefore, (1/(n))(x n + n)n1 e(x n+n) n 1 ex /2 as n . Substituting x = 0 2 yields n! nn+1/2 en 2. 5.37 a. For the exact calculations, use the fact that Vn is itself distributed negative binomial(10r, p). The results are summarized in the following table. Note that the recursion relation of problem 3.48 can be used to simplify calculations. v 0 1 2 3 4 5 6 7 8 9 10 (a) Exact .0008 .0048 .0151 .0332 .0572 .0824 .1030 .1148 .1162 .1085 .0944 P (Vn = v) (b) (c) Normal App. Normal w/cont. .0071 .0056 .0083 .0113 .0147 .0201 .0258 .0263 .0392 .0549 .0588 .0664 .0788 .0882 .0937 .1007 .1100 .1137 .1114 .1144 .1113 .1024 ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics, Probability

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