If show that (a) E[X 2n ] = 1 · 3 · 5 ··· (2n - 1)
Question:
show that
(a) E[X2n] = 1 · 3 · 5 ··· (2n - 1) Ï2n, for n = 1,2,...
(b) E[X2n-1] = 0 for n = 1,2,...
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fx(x) -(2π σ') exp 2σ2
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a The integral evaluates as follows where y x 2 12 ...View the full answer
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Related Book For
Principles of Communications Systems, Modulation and Noise
ISBN: 978-8126556793
7th edition
Authors: Rodger E. Ziemer, William H. Tranter
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