Let {Xn}1 be a sequence of random variables. (i) Prove that if the sequence converges in probability, then the limiting random vari- able is unique P-a.s., i.e., if X, 4 X and Xn 4 Y, then P(X = Y) = 1. 4 X. (ii) Using chebyshev's inequality, show that if E [(Xn – X)²] → 0 then X,,

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 68E
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Let {Xn}1 be a sequence of random variables.
n=1
(i) Prove that if the sequence converges in probability, then the limiting random vari-
able is unique P-a.s., i.e.,
if X, 4 X and X, 4Y, then P(X = Y) = 1.
(11) Using chebyshev's inequality, show that if E [(Xn – X)²] → 0 then X,
4 x.
Transcribed Image Text:Let {Xn}1 be a sequence of random variables. n=1 (i) Prove that if the sequence converges in probability, then the limiting random vari- able is unique P-a.s., i.e., if X, 4 X and X, 4Y, then P(X = Y) = 1. (11) Using chebyshev's inequality, show that if E [(Xn – X)²] → 0 then X, 4 x.
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