Σ 1 2 E X t 1 X t 2 X t 3 e t 1 t 3 e t 2 t 3 Therefore X t 3 X t 1 X t 2 N μ Σ

# Σ 1 2 e x t 1 x t 2 x t 3 e t 1 t 3 e t 2 t 3

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Σ 1 , 2 = E X t 1 X t 2 ( X t 3 ) = e t 1 - t 3 e t 2 - t 3 Therefore, X t 3 | X t 1 , X t 2 ∼ N ( μ 0 , Σ 0 ), where : μ 0 = μ t 3 + Σ 21 Σ - 1 11 ( X (1) ) Σ 0 = Σ 22 - Σ 21 Σ - 1 11 Σ 12 After the terms are replaced we obtain: μ 0 = X t 2 e t 2 - t 3 Σ 0 = 1 - e 2( t 2 - t 3 ) Because X t 3 , X t 1 , X t 2 are Gaussian, than X t 3 given X t 1 , X t 2 is Gaussian with mean μ 0 and variance Σ 0 , which are both not function of t 1 and depend only on t 2 , therefore E [ X t 3 | X t 1 , X t 2 ] depends only on X t 2 and not on X t 1 . b) We will see what is the form of E [ X t 3 | X t 2 ]. Σ = R x ( t 2 , t 2 ) R x ( t 2 , t 3 ) R x ( t 3 , t 2 ) R x ( t 3 , t 3 ) = 1 e t 2 - t 3 e t 2 - t 3 1 , Σ 11 = 1 = Σ 22 and Σ 12 = e t 2 - t 3 = Σ 21 . Page 4 of 5

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ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017 Problem 3 We can conclude that X t 3 | X t 2 ∼ N ( μ 0 , Σ 0 ) where both the mean and variance are independent of t 1 : μ 0 = 0 + Σ 21 Σ - 1 11 ( X t 2 - 0) Σ 0 = Σ 22 - Σ 21 Σ - 1 11 Σ 12 Therefore, f ( X t 3 | X t 2 ) is independent of X t 1 and so E [ X t 3 | X t 2 ] is independent of t 1 and such X t 1 , X t 2 , X t 3 form a Markov Chain. Page 5 of 5
• Spring '14
• Aazhang,Behnaam
• Probability theory, random process Xt, Behnaam Aazhang, different points of time t1

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