Σ1,2=EXt1Xt2(Xt3)=et1-t3et2-t3Therefore,Xt3|Xt1, Xt2∼ N(μ0,Σ0), where :μ0=μt3+ Σ21Σ-111(X(1))Σ0= Σ22-Σ21Σ-111Σ12After the terms are replaced we obtain:μ0=Xt2et2-t3Σ0= 1-e2(t2-t3)BecauseXt3, Xt1, Xt2are Gaussian, thanXt3givenXt1, Xt2is Gaussian with meanμ0and variance Σ0,which are both not function oft1and depend only ont2, thereforeE[Xt3|Xt1, Xt2] depends only onXt2andnotonXt1.b) We will see what is the form ofE[Xt3|Xt2].Σ =Rx(t2, t2)Rx(t2, t3)Rx(t3, t2)Rx(t3, t3)=1et2-t3et2-t31, Σ11= 1 = Σ22and Σ12=et2-t3= Σ21.Page 4 of 5
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ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017Problem 3We can conclude thatXt3|Xt2∼ N(μ0,Σ0) where both the mean and variance are independent oft1:μ0= 0 + Σ21Σ-111(Xt2-0)Σ0= Σ22-Σ21Σ-111Σ12Therefore,f(Xt3|Xt2) is independent ofXt1and soE[Xt3|Xt2] is independent oft1and suchXt1, Xt2, Xt3form a Markov Chain.Page 5 of 5
Probability theory, random process Xt, Behnaam Aazhang, different points of time t1
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