Since by assumption that the process is stationary E(Y)=E(Y-1)=1 herefore =c++0 reproducing the earlier result To find a higher moments of Yt in an analogous manner, we rewrite this AR(1) Yt=1(1-)+Yt-1+Et (Yt-u=o(Yi-1-F)+Et For j20, multiply (Yi-i-u)on both side of (3)and take expectation E E(Y-1-1)(Y=1-1)+E(Y-1-1)et mj-1+ E(Y-,)Et Next we consider the term E(Yt-i-pEt When j=0, multiply Et on both side of (3) and take expectation E(Y1-p)t=E[0(Y1-1-)E+E(e2) Recall that Yt-1-u is a linear function of Et-1, Et-2, 62 E[0(Y1-1-1)2]=0.Since by assumption that the process is stationary, E(Yt) = E(Yt−1) = µ. Therefore, µ = c + φµ + 0 or µ = c 1 − φ , reproducing the earlier result. To find a higher moments of Yt in an analogous manner, we rewrite this AR(1) as Yt = µ(1 − φ) + φYt−1 + εt or (Yt − µ) = φ(Yt−1 − µ) + εt . (3) For j ≥ 0, multiply (Yt−j − µ) on both side of (3) and take expectation: γj = E[(Yt − µ)(Yt−j − µ)] = φE[(Yt−1 − µ)(Yt−j − µ)] + E(Yt−j − µ)εt = φγj−1 + E(Yt−j − µ)εt . Next we consider the term E(Yt−j − µ)εt . When j = 0, multiply εt on both side of (3) and take expectation: E(Yt − µ)εt = E[φ(Yt−1 − µ)εt ] + E(ε 2 t ). Recall that Yt−1 − µ is a linear function of εt−1, εt−2, ... : Yt−1 − µ = εt−1 + φεt−2 + φ 2 εt−3 + ..... we have E[φ(Yt−1 − µ)εt ] = 0. 10