Question

2. Consider the time series X, = 2 + 0.5t +0.8X1-1 + W, where W N(0.1). (a) (8 points) Calculate E(X2) Is this process weakly stationary? Give reasons for your answer. Hint: Find the mean function of {X) and then substitute t = 20. (b) (3 points) Calculate Var(X20) Question 2 continues on the next page... Page 4 of 12 c)(4 points) Consider the first differences of the time series above, that is Is {%) a weakly stationary process. Prove your answer.

Answer 1

Given W N(0,1) Taking Expectations on both sides for(1), we obtain E(X) 2+0.5t+0.8EX)0 Therefore, mean function is E(X) 2+0.5t +0.8E(X ...(2) As the above mean function depends on time t and it is not constant, then the given time series is not weak stationery Now takingt 20 in (2), we have E(X 2+0.5x 20+0.8E(X9) E(X)-12+0.8E(X) (b) Since, X-12+0.8X9+W2 Taking Variances on both sides, then we obtain

ar(X20)-1+0.64Var(X19) (C) Given the difference equation for the above stationery time series as -X-X y-2+0.51+0.8X,-1 +M-2-0.5(-1)-0.8x,,-M-1 Then, 0.5 + 0.8(Xt-1-X,-2) +(M-W,-1) Now, taking expectations on both sides EC)-0.5+0.8E+0-0 E)-0.5+0.8E() As the above mean(Expectation) function does not depend on time t and it gives constant value, then {Yj may become weakly stationery process.

Var(Yt) = E{yt-E(yt)}² and the covariance function is as follows:

COV(Y_t, Y_s) =E[{y_t-E(y_t)}{y_s-E(y_s)} ]

= E[0.8{y_t-1 - E(y_t-1}0.8{y_s-1-E(y_s-1} ]

=0.64E[{y_t-1 - E(y_t-1}{y_s-1-E(y_s-1} ]

=0.64COV(Y_t-1, Y_s-1) which does not depend on time.

Hence, the process {Y_t} is weakly stationery process.