[SOLVED] CS MP, MS, DT.

MP, MS, DT.
F70TS2 Time Series
Solution to Exercise Sheet 1 Stationarity and the autocorrelation function
Solution 1 To calculate the autocorrelation function we first calculate the autocovariance function
and
and
Var(Xt)=E[Xt2]
(0) =
= E[(t1 + t)2]
= E[22t1 + 2t + 2t1t]
= 2E[2t1] + E[2t ]
= (1+2)2,
(1) =
= E[ 2t1 ]
E[(t2 + t1)(t1 + t)] = 2,
(k) = E[(tk1 + tk)(t1 + t)] =0,
for k 2 because of {t} is i.i.d with mean zero. Hence, the autocorrelation function is (1) = 1, (1)= ,and(k)=0fork2. For||<1itholds1+2 >2and1+2 >2.
1+2 So1< <1. 2 1+2 2Solution 2We present a detailed solution for (c), which is the most difficult case. The solutions to (a) and (b) are derived by following the same steps, and are omitted as a result.Note that t are iid with mean zero.(0) = var (Xt) = var (t + 0.6t1 + 0.3t2 0.2t3)= var (t) + var (0.6t1) + var (0.3t2) + var (0.2t3) = 1.492.(1) = cov (Xt, Xt+1) = cov (t + 0.6t1 + 0.3t2 0.2t3, t+1 + 0.6t + 0.3t1 0.2t2)= 1 0.6var (t) + 0.6 0.3var (t1) 0.3 0.2var (t2) = 0.722.(2) = cov (Xt, Xt+2) = cov (t + 0.6t1 + 0.3t2 0.2t3, t+2 + 0.6t+1 + 0.3t 0.2t1)= 1 0.3var (t) 0.6 0.2var (t1) = 0.182.(3) = cov (Xt, Xt+3) = cov (t + 0.6t1 + 0.3t2 0.2t3, t+3 + 0.6t+2 + 0.3t+1 0.2t)= 1 0.2var (t) = 0.22.(k) = 0 for k > 3.
And (0) = 1, (1) = 0.483, (2) = 0.121, (3) = 0.134, (k) = 0 for k > 3. Solution 3
(i) Since E[Zt] = 0, E[Yt] = E[Yt1] for all t, and so the mean of the process is constant. But
Var(Yt) = Var(Yt1) + Var(Zt) + 2Cov(Yt1, Zt) = Var(Yt1) + Var(Zt)
So, for stationarity we would need Var(Zt) = 0, which is not the case. Hence, the process
is not stationary. It is, in fact, a random walk. 1

(ii) E[Yt] = E[Yt1]+ so for the mean to be constant we would need = 0, which is not the case. Hence, the process is not stationary. It is a random walk with deterministic drift.
(iii) E[Yt] = E[Yt1], so for stationarity in mean we have = , so = 0.
Var(Yt) = 2Var(Yt1) + Var(Zt) + 2Cov(Yt1, Zt) = 2Var(Yt1) + Var(Zt)
So for stationarity we have have 0 = 20 + Z2 , so 0 = Z2 /(1 2). Note that the variance 0 increases as || increases and 0 Z2 as || 0. With = 0, Yt is itself just a noise process.
Cov(Yt, Yt+k) = Cov(Yt, Yt+k1 + Zt+k)
= Cov(Yt, Yt+k1) + Cov(Yt, Zt+k) = Cov(Yt, Yt+k1)
fork1. So,forstationaritywehavek =k1 fork1. Thatis,1 =0, 2 = 1 = 20 and generally k = k0. This gives k which depends only on k, not on t, so the process is stationary.
In fact, this is a Markov time series process, an autoregressive process of order 1. (iv) E[Yt] = E[Zt1]E[Yt2] + E[Zt] = 0, so the process is stationary in mean.
WehaveY2=Z2 Y2 +Z2+2ZZ Y ,andso t t1 t2 t t t1 t2
E[Y2]=E[Z2 ]E[Y2 ]+E[Z2]+2E[Z]E[Z ]E[Y ] t t1 t2 t t t1 t2
So for stationarity we would have 0 = 0 + 1, which is not possible for finite 0. Hence, the process is not stationary.
Solution 4 E[Yt] = E[Ut]+E[Vt] = U +V . This does not depend on t and so Y is stationary in mean.
Cov(Yt, Yt+k) = Cov(Ut + Vt, Ut+k + Vt+k)
= Cov(Ut, Ut+k) + Cov(Ut, Vt+k) + Cov(Vt, Ut+k) + Cov(Vt, Vt+k) = kU +0+0+kV
since the U and V processes are independent. So, {Yt} is stationary in secondorder moments and hence (weakly) stationary.
Solution 5 From Question 4, {Yt} is stationary with kY = kU + kZ. Recall that kZ = 0 for k1. Hence,Y2 =0Y =0U +0Z =U2 +Z2 andkY =kU fork1. Therefore
Y kY kU Uk Uk k=Y=2+2= Z2=1+1
0UZ1+U2 SNR
for k 1. In practice we see that Yk < Uk , i.e. the autocorrelation of the observed signal is less than that of the pure signal.If the noise variance is small compared to that of the signal, the signal is not disturbed very much: the SNR is large and we have Yk Uk (the observed signal has much the same autocorrelation structure as the pure signal). If the noise variance is of the same order as that of the signal then Yk 0.5Uk . If the noise variance is large compared to that of the signal, the signal is seriously disturbed: the SNR is small and Yk is much smaller than Uk (the observed signal has much weaker autocorrelation structure than the pure signal). In general, increasing the variance of the noise imposed on the pure signal weakens the autocorrelation structure of the observed signal.2