Expectation, variance and covariance question

whui

New Member
#1
Hi

I have a question which I stuck some point.

Question:
X_t, X_s, Y_t, Y_s, Z_t, Z_s, (~Z_t), (~Z_s) are i.i.d time series r.v. (~ is tilda above the Z)
t,s, are integers

Given,
X_t = Z_t + (theta)*Z_t-1
X_s = Z_s + (theta)*Z_s-1
Y_t = (~Z_t) + 1/(theta)*(~Z_t-1)
Y_s = (~Z_s) + 1/(theta)*(~Z_s-1)

Distribution:
Z_t, Z_s, Z_t-1, Z_s-1 ~ N(0, sigma^2)
~Z_t, ~Z_s, ~Z_t-1, ~Z_s-1 ~ N(0, (sigma*theta)^2)
Also,
E[Z_t * Z_s] = E[~Z_t * ~Z_s]=0, for t not= s

Show that the auto covariance of Y_t and X_t are the same, i.e. show
Cov(Y_t, Y_s) = Cov(X_t, X_s)



My effort: I tried to solve as far as I can go.
I used Xt to stand for X_t, and so on for all X,Y,Z,~Z

I expanded the covariance
Cov(Xt, Xs) = E[Xt * Xs] - E[Xt]E[Xs]
Cov(Yt, Ys) = E[Yt * Ys] - E[Yt]E[Ys]

Also,
E[Xt]=E[Zt] +(theta)*E[Z_t-1].
E[Xs]=E[Zs] +(theta)*E[Z_s-1].
E[Yt]=E[~Zt] + 1/(theta)*E[~Z_t-1].
E[Ys]=E[~Zs] + 1/(theta)*E[~Z_s-1].

Since E[Zt]=E[Z_t-1]=E[Zs]=E[Z_s-1]=0,
and E[~Zt]=E[~Z_t-1]=E[~Zs]=E[~Z_s-1]=0,
so E[Xt]=E[Xs]=0, and E[Yt]=E[Ys]=0

So, Cov(Xt, Xs) = E[Xt * Xs] - 0 = E[Xt * Xs]
Cov(Yt, Ys) = E[Yt * Ys] - 0 = E[Yt * Ys]

Now,
Xt * Xs = Zt*Zs +(theta)(Zs*Z_t-1 + Zt*Z_s-1) + (theta)^2 (Z_s-1)(Z_t-1)
Yt * Ys = ~Zt*~Zs +1/(theta)(~Zs*~Z_t-1 + ~Zt*~Z_s-1) + (theta)^(-2) (~Z_s-1)(~Z_t-1)

E[Xt * Xs] = E[Zt*Zs] +(theta)E[Zs*Z_t-1 + Zt*Z_s-1] +(theta)^2 E[Z_s-1 *Z_t-1]
E[Yt * Ys] = E[~Zt*~Zs] +1/(theta)E[~Zs*~Z_t-1 + ~Zt*~Z_s-1] +(theta)^(-2) E[~Z_s-1 * ~Z_t-1]

If(t!=s), then E[Zt*Zs]=E[~Zt*~Zs]=0

If(t==s), then E[Zt^2]=Var(Zt)+{E[Zt]}^2 = Var(Zt) = (sigma)^2
and E[~Zt^2]=Var(~Zt)+{E[~Zt]}^2 = Var(~Zt) = (sigma * theta)^2

I stuck here. Any help and hints are highly appreciated.
Thanks.