Question

Let Xt[1] = Zt[1] - 3Zt-2[1] + 5Zt-2[2] and Xt[2] = Zt[2] + 4Zt-2[1] + Zt-2[2], where Z is WN(0,C) with covariance matrix C = [1 0.5; 0.5 2]. (Here, for the vector-valued processes X and Z, [1] and [2] denote their first and second components.) Calculate the mean function and the covariance function of X. Is X stationary?

Name: let xt1 zt1 3zt 21 5zt 22 and xt2 zt2 4zt 21 zt 22 where z is wn0c with covariance matrix c1 05 05 2 here for the vector valued processes x and z 1 and 2 denotes their first and second compo 79916
Uploaded: 2023-06-04T01:50:20-08:00
Duration: 3 min 53 s
Channel: Adi S
Description: let xt1 zt1 3zt 21 5zt 22 and xt2 zt2 4zt 21 zt 22 where z is wn0c with covariance matrix c1 05 05 2 here for the vector valued processes x and z 1 and 2 denotes their first and second compo 79916

          Let Xt[1] = Zt[1] - 3Zt-2[1] + 5Zt-2[2] and Xt[2] = Zt[2] + 4Zt-2[1] + Zt-2[2], where Z is WN(0,C) with covariance matrix C = [1 0.5; 0.5 2]. (Here, for the vector-valued processes X and Z, [1] and [2] denote their first and second components.) Calculate the mean function and the covariance function of X. Is X stationary?

Added by Antonia R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/04/2023

Step 1

Since Z is a white noise process with mean 0, the mean function of X is also 0. So, we have: $$ E[X_t] = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Show more…

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Let Xt[1] = Zt[1] - 3Zt-2[1] + 5Zt-2[2] and Xt[2] = Zt[2] + 4Zt-2[1] + Zt-2[2], where Z is WN(0,C) with covariance matrix C = [1 0.5; 0.5 2]. (Here, for the vector-valued processes X and Z, [1] and [2] denote their first and second components.) Calculate the mean function and the covariance function of X. Is X stationary?