Questions 1 consider the following models i zt 06zt 1 at...

Questions 1. Consider the following models: (i) Zt = 0.6Zt-1 + at - 1.2at-1 + 0.5at-2 (ii) Zt = Zt-1 - 0.21Zt-2 + at - 0.7at-1 (iii) Zt = -0.4Zt-1 + 0.32Zt-2 + at + 0.6at-1 - 0.16at-1 a. Write the models in the backshift operator form b. State the name of the process and write the parameters c. Show the Φ(B) and Θ(B) d. Is Z stationary and invertible?

Transcript

00:01 We are given for each models and in the first support we are given here that x t is equal to we can write 0 .3 multi material we can write x t minus 1 plus we can write z z t so this values we are given as it as are three equations so let's solve our first equation so in this case we want to find here that find the equivalent am a representation equivalent the m a representation so we can write here in this question we can write the x of x t minus we can write 0 .3 multiple we can write here x of t minus 1 plus we can write x t now we can write the 1 minus 0.

00:52 38 multiple we can write x t which is equal we can write z here now we know that the polynomial here is the we can write the pi b you can write the 5b equal to 1 minus 0 .3 multiple by b now again we can write the 5 of z which is equal to 0 which is equal to 0 .3 multiple by z which is equal to 0 and here we just simplify so we can write here z which is equal to 0 .33 which is greater than 1 now this is the stationary here this is the stationary so this is stationary now here we can write the model this is a r of one model this is a r of one model we are given here it is a stationary and it will be automatically here invertible is stationary the model here that is the model is stationary then it will be then it will be automatically then it will be automatically invertible here invertible so we can write the the model is invertible so we can write here this x t which is equal to 1 minus 0 .3 multiple we can write b raised to minus 1 multiple we can write here z t now we can write here x t which is equal we can write 1 plus 0 .3 multiple by b plus we can write here 1 multiple by here we can write 1 multiple by 0 we can write divided by 2 multiply we can write 0 .3 multiple by b square and here we can make multiple by xx t now again we can write x t which is equal to answer is 1.

03:01 Here 1 plus 0 .3 multiple b plus we can write here 0 .0 here we can write 0 .09 multiplied value 2 not 0...

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