Question

For each of the following ARMA models, find the roots of the AR and MA polynomials, and identify the values of p and q for which they are ARMA(p,q). Remember to cancel out the common factors of AR and MA polynomials when determining p and q. Assume that {at}t?0 is a zero-mean unit variance white noise sequence. • rt + 0.81rt-2 = at + 1/3at-1. • rt - rt-1 = at - 1/2at-1 - 1/2at-2. • rt - 3rt-1 = at + 2at-1 - 8at-2. • rt - 2rt-1 + 2rt-2 = at - 8/9at-1. • rt - 4rt-2 = at - at-1 + 1/2at-2. • rt - 9/4rt-1 - 9/4rt-2 = at.

          For each of the following ARMA models, find the roots of the AR and MA polynomials, and identify the values of p and q for which they are ARMA(p,q). Remember to cancel out the common factors of AR and MA polynomials when determining p and q. Assume that {at}t?0 is a zero-mean unit variance white noise sequence.

• rt + 0.81rt-2 = at + 1/3at-1.
• rt - rt-1 = at - 1/2at-1 - 1/2at-2.
• rt - 3rt-1 = at + 2at-1 - 8at-2.
• rt - 2rt-1 + 2rt-2 = at - 8/9at-1.
• rt - 4rt-2 = at - at-1 + 1/2at-2.
• rt - 9/4rt-1 - 9/4rt-2 = at.
        
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For each of the following ARMA models, find the roots of the AR and MA polynomials, and identify the values of p and q for which they are ARMA(p,q). Remember to cancel out the common factors of AR and MA polynomials when determining p and q. Assume that att?0 is a zero-mean unit variance white noise sequence.

• rt + 0.81rt-2 = at + 1/3at-1.
• rt - rt-1 = at - 1/2at-1 - 1/2at-2.
• rt - 3rt-1 = at + 2at-1 - 8at-2.
• rt - 2rt-1 + 2rt-2 = at - 8/9at-1.
• rt - 4rt-2 = at - at-1 + 1/2at-2.
• rt - 9/4rt-1 - 9/4rt-2 = at.

Added by Kristi M.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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For each of the following ARMA models, find the roots of the AR and MA polynomials, and identify the values of p and q for which they are ARMA(p,q). Remember to cancel out the common factors of AR and MA polynomials when determining p and q. Assume that {at}t≥0 is a zero-mean unit variance white noise sequence.
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Transcript

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00:01 Here the first equation is x t plus 0 .8 x t plus 2 equal to et plus 1 by 3 et minus 1.
00:14 Now this is of type a rm a of 2 1.
00:19 Now consider 5 of z which is equal to 1 plus 0 .81 z power 4 which is u.
00:27 From this we can which is zero hence we can have 0 .81 z power 4 equal to minus 1 so z power 4 is minus 1 by 0 .81 from this is it equal to square root of minus 1 divided by 0 .81 power 1 by 2 which is i by 0 .9 the second one is x t minus x t minus 1 .1 equal to e .t minus 0 .5e .t minus 1 minus 0 .5e t minus 2.
01:15 This is of type a or ma of 1 comma 2...
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