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5.6 Determine whether the following models are stationary and/or invertible and explain your answers. (a) $X_t - 1.6X_{t-1} + X_{t-2} = (1 - .9B)a_t$ (b) $X_t - 2.6X_{t-1} + 1.5X_{t-2} - .38X_{t-3} - .72X_{t-4} = a_t - 1.9a_{t-1} + 1.7a_{t-2} - .72a_{t-3}$ (c) $X_t = a_t - 2.9a_{t-1} + 2.7a_{t-2} - 1.52a_{t-3}$ (d) $X_t - 1.9X_{t-1} + 2.5X_{t-2} - 2.24X_{t-3} + 1.36X_{t-4} - .576X_{t-5} = a_t - .9a_{t-1}$ 

          5.6 Determine whether the following models are stationary and/or invertible and explain your answers.
(a) $X_t - 1.6X_{t-1} + X_{t-2} = (1 - .9B)a_t$
(b) $X_t - 2.6X_{t-1} + 1.5X_{t-2} - .38X_{t-3} - .72X_{t-4} = a_t - 1.9a_{t-1} + 1.7a_{t-2} - .72a_{t-3}$
(c) $X_t = a_t - 2.9a_{t-1} + 2.7a_{t-2} - 1.52a_{t-3}$
(d) $X_t - 1.9X_{t-1} + 2.5X_{t-2} - 2.24X_{t-3} + 1.36X_{t-4} - .576X_{t-5} = a_t - .9a_{t-1}$
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5.6 Determine whether the following models are stationary and/or invertible and explain your answers. (a) Xt - 1.6Xt-1 + Xt-2 = (1 - .9B)at (b) Xt - 2.6Xt-1 + 1.5Xt-2 - .38Xt-3 - .72Xt-4 = at - 1.9at-1 + 1.7at-2 - .72at-3 (c) Xt = at - 2.9at-1 + 2.7at-2 - 1.52at-3 (d) Xt - 1.9Xt-1 + 2.5Xt-2 - 2.24Xt-3 + 1.36Xt-4 - .576Xt-5 = at - .9at-1

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Computer Science and Information Technology
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Trishna Knowledge Systems 2018 Edition
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5.6 Determine whether the following models are stationary and/or invertible and explain your answers. aX,-1.6X-1+X-2=(1-.9B a bX,-2.6X,-+1.5X,-2-.38X,-3-.72X,-4=a,-1.9a-1+1.7a-2-.72a-3 cX,=a,-2.9a,-1+2.7a,-2-1.52a-3 dX,-1.9X,-1+2.5X,-2-2.24X,-3+1.36X,-4-.576X,-s=a,-.9a,-1
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Transcript

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00:01 Hello, so here you consider our given matrix.
00:03 The matrix, matrix, one, two, three, four, as the first row.
00:06 And then we have two, four, seven, eleven, and then we have three, seven, fourteen, twenty five, and then we have four, eleven, twenty five, and fifty.
00:20 So now we just go ahead and do, um, do, um, row operations and we, um, are invertible.
00:28 If and only if we do the row reduced echelon form, so the row reduced echelon form of our matrix a is going to be equal to the identity, i...
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