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Can $$ X_t - a X_{t-1} = Z_t + Z_{t-1} $$ be written with recursion as $$ X_t - a^2 X_{t-2} = Z_t + (1-a) Z_{t-1} $$ where $a = \begin{pmatrix} 0 & 1/3 \\ 1/3 & 0 \end{pmatrix} $?

I wasn't sure of how to find the autocovariance when the MA(1) did not have a companion matrix.

Richard Hardy
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Benis Perovskias
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1 Answers1

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No, it generally cannot, unless $Z_{t-2} = 0$ for all $t$: Define the lag operator $L$ such that $LX_t = X_{t-1}$, and similarly for $Z_t$. Then the equation you start with above can be rewritten as a lag polynomial of form \begin{align} (1-aL)X_t = (1-L)Z_t. \end{align} Using the third binomial formula, it is obvious that $(1+aL)(1-aL) = (1-a^2L^2)$, and as $(1-a^2L^2)X_t = X_t - a^2X_{t-2}$ by definition of the lag operator, one can now write \begin{align} (1+aL)(1-aL)X_t & = (1+aL)(1-L)Z_t\\ (1-a^2L^2)X_t & = (1 + (a-1)L - aL^2)Z_t \\ X_t - a^2X_{t-2} & = Z_t + (a-1)Z_{t-1} - aZ_{t-2}. \end{align} So in particular, $X_t - a^2X_{t-2} = Z_t + (1-a)Z_{t-1}$ if and only if $(a-1)Z_{t-1} - aZ_{t-2} = (1-a)Z_{t-1}$, which obviously only holds if $Z_{t-2} = 0$.

Jeremias K
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    (+1) well the last part could hold for other situations, but ${Z_t}$ wouldn't stationary. – Taylor Dec 10 '17 at 20:56
  • Oh, I think I see what you mean! That is actually a very good point - so you are saying that defining $W_{t-1} = (1-a)Z_{t-1}$, one could write the last part as $2W_{t-1} = -a Z_{t-2}$, which would then imply that we have the relationship $-(2 \cdot a)^{-1}W_{t-1} = Z_{t-2}$, i.e. $-2(1-a)a^{-1}Z_{t-1} = Z_{t-2}$. Now noticing that $a^{-1}Z_{t-1} = 3 Z_{t-1} $ and that $-2(1-a) = ((2,-2/3), (-2/3, 2))$, it becomes clear that indeed as you say, the resulting process ${Z_t}$ is non-stationary. Thanks, that was a great remark! – Jeremias K Dec 10 '17 at 21:43