Oct 17

Canonical form for time series

Canonical form for time series

We start with a doubly-infinite time series \{ Y_t:t=0, \pm 1, \pm 2,... \} . At each point in time t, in addition to Y_t, we are given an information set I_t. It is natural to assume that with time we know more and more: I_t\subset I_{t+1} for all t. We want to apply the idea used in two simpler situations before:

1) Mean-plus-deviation-from-mean representation: Y=\mu+\varepsilon, where \mu=EY, \varepsilon=Y-EY, E\varepsilon=0.

2) Conditional-mean-plus-remainder representation: having some information set I, we can write Y=E_IY+\varepsilon, where E_IY=E(Y|I), \varepsilon=Y-E_IY, E_I\varepsilon=0.

Notation: for any random variable X, the conditional mean E(X|I_t) will be denoted E_{t}X.

Following the above idea, we can write Y_{t+1}=Y_{t+1}-E_tY_{t+1}+E_tY_{t+1}. Hence, denoting

\mu_{t+1}=E_tY_{t+1}, \varepsilon_{t+1}=Y_{t+1}-E_tY_{t+1}, we get the canonical form

(1) Y_{t+1}=\mu_{t+1}+\varepsilon_{t+1}.


a) Conditional mean of the remainder: E_t\varepsilon_{t+1}=E_t(Y_{t+1}-E_tY_{t+1})=0, because E_tE_t=E_t. This implies for the unconditional mean E\varepsilon _{t+1}=0 by the LIE.

b) Conditional variances of Y_{t+1} and \varepsilon _{t+1} are the same:


c) The two terms in (1) are conditionally uncorrelated:


(\mu_{t+1} is known at time t).

They are also unconditionally uncorrelated: by the LIE


d) Full (long-term) variance of Y_{t+1}, in addition to V(\varepsilon_{t+1}), includes variance of the conditional mean \mu _{t}:


e) The remainders are uncorrelated. When considering Cov(\varepsilon_t,\varepsilon_s) for s\neq t, by symmetry of covariance we can assume that t\leq s-1. Then, remembering that \varepsilon_t is known at time s-1, by the LIE we have

Cov(\varepsilon_t,\varepsilon_s)=E[E_{s-1}(\varepsilon_t\varepsilon_s)]=E[\varepsilon_tE_{s-1}\varepsilon_s] =0.

Question: do the remainders represent white noise?

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