Raising the bar

Canonical form for time series

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

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

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

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

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

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

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

Properties

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

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

$latex V_t(\varepsilon_{t+1})=E_t(\varepsilon_{t+1}-E_t\varepsilon_{t+1})^2=E_t\varepsilon_{t+1}^2=E_t(Y_{t+1}-E_tY_{t+1})^2=V_t(Y_{t+1}).$

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

$latex Cov_t(\mu_{t+1},\varepsilon_{t+1})=E_t[(\mu_{t+1}-E_t\mu_{t+1})\varepsilon_{t+1}]=E_t[(\mu_{t+1}-\mu_{t+1})\varepsilon_{t+1}]=0$

($latex \mu_{t+1}$ is known at time $latex t$).

They are also unconditionally uncorrelated: by the LIE

$latex Cov(\mu_{t+1},\varepsilon_{t+1})=E[(\mu_{t+1}-E\mu_{t+1})\varepsilon_{t+1}]=EE_t[(\mu_{t+1}-E\mu_{t+1})\varepsilon_{t+1}]=E[(\mu_{t+1}-E\mu_{t+1})E_t\varepsilon_{t+1}]=0.$

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

$latex V(Y_{t+1})=V(\mu_{t+1}+\varepsilon_{t+1})=V(\mu_{t+1})+2Cov(\mu_{t+1},\varepsilon_{t+1})+V(\varepsilon_{t+1})=V(\mu_{t+1})+V(\varepsilon_{t+1}).$

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

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

Question: do the remainders represent white noise?