Backward residuals

We define the vectors of backward residuals r_k,T-1 with the recursions:  

$$\!\!\left\{\begin{array} {lll} \!r_{0,T-1}&=&Zy_T \\ \!r_{k,... ...over r'_{k-1,T-1}r_{k-1,T-1}}}r_{k-1,T-1}\;. \end{array}\right.$$ (1)

The index $T\!-\!1$ appears because these residuals can all be expressed as Z applied to a known vector: r_0,T=y_T for example. These recursions are in fact equivalent to performing a Gram-Schmidt orthogonalization of the space $(Zy_T,\cdots,Z^ny_T)$. Then, for any k, the space $(r_{0,T-1},\cdots,r_{k,T-1})$ is equal to $(Zy_T,\cdots,Z^{k+1}y_T)$; also, r_k,T-1 is orthogonal to $(r_{0,T-1},\cdots,r_{k-1,T-1})$.So, with these recursions, we see that r_k,T-1 is the orthogonal complement of the shifted data Z^k+1y_T on $(Zy_T,\cdots,Z^ky_T)$:

$$r_{k,T-1}=P^{\perp}_{1,k,T}(Z^{k+1}y_T) \mbox{\hspace{1.0cm} for all } k=0,\cdots,n-1\;.$$

In other words, r_k,T-1(t) is the residual obtained from the optimal prediction of y(t-(k+1)) with the future values $y(t-1),\cdots,y(t-k)$;this explains the term ``backward'' residuals. The geometrical interpretations of $\varepsilon_{k,T}$ and r_k,T-1 appear together on Figure .

Geometrical interpretation of the residuals $\varepsilon_{k,T}$ and r_k,T-1.
$\varepsilon_{k,T}$ is the orthogonal complement of the data y_T on Y_1,k,T.
The predictable part of y_T of order k is the orthogonal projection of y_T on Y_1,k,T.
r_k,T-1 is the orthogonal complement of the shifted data Z^k+1y_T on Y_1,k,T.
The new space of regressors Y_1,k+1,T is $Y_{1,k,T}\oplus Z^{k+1}y_T$.
$\varepsilon_{k+1,T}$ is the orthogonal complement of y_T on Y_1,k+1,T.
$\varepsilon_{k+1,T}-\varepsilon_{k,T}$ is equal to r_k,T-1 multiplied by the reflection coefficient K^r_k+1.

Finally, knowing that the forward residuals $\varepsilon_{k,T}$ are the orthogonal complement of y_T on $(Zy_T,\cdots,Z^ky_T)$, so on $(r_{0,T-1},\cdots,r_{k-1,T-1})$, we have the recursive formulas:  

$$\!\!\left\{\begin{array} {lll} \!\varepsilon_{0,T}&=&y_T \no... ...\over r'_{k-1,T-1}r_{k-1,T-1}}}r_{k-1,T-1}\;.\end{array}\right.$$