Consider the feature stream $ {\bf X}= \{{\bf x}_1, {\bf x}_2 \ldots {\bf x}_{\mbox{\tiny $T$}}\}$ , where $ {\bf x}_t \in {\cal R}^D$. For simplicity, we assume that the first derivatives are obtained by the first-order difference: $ {\bf d}_t = {\bf x}_t - {\bf x}_{t-1}$ and define the DAF as

$\displaystyle {\bf z}_t = \left[ \begin{array}{c} {\bf x}_t  {\bf d}_{t} \end...
...t[ \begin{array}{c} {\bf x}_t {\bf x}_{t} - {\bf x}_{t-1} \end{array}\right].$

Note that we are forced to eliminate one time step, thus $ {\bf Z}= \{{\bf z}_2, {\bf z}_3 \ldots {\bf z}_{\mbox{\tiny $T$}}\}.$ For analysis, it is more convenient to work with the equivalent history form of the DAF defined by $ {\bf Y}= \{{\bf y}_2, {\bf y}_3 \ldots {\bf y}_{\mbox{\tiny $T$}}\},$ where

$\displaystyle {\bf y}_t = \left[ \begin{array}{c} {\bf x}_{t-1}  {\bf x}_{t} \end{array}\right].$ (15.1)

From a theoretical point of view, features $ {\bf y}_t$ and $ {\bf z}_t$ are equivalent since we can obtain $ {\bf z}$ from $ {\bf y}$ by linear transformation $ {\bf T}$ with determinant 1.

Baggenstoss 2017-05-19