A general framework for trajectory data warehousing and visual OLAP

Abstract

In this paper we present a formal framework for modelling a trajectory data warehouse (TDW), namely a data warehouse aimed at storing aggregate information on trajectories of moving objects, which also offers visual OLAP operations for data analysis. The data warehouse model includes both temporal and spatial dimensions, and it is flexible and general enough to deal with objects that are either completely free or constrained in their movements (e.g., they move along a road network). In particular, the spatial dimension and the associated concept hierarchy reflect the structure of the environment in which the objects travel. Moreover, we cope with some issues related to the efficient computation of aggregate measures, as needed for implementing roll-up operations. The TDW and its visual interface allow one to investigate the behaviour of objects inside a given area as well as the movements of objects between areas in the same neighbourhood. A user can easily navigate the aggregate measures obtained from OLAP queries at different granularities, and get overall views in time and in space of the measures, as well as a focused view on specific measures, spatial areas, or temporal intervals. We discuss two application scenarios of our TDW, namely road traffic and vessel movement analysis, for which we built prototype systems. They mainly differ in the kind of information available for the moving objects under observation and their movement constraints.

A Appendix

This appendix includes the proofs of the propositions presented in Section 5, as well as some formal definitions and lemmas that are only referred inside proofs.

The spatio-temporal operators provided by the MOD allow to load into the TDW the correct values for the measures at the base granularity. Here we prove that they remain correct also when roll-up operations are performed, as stated by Proposition 1. The proof of this proposition is divided into two lemmata, i.e, Lemmas 1 and 4.

We start with a lemma which takes into account the aggregate functions for all the measures except $\mathcal{V}$.

Lemma 1

Let $\mathcal{H}_{TS}$ be a hierarchy, let $\ensuremath{\textsf{T}}$ be a set of trajectories and U be an object group. For any $G\in \mathcal{H}_{TS}$ with $G\neq \ensuremath{G_\bot}$ and g, g′ ∈ G with g ≠ g′

$$\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = \Sigma_{g_p \subseteq g}\mathcal{S}^{\ensuremath{\textsf{T}}}(g_p, U) \\ $$

(2)

$$\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) = \Sigma_{g_p \subseteq g}\mathcal{E}^{\ensuremath{\textsf{T}}}(g_p, U) \\ $$

(3)

$${\ensuremath{\mathit{dist}}}^{\ensuremath{\textsf{T}}}(g, U) = \Sigma_{g_p \subseteq g} {\ensuremath{\mathit{dist}}}^{\ensuremath{\textsf{T}}}(g_p, U) \\ $$

(4)

$${\ensuremath{\mathit{trav\_t}}}^{\ensuremath{\textsf{T}}}(g, U) = \Sigma_{g_p \subseteq g} {\ensuremath{\mathit{trav\_t}}}^{\ensuremath{\textsf{T}}}(g_p, U)\\ $$

(5)

$$\mathcal{C}^{\ensuremath{\textsf{T}}}(g, g', U) = \Sigma_{g_p \subseteq g, g_p' \subseteq g'}\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p, g_p', U) \label{eq:cross} $$

(6)

where g _p , g′ _p  ∈ G _p with g _p  ≠ g′ _p and $G_p \preceq G$ and G _p  ≠ G.

Proof

$$ \begin{array}{lll} &&\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) =\\[1mm] && \qquad {\rm by~Definition~of}~\mathcal{S} \\[1mm] &=& |\{id\ |\ id \in U\, \wedge \delta(\ensuremath{\textsf{T}}_{id}) = \langle s_{id}^1, \ldots, s_{id}^n\rangle\, \wedge\, s_{id}^1 \subseteq g \}| \\[1mm] && \qquad {\rm by~Definition~3},~\ensuremath{G_\bot} \preceq G_P~{\rm and}~ G_p \preceq G \\[1mm] &=& |\{id\ |\ id \in U\, \wedge \, \delta(\ensuremath{\textsf{T}}_{id}) = \langle s_{id}^1, \ldots, s_{id}^n\rangle\, \wedge\, s_{id}^1 \subseteq g_p\\ && \qquad \qquad \qquad \qquad \mathit{for \ some} \ g_p \subseteq g \}| \\[1mm] && \qquad {\rm since~G_p~is~a~partition~g_p~is~unique} \\[1mm] &=& \Sigma_{g_p \subseteq g} |\{id\ |\ id \in U\, \wedge \delta(\ensuremath{\textsf{T}}_{id}) = \langle s_{id}^1, \ldots, s_{id}^n\rangle\, \wedge\, s_{id}^1 \subseteq g_p\}| \\[1mm] &&\qquad {\rm by~Definition~of}~\mathcal{S} \\[1mm] &=& \Sigma_{g_p \subseteq g} \mathcal{S}^{\ensuremath{\textsf{T}}}(g_p, U) \end{array} $$

The remaining statements are proved in a similar way.□

The case of measure $\mathcal{V}$ is more complex and requires the introduction of an auxiliary measure B, counting the number of times trajectories cross the border of a granule.

Definition 8

Let G be a spatio-temporal granularity and g ∈ G be a granule, let $\ensuremath{\textsf{T}}$ be a set of trajectories and let U be an object group. We denote by ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g,U)$ the number of intersections between the trajectories of object group U and the border of the spatio-temporal granule g.

$$ B^{\ensuremath{\textsf{T}}}(g, U) = \Sigma_{g' \in G, g'\neq g}\left(\mathcal{C}^{\ensuremath{\textsf{T}}}(g,g',U) + \mathcal{C}^{\ensuremath{\textsf{T}}}(g',g,U)\right) $$

The following lemma states how measure B can be computed by using sub-aggregates of $\mathcal{C}$.

Lemma 2

Let $\mathcal{H}_{TS}$ be a hierarchy, let $\ensuremath{\textsf{T}}$ be a set of trajectories and U be an object group. For any $G\in \mathcal{H}_{TS}$ with $G\neq \ensuremath{G_\bot}$ and g ∈ G

$$ B^{\ensuremath{\textsf{T}}}(g, U) = \sum\limits_{g_p\subseteq g, \, g_p'\not \subseteq g}\left(\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g_p', U) + \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p',g_p, U)\right) $$

where g _p , g′ _p  ∈ G _p with $G_p \preceq G$ and G _p  ≠ G.

Proof

$$ \begin{array}{rll} && B^{\ensuremath{\textsf{T}}}(g,U) = \\[1mm] && \qquad \mathrm{by\, definition \,of\,} B \\[1mm] &&= \sum_{g'\in G, \, g'\neq g}(\mathcal{C}^{\ensuremath{\textsf{T}}}(g,g',U)+\mathcal{C}^{\ensuremath{\textsf{T}}}(g',g,U)) \\[1mm] &&\qquad \mathrm{by\, statement~\,(6) \,of \,Lemma~1} \\[1mm] &&= \sum_{g'\in G, \, g'\neq g} (\sum_{g_p \subseteq g, g_p' \subseteq g'}\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g'_p,U) +\\ &&\qquad\qquad\qquad \sum_{g_p \subseteq g, g_p' \subseteq g'}\mathcal{C}^{\ensuremath{\textsf{T}}}(g'_p,g_p,U)) \\[1mm] &&= \sum_{g'\in G, \, g'\neq g} (\sum_{g_p \subseteq g, g_p' \subseteq g'}\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g'_p,U) + \mathcal{C}^{\ensuremath{\textsf{T}}}(g'_p,g_p,U)) \\[1mm] && \qquad G \,\mathrm{is\, a \,partition} \\[1mm] &&= \sum_{g_p \subseteq g, g_p' \not\subseteq g}(\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g'_p,U) + \mathcal{C}^{\ensuremath{\textsf{T}}}(g'_p,g_p,U)) \end{array}$$

□

The next lemma provides an inductive characterisation of measure ${\ensuremath{B}}$.

Lemma 3

Let $\mathcal{H}_{TS}$ be a hierarchy, let $\ensuremath{\textsf{T}}$ be a set of trajectories and U be an object group. For any $G\in \mathcal{H}_{TS}$ with $G\neq \ensuremath{G_\bot}$ and g ∈ G

$$ B^{\ensuremath{\textsf{T}}}(g, U) = \Sigma_{g_p \subseteq g}{\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g_p, U) -2 \Sigma_{g_p, g_p' \subseteq g, g_p \neq g_p'}\left(\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g_p', U) + \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p',g_p, U)\right) $$

where g _p , g′ _p  ∈ G _p with $G_p \preceq G$ and G _p  ≠ G.

Proof

This property easily follows by definition of ${\ensuremath{B}}$. More in detail, for the sake of simplicity assume g = g _p  ∪ g _p ′. Then the border of the granule g consists of the union of the borders of g _p and g _p ′ minus the common border between the granules g _p and g _p ′. As a consequence trajectories crossing the common border remain inside the granule g and they should not be counted in ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U)$. In formulas:

$$\begin{array}{lll} {\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) &=& {\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g_p, U) - \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g_p',U) - \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p',g_p,U) \\ &&+{\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g_p',U) - \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p',g_p,U) - \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p,g_p',U) \end{array}$$

This is exactly the desired result.□

Now we show how the measure $\mathcal{V}$ can be expressed analytically in terms of ${\ensuremath{B}}$, $\mathcal{S}$ and $\mathcal{E}$.

Proposition 4

Let G be a spatio-temporal granularity, let $\ensuremath{\textsf{T}}$ be a set of trajectories and let U be an object group. Then for each g ∈ G the following statement holds:

$$ \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U)=\frac{B^{\ensuremath{\textsf{T}}}(g, U)+\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U)+\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U)}{2} $$

Proof

We observe that, as obvious from its definition, $\mathcal{V}$ can be computed by summing up the contributions given to such a measure separately by each trajectory. More precisely, let g be a granule and $\ensuremath{\textsf{T}} = \{\ensuremath{\textsf{T}}_{id}\}_{id\in\mathcal{I}}$ a set of trajectories. Then $\mathcal{V}^{\ensuremath{\textsf{T}}}(g,U) = \sum_{id\in \mathcal{I}} \mathcal{V}^{\{ \ensuremath{\textsf{T}}_{id} \}}(g,U)$.

Therefore, we can prove the proposition for a single trajectory $\ensuremath{\textsf{T}}_{id}$ and thesis will trivially extend to a generic set of trajectories.

Let $\delta(\ensuremath{\textsf{T}}_{id}) = \langle s_{id}^1, \ldots, s_{id}^n\rangle$, we prove the thesis by induction on n (the number of sub-trajectories of the trajectory decomposition).

• [ n = 1]   In this case $\delta(\ensuremath{\textsf{T}}_{id}) = \langle s_{id}^1\rangle$. If $s_{id}^1 \subseteq g$, then $\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = 1$, ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) = 0$, $\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = 1$, and $\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U)= 1$. Thus the thesis holds.

  If, instead, $s_{id}^1 \not \subseteq g$, then $\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = 0$, ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) = 0$, $\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = 0$, and $\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) = 0$ and the thesis holds.

  Notice that in both cases the measure ${\ensuremath{B}}$ is equal to 0 since at least two sub-trajectories are necessary to have ${\ensuremath{B}}>0$ because each sub-trajectory belongs to exactly one granule.

• [ $n\Rightarrow n+1$ ]    In this case $\delta(\ensuremath{\textsf{T}}_{id}) = \langle s_{id}^1,\ldots,s_{id}^n, s_{id}^{n+1}\rangle$. Let $\delta({\ensuremath{\textsf{T}}}_{id}')= \langle s_{id}^1,\ldots,s_{id}^n\rangle$. We consider four cases:

  • [case $s_{id}^n \subseteq g$ and $s_{id}^{n+1} \subseteq g$ ]   By definition $\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{S}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{E}^{\ensuremath{\textsf{T}}'}(g, U)$, and ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U)={\ensuremath{B}}^{\ensuremath{\textsf{T}}'}(g, U)$. Thus, the thesis holds by inductive hypothesis.

  • [case $s_{id}^n \not \subseteq g$ and $s_{id}^{n+1} \subseteq g$ ]   By definition $\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}'}(g, U) + 1$, $\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{S}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{E}^{\ensuremath{\textsf{T}}'}(g, U) + 1$, and ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) = {\ensuremath{B}}^{\ensuremath{\textsf{T}}'}(g, U)+1$. Thus by using the inductive hypothesis

    $$\begin{array}{lll} \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) &=& \mathcal{V}^{\ensuremath{\textsf{T}}'}(g, U) + 1 \\ &=& \displaystyle \frac{{\ensuremath{B}}^{\ensuremath{\textsf{T}}'}(g, U) + \mathcal{S}^{\ensuremath{\textsf{T}}'}(g, U)+\mathcal{E}^{\ensuremath{\textsf{T'}}}(g, U)}{2} + 1 \\ &=&\displaystyle \frac{{\ensuremath{B}}^{\ensuremath{\textsf{T}}'}(g, U) + 1 +\mathcal{S}^{\ensuremath{\textsf{T}}'}(g, U)+\mathcal{E}^{\ensuremath{\textsf{T}}'}(g, U) + 1}{2} \\ &=&\displaystyle \frac{{\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U)+\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U)+\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U)}{2} \end{array}$$

  • [case $s_{id}^n \subseteq g$ and $s_{id}^{n+1} \not \subseteq g$ ]    By definition $\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{S}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{E}^{\ensuremath{\textsf{T}}'}(g, U) - 1$, and $B^{\ensuremath{\textsf{T}}}(g, U) = {\ensuremath{B}}^{\ensuremath{\textsf{T}}'}(g, U) + 1$. As in the previous case we can conclude.

  • [case $s_{id}^n \not \subseteq g$ and $s_{id}^{n+1}\not \subseteq g$ ]   By definition $\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{S}^{\ensuremath{\textsf{T}}'}(g, U)$, $\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{E}^{\ensuremath{\textsf{T}}'}(g, U)$, and ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) = {\ensuremath{B}}^{\ensuremath{\textsf{T}}'}(g, ,U)$. Thus the thesis holds by inductive hypothesis.□

The following lemma concludes the proof for measure $\mathcal{V}$.

Lemma 4

Let $\mathcal{H}_{TS}$ be a hierarchy, let $\ensuremath{\textsf{T}}$ be a set of trajectories and U be an object group. For any $G\in \mathcal{H}_{TS}$ with $G\neq \ensuremath{G_\bot}$ and g ∈ G

$$ \mathcal{V}^{\ensuremath{\textsf{T}}}(g,U) = \Sigma_{g_p \subseteq g} \left(\mathcal{V}^{\ensuremath{\textsf{T}}}(g_p, U) - \Sigma_{g_p' \subseteq g}\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p, g_p', U)\right) $$

where g _p  ∈ G _p with $G_p \preceq G$ and G _p  ≠ G.

Proof

For the sake of simplicity, let g = g _p  ∪ g′ _p with g _p , g′ _p  ∈ G _p and g _p  ≠ g′ _p .

$$\begin{array}{lll} &&\mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) =\\[1mm] &&\qquad {\rm by~Proposition~4} \\[1mm] &=& \displaystyle \frac{B^{\ensuremath{\textsf{T}}}(g,U)+\mathcal{S}^{\ensuremath{\textsf{T}}}(g,U) +\mathcal{E}^{\ensuremath{\textsf{T}}}(g,U)}{2} \\[1mm] &&\qquad {\rm by~Lemma~3~and}~~g = g_p \cup g'_p \\[1mm] &=& \dfrac{{\ensuremath{B}}^{\ensuremath{\textsf{T}}}\!(g_p,U) \!+\! {\ensuremath{B}}^{\ensuremath{\textsf{T}}}\!(g'_p,U) {\kern-1.5pt} -{\kern-1.5pt} 2 \left(\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p, g_p', U){\kern-1.5pt} +{\kern-1.5pt} \mathcal{C}^{\ensuremath{\textsf{T}}}(g'_p, g_p, U)\right) {\kern-1.5pt} \!+\! \mathcal{S}^{\ensuremath{\textsf{T}}}(g,U)\!+\!\mathcal{E}^{\ensuremath{\textsf{T}}}(g,U)}{2} \end{array}$$

By Lemma 1, we have

$$ \mathcal{S}^{\ensuremath{\textsf{T}}}(g,U) = \mathcal{S}^{\ensuremath{\textsf{T}}}(g_p,U) \!+\! \mathcal{S}^{\ensuremath{\textsf{T}}}(g'_p,U) $$

and

$$ \mathcal{E}^{\ensuremath{\textsf{T}}}(g,U) = \mathcal{E}^{\ensuremath{\textsf{T}}}(g_p,U) + \mathcal{E}^{\ensuremath{\textsf{T}}}(g'_p,U) $$

By Proposition 4

$$ \mathcal{V}^{\ensuremath{\textsf{T}}}(g_p, U) = \displaystyle \frac{B^{\ensuremath{\textsf{T}}}(g_p,U)+\mathcal{S}^{\ensuremath{\textsf{T}}}(g_p,U)+\mathcal{E}^{\ensuremath{\textsf{T}}}(g_p,U)}{2} $$

and

$$ \mathcal{V}^{\ensuremath{\textsf{T}}}(g'_p, U) = \displaystyle \frac{B^{\ensuremath{\textsf{T}}}(g'_p,U)+\mathcal{S}^{\ensuremath{\textsf{T}}}(g'_p,U)+\mathcal{E}^{\ensuremath{\textsf{T}}}(g'_p,U)}{2} $$

Hence we can conclude

$$ \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}}(g_p, U) + \mathcal{V}^{\ensuremath{\textsf{T}}}(g'_p, U) - \mathcal{C}^{\ensuremath{\textsf{T}}}(g_p, g_p', U) - \mathcal{C}^{\ensuremath{\textsf{T}}}(g'_p, g_p, U) $$

This is the thesis since $\mathcal{C}^{\ensuremath{\textsf{T}}}(g, g, U) = 0$ for any granule g.□

The join of Lemmas 1 and 4 is exactly Proposition 1.

Proposition 1

Let $\mathcal{H}_{TS}$ be a hierarchy, let $\ensuremath{\textsf{T}}$ be a set of trajectories and U be an object group. For any $G\in \mathcal{H}_{TS}$ with $G\neq \ensuremath{G_\bot}$ , g, g′ ∈ G with g ≠ g′

$$\begin{array}{rll} \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) \> &=& \Sigma_{g_p \subseteq g} \left(\mathcal{V}^{\ensuremath{\textsf{T}}}(g_p, U) - \Sigma_{g_p' \subseteq g}\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p, g_p', U)\right) \\[1mm] \mathcal{S}^{\ensuremath{\textsf{T}}}(g, U) \> &=& \Sigma_{g_p \subseteq g}\mathcal{S}^{\ensuremath{\textsf{T}}}(g_p, U) \\[1mm] \mathcal{E}^{\ensuremath{\textsf{T}}}(g, U) \> &=& \Sigma_{g_p \subseteq g}\mathcal{E}^{\ensuremath{\textsf{T}}}(g_p, U)\\[1mm] \mathcal{C}^{\ensuremath{\textsf{T}}}(g, g', U) &=& \Sigma_{g_p \subseteq g, g_p' \subseteq g'}\mathcal{C}^{\ensuremath{\textsf{T}}}(g_p, g_p', U) \\[1mm] {\ensuremath{\mathit{dist}}}^{\ensuremath{\textsf{T}}}(g, U)&=& \Sigma_{g_p \subseteq g} {\ensuremath{\mathit{dist}}}^{\ensuremath{\textsf{T}}}(g_p, U)\\[1mm] {\ensuremath{\mathit{trav\_t}}}^{\ensuremath{\textsf{T}}}(g, U) \> &=& \Sigma_{g_p \subseteq g} {\ensuremath{\mathit{trav\_t}}}^{\ensuremath{\textsf{T}}}(g_p, U) \end{array}$$

where g _p , g _p ′ ∈ G _p with g _p  ≠ g _p ′ and G _P is a predecessor of G , i.e., $G_p \preceq G$ and G _p  ≠ G.

We finally prove the assertion regarding the relation between visits and presence.

Proposition 3

Let G be a spatio-temporal granularity and g ∈ G, let $\ensuremath{\textsf{T}}$ be a set of trajectories and let U be an object group, then

• 1. if each trajectory visits g at most once then $\mathcal{P}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U)$

• 2. if ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) = 0$ then $\mathcal{P}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U)$ .

Proof

The first statement is a straightforward consequence of definitions $\mathcal{V}$ and $\mathcal{P}$.

In order to prove the second statement observe that the hypothesis ${\ensuremath{B}}^{\ensuremath{\textsf{T}}}(g, U) = 0$ means that all the trajectories are completely contained in a granule g. Hence ${\ensuremath{\cal S}}^{\ensuremath{\textsf{T}}}(g, U) = {\ensuremath{\cal E}}^{\ensuremath{\textsf{T}}}(g, U) = \mathcal{P}^{\ensuremath{\textsf{T}}}(g, U)$. Then, by Proposition 4, we have that

$$ \mathcal{V}^{\ensuremath{\textsf{T}}}(g, U) = \frac{\mathcal{S}^{\ensuremath{\textsf{T}}}(g, U)+\mathcal{E}^{\ensuremath{\textsf{T}}}(g, U)}{2} = \frac{2 \mathcal{S}^{\ensuremath{\textsf{T}}}(g, U)}{2} = \mathcal{P}^{\ensuremath{\textsf{T}}}(g, U) $$

□