# Modeling the Accuracy of Estimating a Neighbor's Evolving Position in VANET
## 2 Analytic Model for the Estimation Error of a Neighbor's Evolving Position
### Estimation Error of a Neighbor's Evolving Position  
The procedure for deriving an estimation error of a neighbor's evolving position consists of two steps. In the first step, we derive an estimation error according to two variables: (1) an acceleration of a neighbor (\(a\)) and (2) \(n_{f}\). Here, \(n_{f}\) represents the number of failures in receiving a beacon by an ego vehicle from the most recent reception of a beacon (e.g., \(n_{f}=1\) at the second estimation and \(n_{f}=2\) at the third estimation in Figure 2). In the second step, we average the estimation error obtained in step one over the two variables.  
First, we derive the estimation error according to two variables, which can be obtained by  
\[e_{n_{f},a}^{est}=x_{cur}-\widetilde{x}_{n_{f},a} \tag{1}\]  
where \(x_{cur}\) is the current position of a neighbor and \(\widetilde{x}_{n_{f},a}\) is an estimated position according to two variables, \(n_{f}\) and \(a\). \(x_{cur}\) and \(\widetilde{x}_{n_{f},a}\) can be described as  
\[\begin{split} x_{cur}&=x_{0}+\int_{0}^{n_{f}T_{BI}}v (\tau)d\tau\\ \widetilde{x}_{n_{f},a}&=x_{0}+v(0)\cdot n_{f}T_{BI} \end{split} \tag{2}\]  
where \(x_{0}\) is an actual position of a neighbor at the time when an ego vehicle receives a beacon from a neighbor. Here, we set the time of receiving a beacon by an ego vehicle to 0 for the ease of explanation. However, we can set this time to another value without reducing the accuracy of our model. \(v(\tau)\) is  
Figure 3: Average acceleration according to time in 2 by 2 grid obtained by SUMO: (**a**) Krauss model without lane changing model and (**b**) Krauss model with lane changing model.  
the velocity of a neighbor at time \(\tau\); \(T_{BI}\) is a beacon interval. By applying Equation (2) to Equation (1), we can express \(e_{n_{f},a}^{est}\) as  
\[\begin{split} e_{n_{f},a}^{est}&=\int_{0}^{n_{f}T_{ BI}}(v(\tau)-v(0))d\tau\\ &=\int_{0}^{n_{f}T_{BI}}\big{(}\int_{0}^{\tau}a(u)du\big{)}d\tau \end{split} \tag{3}\]  
where \(a(u)\) is an acceleration at time \(u\).  
Second, we average \(e_{n_{f},a}^{est}\) over \(a\) and can obtain the estimation error according to \(n_{f}\) (\(e_{n_{f}}^{est}\)) as follows.  
\[\begin{split} e_{n_{f}}^{est}&=E_{a}[e_{n_{f},a}^{ est}]\\ &=\int_{0}^{n_{f}T_{BI}}\big{(}\int_{0}^{\tau}E_{a}[a(u)]du\big{)}d \tau\\ &=\int_{0}^{n_{f}T_{BI}}(a_{m}\tau)d\tau\\ &=\frac{a_{m}}{2}\cdot T_{BI}^{2}\cdot n_{f}^{2}\end{split} \tag{4}\]  
where \(a_{m}\) is an average acceleration.  
Finally, we average \(e_{n_{f}}^{est}\) over \(n_{f}\) to get an estimation error \(e^{est}\), which can be expressed as  
\[\begin{split} e^{est}&=E_{n_{f}}[e_{n_{f}}^{est}]\\ &=\sum_{n=0}^{n_{f}^{max}}\frac{a_{m}}{2}\cdot T_{BI}^{2}\cdot n ^{2}\cdot p_{n|n_{f}^{max}}\\ &=\sum_{n=0}^{n_{f}^{max}}\frac{a_{m}}{2}\cdot T_{BI}^{2}\cdot n ^{2}\cdot\frac{(1-p_{s})^{n}\cdot p_{s}}{1-(1-p_{s})^{n_{f}^{max}+1}}\end{split} \tag{5}\]  
where \(n_{f}^{max}\) is the maximum number of \(n_{f}\), and \(p_{n|n_{f}^{max}}\) is a probability that \(n_{f}\) is equal to \(n\) on the condition that \(n_{f}\leq n_{f}^{max}\). \(n_{f}\) is equal to \(n\) when the following two events happen simultaneously. First, an ego vehicle must receive a beacon from the neighbor. Second, after the beacon reception, the ego vehicle must not have received a beacon \(n\) times from the neighbor. Thus, \(p_{n|n_{f}^{max}}\) can be expressed as \(\frac{(1-p_{s})^{n}\cdot p_{s}}{1-(1-p_{s})^{n_{f}^{max}+1}}\) where \(p_{s}\) is a probability of successful reception. We derive \(p_{s}\) in the next subsection.