2.3.2. Prove of ${Y}_{n}^{i}$ Being a Martingale Process

At first, it is known that:

$E\left[{z}_{n+1}^{i\left(l\right)}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]={z}_{n}^{i\left(l-1\right)},\text{\hspace{0.17em}}l=1,\cdots ,L,\text{\hspace{0.17em}}n\ge l-1,\text{\hspace{0.17em}}i=1,\cdots ,5$ (5)

According to literature [3] , ${z}_{n}^{i\left(0\right)}=0$ . Thus, regulate that ${z}_{n}^{i\left(0\right)}=0$ .

Then it can be proved that ${Y}_{n}^{i}\left(i=1,\cdots ,5\right)$ is a martingale process, that is to prove

$E\left[{Y}_{n+1}^{i}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]={Y}_{n}^{i},\text{\hspace{0.17em}}n\ge L,\text{\hspace{0.17em}}i=1,\cdots ,5$ (6)

However,

$\begin{array}{l}E\left[{Y}_{n+1}^{i}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]\\ =E\left[{\displaystyle \underset{j=L}{\overset{n+1}{\sum}}{z}_{j}^{i\left(L\right)}}+{\displaystyle \underset{l=1}{\overset{L-1}{\sum}}{z}_{n+1}^{i\left(l\right)}}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]\\ =E\left[{\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{i\left(L\right)}}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]+E\left[{z}_{n+1}^{i\left(L\right)}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]+E\left[{\displaystyle \underset{l=1}{\overset{L-1}{\sum}}{z}_{n+1}^{i\left(l\right)}}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]\\ =E\left[{\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{i\left(L\right)}}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]+E\left[{\displaystyle \underset{l=1}{\overset{L}{\sum}}{z}_{n+1}^{i\left(l\right)}}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]\end{array}$

According to (5), it is clear that:

$E\left[{\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{n\left(L\right)}}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]={\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{n\left(L\right)}}$ (7)

while it is known that:

$E\left[{z}_{n+1}^{i\left(l\right)}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]={z}_{n}^{i\left(l-1\right)}$ (8)

Hence, it can be concluded that:

$E\left[{Y}_{n+1}^{i}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]={\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{i\left(L\right)}}+{\displaystyle \underset{l=1}{\overset{L}{\sum}}{z}_{n}^{i\left(l-1\right)}}$ (9)

As ${z}_{n}^{i\left(0\right)}=0$ , it is clear that:

$\begin{array}{c}E\left[{Y}_{n+1}^{i}|{X}_{1}^{i},\cdots ,{X}_{n}^{i}\right]={\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{i\left(L\right)}}+{\displaystyle \underset{l=1}{\overset{L}{\sum}}{z}_{n}^{i\left(l-1\right)}}={\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{i\left(L\right)}}+{\displaystyle \underset{l=2}{\overset{L}{\sum}}{z}_{n}^{i\left(l-1\right)}}\\ ={\displaystyle \underset{j=L}{\overset{n}{\sum}}{z}_{j}^{i\left(L\right)}}+{\displaystyle \underset{l=2}{\overset{L-1}{\sum}}{z}_{n}^{i\left(l\right)}}={Y}_{n}^{i}\end{array}$

From what have been discussed above, ${Y}_{n}^{i}\left(i=1,\cdots ,5\right)$ is proved to be a mar-

tingale process and $E\left[{Y}_{L}^{i}\right]=E\left[{\displaystyle \underset{l=1}{\overset{L}{\sum}}{z}_{L}^{i\left(l\right)}}\right]={\displaystyle \underset{l=1}{\overset{L}{\sum}}E\left[{z}_{L}^{i\left(l\right)}\right]}=0$ .

2.3.3. ${{\rm N}}_{\iota}$ is a Stopping Time of $\left\{{X}_{n}^{i},n\ge L,i=1,\cdots ,5\right\}$ and Suits What Stopping Time Theory Require

The demonstration can be founded in literature [3] .

2.3.4. Prove $E\left[{N}_{i}\right]=\frac{{p}_{i}^{-L}-1}{1-{p}_{i}},i=1,\cdots ,5$

At first, proof of ${Y}_{{N}_{i}}^{i}={N}_{i}+\frac{1-{p}_{i}^{-L}}{1-{p}_{i}}$ is needed.

According to (4), it can be concluded that:

${Y}_{{N}_{i}}^{i}={\displaystyle \underset{{n}_{i}=L}{\overset{{N}_{i}}{\sum}}{z}_{n}^{\left(L\right)}}+{\displaystyle \underset{l=1}{\overset{L-1}{\sum}}{z}_{N}^{\left(l\right)}}$ (10)

1) For ${z}_{n}^{\left(L\right)},n=L,L+1,\cdots ,{N}_{i}$

As
$\underset{j=n-L+1}{\overset{n}{\sum}}{X}_{j}^{i}}=L{a}_{i$ means there are continuously L units of a_{i}, N_{i} is the first

time of a day that have continuously L units of a_{i} on the ith day of a week. Therefore, only when
$n={N}_{i}$ , it can concluded that
$\underset{j=n-L+1}{\overset{n}{\sum}}{X}_{j}^{i}}=L{a}_{i$ .

As for ${z}_{n}^{\left(L\right)},n=L,L+1,\cdots ,{N}_{i}-1$ , as $\underset{j=n-L+1}{\overset{n}{\sum}}{X}_{j}^{i}}\ne L{a}_{i$ , it is clear that:

${z}_{n}^{\left(L\right)}=\{\begin{array}{l}1-{p}_{i}^{-L}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}n={N}_{i}\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}L\le n\le {N}_{i}-1\end{array}$ (11)

which means:

$\underset{n=L}{\overset{{N}_{i}}{\sum}}{z}_{n}^{\left(L\right)}}={N}_{i}-L+1-{p}_{i}^{-L$ (12)

2) For ${z}_{{N}_{i}}^{\left(l\right)},l=1,\cdots ,L-1$

According to definition of ${z}_{{N}_{i}}^{\left(l\right)},l=1,\cdots ,L-1$ and ${N}_{i}$ , it is obvious that ${z}_{{N}_{i}}^{\left(l\right)}=1-{p}_{i}^{-l},l=1,\cdots ,L-1,i=1,\cdots ,5$ . Thus, it is clear that:

$\underset{l=1}{\overset{L-1}{\sum}}{z}_{{N}_{i}}^{\left(l\right)}}={\displaystyle \underset{l=1}{\overset{L-1}{\sum}}\left(1-{p}_{i}^{-l}\right)}=L-1-\frac{{p}_{i}^{-1}\left(1-{p}_{i}^{-\left(L-1\right)}\right)}{1-{p}_{i}^{-1}}=L-1+\frac{1+{p}_{i}^{1-L}}{1-{p}_{i}$ (13)

According to what have been discussed above, it is known that:

${Y}_{{N}_{i}}^{i}={N}_{i}-L+1-{p}_{i}^{-L}+L-1+\frac{1-{p}_{i}^{1-L}}{1-{p}_{i}}={N}_{i}+\frac{1-{p}_{i}^{-L}}{1-{p}_{i}}$ (14)

As ${N}_{i}$ is a stopping time of $\left\{{X}_{n}^{i},n\ge L,i=1,\cdots ,5\right\}$ and suits what stopping time theory require, it is obvious that:

$E\left[{Y}_{{N}_{i}}^{i}\right]=E\left[{Y}_{L}^{i}\right]$ (15)

That is:

$E\left[{N}_{i}\right]=\frac{{p}_{i}^{-L}-1}{1-{p}_{i}},i=1,\cdots ,5$ (16)

3. Empirical Analysis

3.1. Taking Wuhan University of Technology for Example

Taking Wuhan University of Technology for example, it has 4 districts in Ma Fangshan district, which are Nan Hu district, Jian Hu district, Dong Yuan district and Xi Yuan district, and it also contains an accommodation area for students outside of the campus named Shengsheng. It can be clearly seen from the distribution of the school that the need for students to take the school bus is very urgent as there are 5 districts. Therefore, a good dispatch of the school bus is of a great importance.

To collect data, the number of people those take the school bus in every 3 minutes, from Monday to Friday, 7:30 a.m. to 9:00 a.m. is recorded. By using martingale process to predict peak time for each day, the schedule of the departure time can be optimized.

The data was collected in Nan Hu district, as it is the biggest district of the 5 ones and holds the largest population. There are 14 school buses in Wuhan University of Technology. According to our research, the average traffic for each day can be obtained. After studying on the data, it appears to relate to the total number of courses of the whole school, which makes sense.

3.2. The Determine of a_{i}

Analyzing the data, a_{i} is related to the real situations. Therefore, the values of a_{i} combine with the real situations are determined.

As the number of courses is of a large amount in Monday, the traffic should be largest of the weak. Thus, a peak time for Monday is a period of time when there are more than 9 people arriving at the station continuously in adjacent L unit of time.

As the number of courses is not very crowded in Tuesday and Wednesday, a peak time for Tuesday and Wednesday is a period of time when there are more than 7 people arriving at the station continuously in adjacent L unit of time.

As most professors have their meetings on Thursday, the number of courses is not very large. Thus, a peak time for Thursday is a period of time when there are more than 5 people arriving at the station continuously in adjacent L unit of time.

As there are usually have practical courses on Friday, traffic is a little larger than Thursday. Thus, a peak time for Friday is a period of time when there are more than 6 people arriving at the station continuously in adjacent L unit of time.

That is Table 2.

3.3. The Determine of L

The value of a_{i} has been determined by the feature of the data in section 3.2 while the value of L is not. According to what have been observed in each school bus station in Wuhan University of Technology, it will significantly be crowded when there are a_{i} person arriving at the station to wait for the bus continuously in adjacent 3 unit of time. Thus L is determined as:

$L=3$ . (17)

3.4. Results and Conclusions

Applying the model in section 2, the peak time for each day is predicted as shown in the Table 3.

It can be seen from the table that the peak time for Monday is around 12:05 and 17:25. T the peak time for Tuesday is around 10:33, the peak time for Wednesday is around 14:33, the peak time for Thursday is around 17:25 and the peak time for Friday is around 11:23. According to our research, it is corresponded to the real situation.

As for the occurrence of 2 peak times on Monday, according to the collected data, there are usually more people taking the bus on Monday in Wuhan University of Technology, thus there are 2 peak times that 9 person arrives at the station continuously in adjacent 3 unit of time in the original data. In this way, there are 2 peak times predicted by the model.

4. Conclusions and Suggestions

A peak time prediction model based on martingale process is established and solved with the stopping time theory. Taking Wuhan University of Technology for example, the peak time for each day is predicted by applying the model.

The suggestion is to reduce the time for departure to increase the trips when it comes to peak times. In this way, the efficiency of the school bus will increase and the cost will decrease at the same time. It will also provide a lot of convenience for students and professors. Besides, it is also suggested that students should prevent these time to take the school bus as far as possible.

Acknowledgements

The paper is financially supported by Students innovation and entrepreneurship

Table 2. The determine of a_{i}.

Table 3. The prediction of the peak time for each day.

training program, Wuhan University of Technology, China (No.166814007).

Cite this paper

Zhou, Y. , Liu, S. and Gui, Y. (2017) The Prediction of the Peak Time of People Taking School Bus Based on Martingale Process.*Applied Mathematics*, **8**, 630-636. doi: 10.4236/am.2017.85049.

Zhou, Y. , Liu, S. and Gui, Y. (2017) The Prediction of the Peak Time of People Taking School Bus Based on Martingale Process.

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