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研究日志

20250304

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\univlogo\title{Research Diary }{\Huge \today}\\[5mm]\textit{This is the research journal of Gu Yubo, an undergraduate. I dream of becoming a scientist and contributing to human civilization.}
\section*{Estimate the total number of taxis in a city}
\subsection*{Background}
People randomly see car numbers on city streets, and taxi license plate numbers are issued sequentially from a certain number. We consider estimating the total number of taxis by noting down license plate numbers. 
\subsection*{Assumptions}
\begin{enumerate}\item Each number in the sample has an equal sampling probability.\item The data follows some existing distributions.
\end{enumerate}\subsection*{Model - building}
Taxi license plate numbers start from 0001 and are issued sequentially. Randomly select \(n\) numbers to form a sample \(\{x_1, x_2, \cdots, x_n\}\) and sort it in ascending order. Let \(x\) be the terminal number, which is also the total number of taxis.\subsubsection*{Model 1 - Mean Value Model}
\begin{enumerate}\item Sample Mean: \(\bar{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i\).\item Population Mean: \(\overline{X}=\frac{1 + x}{2}\).\item Estimation Formula: Let \(\overline{X}=\bar{x}\), then \(x = 2\bar{x}-1\).
\end{enumerate}
\begin{align*}
\frac{1 + x}{2}&=\bar{x}\\
1 + x&=2\bar{x}\\
x&=2\bar{x}-1
\end{align*}\subsubsection* {Model 2 - Median Model}
\begin{enumerate}\item Sample Median: \(\tilde{x}\).\item Population Median: \(\widetilde{X}=\frac{x + 1}{2}\).\item Estimation Formula: Let \(\widetilde{X}=\tilde{x}\), then \(x = 2\tilde{x}-1\).
\end{enumerate}\begin{align*}
\frac{x + 1}{2}&=\tilde{x}\\
x + 1&=2\tilde{x}\\
x&=2\tilde{x}-1
\end{align*}\subsubsection* {Model 3 - Symmetric Interval at Both Ends Model}
\begin{enumerate}\item Sample Interval: Intervals are \(x_1 - 1\) and \(x - x_n\).\item Symmetric Assumption: \(x - x_n = x_1 - 1\).\item Estimation Formula: From \(x - x_n = x_1 - 1\), we get \(x = x_n + x_1 - 1\).
\end{enumerate}
\begin{align*}
x - x_n &= x_1 - 1\\
x&=x_n + x_1 - 1
\end{align*}\subsubsection* {Model 4 - Average Interval Model}
\begin{enumerate}\item Sample Interval Calculation: Arrange \(1\) and the sample as \(\{1, x_1, x_2, \cdots, x_n\}\). The average interval \(\bar{d}=\frac{1}{n}[(x_1 - 1)+\sum_{i = 2}^{n}(x_i - x_{i - 1}-1)]=\frac{1}{n}(x_n - 1)\), estimating \(x - x_n\).\item Estimation Formula: Let \(x - x_n=\frac{1}{n}(x_n - 1)\), then \(x=(1 + \frac{1}{n})x_n - 1\).
\end{enumerate}
\begin{align*}
x - x_n&=\frac{1}{n}(x_n - 1)\\
x&=x_n+\frac{1}{n}(x_n - 1)\\
x&=(1 + \frac{1}{n})x_n - 1
\end{align*}\subsubsection* {Model 5 - Interval Equal Division Model}
\begin{enumerate}\item Interval Division: Divide \([1, x]\) into \(n\) equal parts, with sub - interval length \(l = \frac{x - 1}{n}\).\item Sample Position Assumption: Each \(x_i\) is at the sub - interval midpoint.\item Interval Derivation: \(x - x_n=\frac{x - 1}{2n}\).\item Estimation Formula: From \(x - x_n=\frac{x - 1}{2n}\), we have:
\end{enumerate}
\begin{align*}
2n(x - x_n)&=x - 1\\
2nx-2nx_n&=x - 1\\
2nx-x&=2nx_n - 1\\
x(2n - 1)&=2nx_n - 1\\
x&=(1+\frac{1}{2n - 1})(x_n-\frac{1}{2n})
\end{align*}
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