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\documentclass[a4paper,12pt]{report}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[round]{natbib}
\usepackage{pgf}
\usepackage{url}
\usepackage[english]{babel}
\usepackage{multirow}
\usepackage{fancyhdr}
\usepackage{anysize}
\usepackage{amsmath, mathtools}
\begin{document}
\begin{equation}\label{eq:20}
\begin{split} \
\MoveEqLeft[3] \{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\} = \\
&\{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{split}
\end{equation}
Taking probabilities on both sides and using the property of independen increments of the Poisson Process N, we obtain:
\begin{equation} \label{eq:21}
\begin{split}
\MoveEqLeft[6] P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) \\ P(N(0,x_{1}]=0)P(N(x_{1},x_{1}+h_{1}]=1)P(N(x_{1}+h_{1},x_{2}]=0)= \\
& P(N(x_{2},x_{2}+h{2}]=1) \dots P(N(x_{n-1}+h_{n-1},x_{n}]=0) \\
&P(N(x_{n}, x_{n}+h_{n}]=1) P(N(x_{n}+h_{n},t]=0)= \\
&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]e^{-\mu(x_{1},x_{1}+h_{1}]}]e^{-\mu(x_{1}+h_{1},x_{2}]} \\
&[\mu(x_{2},x_{2}+h_{2}]e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dots e^{-\mu(x_{n-1}+h_{n-1},x_{n}]} \\
&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}= \\
&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dots \mu(x_{n}, x_{n}+h_{n}]
\end{split}
\end{equation}
\end{document}
我用 LaTeX 写这个但出现了错误。
因为我之前有过其他拆分,所以出现错误?
可以帮我解方程吗?
答案1
正如 David Carlisle 所指出的,第二个equation
环境中语法错误的直接来源是 a}
被错误地写成]
。
此外,似乎还有一些数学错误内容,例如方括号放置不正确。我已尽力清理这两个方程的外观。希望您可以使用此代码作为清理剩余问题的中途点。
请注意,我用(“点二进制”)替换了一些\dots
指令\dotsb
,因为它们(即印刷省略号)似乎表示乘法省略。
\documentclass[a4paper,12pt]{report}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
%% Commented out unneeded \usepackage statements:
%\usepackage[round]{natbib}
%\usepackage{pgf}
%\usepackage{url}
%\usepackage[english]{babel}
%\usepackage{multirow}
%\usepackage{fancyhdr}
%\usepackage{anysize}
\usepackage{mathtools} % 'mathtools' loads 'amsmath' automatically
\begin{document}
\setcounter{equation}{19} % just for this example
\begin{equation}\label{eq:20}
\begin{split}
\MoveEqLeft[3]
\bigl\{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n \bigr\} \\
{}=\bigl\{&N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0 \bigr\}
\end{split}
\end{equation}
Taking probabilities on both sides and using the property of independent increments of the Poisson Process $N$, we obtain:
\begin{equation} \label{eq:21}
\begin{split}
\MoveEqLeft[3]
P\bigl(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\bigr) \\
{}={}&P(N(0,x_{1}]=0)\,
P(N(x_{1},x_{1}+h_{1}]=1)\,
P(N(x_{1}+h_{1},x_{2}]=0) \\
&P(N(x_{2},x_{2}+h_{2}]=1) \dotsb
P(N(x_{n-1}+h_{n-1},x_{n}]=0) \\
&P(N(x_{n}, x_{n}+h_{n}]=1)\,
P(N(x_{n}+h_{n},t]=0) \\
{}={}&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]\,
e^{-\mu(x_{1},x_{1}+h_{1}]}]\,
e^{-\mu(x_{1}+h_{1},x_{2}]}[\mu(x_{2},x_{2}+h_{2}] \\
&e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dotsm
e^{-\mu(x_{n-1}+h_{n-1},x_{n}]}[\mu(x_{n}, x_{n}+h_{n}] \\
&e^{-\mu(x_{n}, x_{n}+h_{n}}]\,
e^{\mu(x_{n}+h_{n},t]} \\
{}={}&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dotsb \mu(x_{n}, x_{n}+h_{n}]
\end{split}
\end{equation}
\end{document}
答案2
该行&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}=
缺少一行\}
(第二段倒数第二行)。
这是可以工作的 LaTeX,但要注意,它看起来很糟糕:
\begin{equation}\label{eq:20}
\begin{split} \
\MoveEqLeft[3] \{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\} = \\
&\{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{split}
\end{equation}
Taking probabilities on both sides and using the property of independen increments of the Poisson Process N, we obtain:
\begin{equation} \label{eq:21}
\begin{split}
\MoveEqLeft[6] P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n)
\\
P(N(0,x_{1}]=0)P(N(x_{1},x_{1}+h_{1}]=1)P(N(x_{1}+h_{1},x_{2}]=0)=
\\
& P(N(x_{2},x_{2}+h{2}]=1) \dots P(N(x_{n-1}+h_{n-1},x_{n}]=0)
\\
&P(N(x_{n}, x_{n}+h_{n}]=1) P(N(x_{n}+h_{n},t]=0)=
\\
&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]e^{-\mu(x_{1},x_{1}+h_{1}]}]e^{-\mu(x_{1}+h_{1},x_{2}]}
\\
&[\mu(x_{2},x_{2}+h_{2}]e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dots e^{-\mu(x_{n-1}+h_{n-1},x_{n}]}
\\
&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}}=
\\
&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dots \mu(x_{n}, x_{n}+h_{n}]
\end{split}
\end{equation}
但我必须建议您使用\begin{align}
或\begin{align*}
代替,\begin{equation}\begin{split}
因为它现在已成为常态。
以下是一个例子:
\begin{align*}
&P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) =
\\
& \hspace{1cm} \{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0,
\\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0,
\\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{align*}
请注意\hspace
使行稍晚一些开始,以使其看起来像换行符而不是新行。