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较长的公式导致内容溢出到页脚

tag-icon • tag-icon amsmath
较长的公式导致内容溢出到页脚

我需要将以下方程（所有方程编号相同）放在两页中​​，而不是将其拆分到下一页，它会溢出到页脚中，其余部分根本不显示。我尝试过类似问题的解决方案，但对我来说不起作用。我还包括了方程的所有元素，这样就不会混淆了。

\documentclass{report}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{amssymb,amsfonts,amsmath,mathtools,amsthm,caption}
\DeclareMathOperator{\var}{\mathbb{V}ar}
\DeclareMathOperator{\cov}{\mathbb{C}ov}
\DeclareMathOperator{\expect}{\mathbb{E}}
\begin{document}
\begin{equation}
\begin{aligned}
X_{t}-X_{t-1}&= (1-\alpha_{1}- \alpha_{2})X_{t}+ \alpha_{2}X_{t-1}+X_{t-   2}+\varepsilon_{t}\\
\rho &=\dfrac{\cov(X_{t},X_{t+k})}{\sqrt{\var(X_{t}).\var(X_{t+k})}}\\  
\cov(X_{t},X_{t+k}) &= \expect(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})\\
&= \sum_{X_{t}} \sum_{X_{t+k}}(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})p(X_{t},X_{t+k}) \text{ for  $(X_{t},X_{t+k})$ discrete}\\
&= \int\limits_{- \infty}^{ \infty} \int\limits_{-\infty}^{\infty} (X- \mu_{X})(Y-\mu_{Y})f(x,y)dxdy \text{ for $(X,Y)$ continuous}\\ 
\text{Racall that }\\ 
\cov(X_{t},X_{t+k}) = \expect(X_{t}-\mu_{x_t})(X_{t+k}- \mu_{x_{t+k}})&=\expect(X_{t}(X_{t+k}-\mu_{x_{t+k}})-\mu_{x_t}(X_{t+k}-\mu_{x_{t+k}}))\\
&=\expect(X_{t}X_{t+k}-\mu_{x_{t+k}}X_{t}-\mu_{x_t}X_{t+k}+\mu_{x_t}\mu_{x_{t+k}})\\
&=\expect(X_{t}X_{t+k})-\mu_{x_{t+k}}\expect(X_{t})-\mu_{X}\expect(X_{t+k})+\mu_{x_t}\mu_{x_{t+k}}\\
&=\expect(X_{t}X_{t+k})-\mu_{x_{t+k}}\mu_{x_t}-    \mu_{X}\mu_{x_{t+k}}+\mu_{x_t}\mu_{x_{t+k}}\\
&=\expect(X_{t}X_{t+k})-2\mu_{x_t}\mu_{x_{t+k}}+\mu_{x_t}\mu_{x_{t+k}}\\
\cov(X_{t},X_{t+k}) = \expect(X_{t}-\mu_{x_t})(X_{t+k}- \mu_{x_{t+k}})&=\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}\\
\text{Recall also that }\\
\var(X_{t})&=\expect(X_{t}-\expect(X_{t})^2)\\
&=\expect(X_{t}-\expect(X_{t}))(X_{t}-\expect(X_{t}))\\
&=\expect(X_{t}^2-2X\expect(X_{t})+(\expect(X_{t})^2))\\
&=\expect(X_{t}^2)-\expect(2X_{t}\expect(X_{t}))+(\expect(X_{t}))^2\\
&=\expect(X_{t}^2)-2\expect(X_{t})\expect(X_{t})+(\expect(X_{t}))^2\\
&=\expect(X_{t}^2)-2(\expect(X_{t}))^2+(\expect(X_{t}))^2\\
&=\expect(X_{t}^2)-(\expect(X_{t}))^2\\
\var(X_{t})&=\sigma^2\\
\text{Similarly}\\
\var(X_{t+k})&=\expect(X_{t+k}-\expect(X_{t+k})^2)\\
&=\expect(X_{t+k}-\expect(X_{t+k}))(X_{t+k}-\expect(X_{t+k}))\\
&=\expect(X_{t+k}^2-2X\expect(X_{t+k})+(\expect(X_{t+k})^2))\\
&=\expect(X_{t+k}^2)-\expect(2X_{t+k}\expect(X_{t+k}))+(\expect(X_{t+k}))^2\\
&=\expect(X_{t+k}^2)-2\expect(X_{t+k})\expect(X_{t+k})+(\expect(X_{t+k}))^2\\
&=\expect(X_{t+k}^2)-2(\expect(X_{t+k}))^2+(\expect(X_{t+k}))^2\\
&=\expect(X_{t+k}^2)-(\expect(X_{t+k}))^2\\
\var(X_{t+k}) &=\sigma^2\\
\text{From }\rho_{t,t+k} &=\dfrac{\cov(X_{t},X_{t+k})} {\sqrt{\var(X_{t})\cdot\var(X_{t+k})}}\\  
\rho_{t,t+k} &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2\cdot\sigma_{x_{t+k}}^2}}\\
&=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2}\cdot\sqrt{\sigma_{x_{t+k}}^2}}\\
&=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}   {\sigma_{x_{t}}\cdot\sigma_{x_{t+k}}}\\
&=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}    {\sigma_{x}\cdot\sigma_x}\\
&=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sigma_x^2}\\
\label{eq3.5}
\end{aligned}
\end{equation}
\end{document}

请帮忙。

答案1

像这样？

在此处输入图片描述

基于猜测并且不知道，你喜欢对这个巨大推导的哪一部分进行编号：

\documentclass{report}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{amssymb,amsfonts,amsthm,mathtools}
\usepackage{changepage}
\usepackage{caption,subcaption}
\usepackage{enumitem,subcaption,tocbibind}
\usepackage{enumitem,subcaption}

\DeclareMathOperator{\cov}{cov}
\DeclareMathOperator{\var}{var}
\DeclareMathOperator{\expect}{E}

\begin{document}
\begin{align*}
X_{t}-X_{t-1}
    &= (1-\alpha_{1}- \alpha_{2})X_{t}+ \alpha_{2}X_{t-1}+X_{t-   2}+\varepsilon_{t}\\
\rho &=\dfrac{\cov(X_{t},X_{t+k})}{\sqrt{\var(X_{t}).\var(X_{t+k})}}\\
\cov(X_{t},X_{t+k}) &= \expect(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})\\
    &= \sum_{X_{t}} \sum_{X_{t+k}}(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})p(X_{t},X_{t+k}) \text{ for  $(X_{t},X_{t+k})$ discrete}\\
    &= \int\limits_{- \infty}^{ \infty} \int\limits_{-\infty}^{\infty} (X- \mu_{X})(Y-\mu_{Y})f(x,y)dxdy 
    \text{ for $(X,Y)$ continuous}
\intertext{Racall that}
\cov(X_{t},X_{t+k}) 
    & = \expect(X_{t}-\mu_{x_t})(X_{t+k}- \mu_{x_{t+k}})    \\
    &=\expect(X_{t}(X_{t+k}-\mu_{x_{t+k}})-\mu_{x_t}(X_{t+k}-\mu_{x_{t+k}}))\\
    &=\expect(X_{t}X_{t+k}-\mu_{x_{t+k}}X_{t}-\mu_{x_t}X_{t+k}+\mu_{x_t}\mu_{x_{t+k}})\\
    &=\expect(X_{t}X_{t+k})-\mu_{x_{t+k}}\expect(X_{t})-\mu_{X}\expect(X_{t+k})+\mu_{x_t}\mu_{x_{t+k}}\\
    &=\expect(X_{t}X_{t+k})-\mu_{x_{t+k}}\mu_{x_t}-    \mu_{X}\mu_{x_{t+k}}+\mu_{x_t}\mu_{x_{t+k}}\\
    &=\expect(X_{t}X_{t+k})-2\mu_{x_t}\mu_{x_{t+k}}+\mu_{x_t}\mu_{x_{t+k}}\\
\cov(X_{t},X_{t+k}) 
    &= \expect(X_{t}-\mu_{x_t})(X_{t+k}- \mu_{x_{t+k}})\\
    &=\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}
\intertext{Recall also that }
\var(X_{t})
    &=\expect(X_{t}-\expect(X_{t})^2)\\
    &=\expect(X_{t}-\expect(X_{t}))(X_{t}-\expect(X_{t}))\\
    &=\expect(X_{t}^2-2X\expect(X_{t})+(\expect(X_{t})^2))\\
    &=\expect(X_{t}^2)-\expect(2X_{t}\expect(X_{t}))+(\expect(X_{t}))^2\\
    &=\expect(X_{t}^2)-2\expect(X_{t})\expect(X_{t})+(\expect(X_{t}))^2\\
    &=\expect(X_{t}^2)-2(\expect(X_{t}))^2+(\expect(X_{t}))^2\\
    &=\expect(X_{t}^2)-(\expect(X_{t}))^2\\
\var(X_{t})
    &=\sigma^2
\intertext{Similarly}
\var(X_{t+k})
    &=\expect(X_{t+k}-\expect(X_{t+k})^2)\\
    &=\expect(X_{t+k}-\expect(X_{t+k}))(X_{t+k}-\expect(X_{t+k}))\\
    &=\expect(X_{t+k}^2-2X\expect(X_{t+k})+(\expect(X_{t+k})^2))\\
    &=\expect(X_{t+k}^2)-\expect(2X_{t+k}\expect(X_{t+k}))+(\expect(X_{t+k}))^2\\
    &=\expect(X_{t+k}^2)-2\expect(X_{t+k})\expect(X_{t+k})+(\expect(X_{t+k}))^2\\
    &=\expect(X_{t+k}^2)-2(\expect(X_{t+k}))^2+(\expect(X_{t+k}))^2\\
    &=\expect(X_{t+k}^2)-(\expect(X_{t+k}))^2\\
\var(X_{t+k}) 
    &=\sigma^2
\intertext{From}
\rho_{t,t+k} 
    &=\dfrac{\cov(X_{t},X_{t+k})} {\sqrt{\var(X_{t})\cdot\var(X_{t+k})}}\\
\rho_{t,t+k}
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2\cdot\sigma_{x_{t+k}}^2}}\\
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2}\cdot\sqrt{\sigma_{x_{t+k}}^2}}\\
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}   {\sigma_{x_{t}}\cdot\sigma_{x_{t+k}}}\\
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}    {\sigma_{x}\cdot\sigma_x}\\
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sigma_x^2}
\label{eq3.5}
\end{align*}
\end{document}

附录

一些（非主题的）评论：

  • 对于方差、协方差和期望，通常使用符号 Var、Cov 和 E，即其定义应为：

\DeclareMathOperator{\var}{Var}
\DeclareMathOperator{\cov}{Cov}
\DeclareMathOperator{\expect}{E}
  • 方程式只有一个数字，而方程式实际上是由文本分隔并分布在两页上的方程式块，这种做法很不可靠，可能会导致得出这样的结论：该数字属于方程式的特定部分。由于解决方案可以使用子方程编号，例如

\begin{subequations}\label{eq3.5}
\begin{align}
    \begin{split}\label{eq3.5a}
...
    \end{split}
\intertext{Racall that}
    \begin{split}\label{eq3.5b}
...
    \end{split}
\intertext{Recall also that}
    \begin{split}\label{eq3.5c}
...
    \end{split}
\intertext{Similarly}
    \begin{split}\label{eq3.5d}
...
    \end{split}
\intertext{From}
    \begin{split}\label{eq3.5e}
...
    \end{split}
\end{align}
\end{subequations}

得出（仅显示最后三个方程式）：

在此处输入图片描述

  • 使用上述方程编号可以将“方程”分解为具有共同主数字的五个方程的简单序列。在这种情况下，文本\intertext{...}将转换为文档中的普通文本。

  • 重新考虑一下，是否真的有必要将推导的每一步都写在新的一行中。将一些步骤合并成一行，“方程式”会变得更短。例如

\intertext{From}
    \begin{split}\label{eq3.5e}
\rho_{t,t+k} 
    &=\dfrac{\cov(X_{t},X_{t+k})} {\sqrt{\var(X_{t})\cdot\var(X_{t+k})}}
     =\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2\cdot\sigma_{x_{t+k}}^2}}\\
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2}\cdot\sqrt{\sigma_{x_{t+k}}^2}}
     =\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}   {\sigma_{x_{t}}\cdot\sigma_{x_{t+k}}}
     =\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}    {\sigma_{x}\cdot\sigma_x}\\
\rho_{t,t+k}
    &=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sigma_x^2}
    \end{split}

这使

在此处输入图片描述

  • 你的“方程式”的第一部分可能最好写成

    \begin{split}\label{eq3.5a}
X_{t}-X_{t-1}
    &= (1-\alpha_{1}- \alpha_{2})X_{t}+ \alpha_{2}X_{t-1}+X_{t-   2}+\varepsilon_{t}\\
\rho&=\dfrac{\cov(X_{t},X_{t+k})}{\sqrt{\var(X_{t}).\var(X_{t+k})}}\\
\cov(X_{t},X_{t+k}) 
    &= \expect(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})\\
    &=  \begin{dcases}
    \sum_{X_{t}}\sum_{X_{t+k}}(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})p(X_{t},X_{t+k}) 
        &   \text{ for  $(X_{t},X_{t+k})$ discrete}\\
    \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} (X - \mu_{X})(Y-\mu_{Y})f(x,y)dxdy 
        &   \text{ for $(X,Y)$ continuous}
        \end{dcases}
    \end{split}

这使

在此处输入图片描述

答案2

希望 autobreak.sty 可以帮到你，这里是代码：

    \documentclass{report}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{amssymb,amsfonts,amsmath,mathtools,amsthm,caption}
\usepackage{autobreak}
\DeclareMathOperator{\var}{\mathbb{V}ar}
\DeclareMathOperator{\cov}{\mathbb{C}ov}
\DeclareMathOperator{\expect}{\mathbb{E}}
\begin{document}

\begin{align}
\everyafterautobreak{\quad}
\begin{autobreak}
X_{t}-X_{t-1}= (1-\alpha_{1}- \alpha_{2})X_{t}+ \alpha_{2}X_{t-1}+X_{t- 2}+\varepsilon_{t} \rho
=\dfrac{\cov(X_{t},X_{t+k})}{\sqrt{\var(X_{t}).\var(X_{t+k})}} \cov(X_{t},X_{t+k})
= \expect(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})
= \sum_{X_{t}} \sum_{X_{t+k}}(X_{t}-\mu_{X_{t}})(X_{t+k}-\mu_{X_{t+k}})p(X_{t},X_{t+k})
\text{ for $(X_{t},X_{t+k})$ discrete} = \int\limits_{- \infty}^{ \infty} \int\limits_{-\infty}^{\infty} (X- \mu_{X})(Y-\mu_{Y})f(x,y)dxdy
\text{ for $(X,Y)$ continuous} \text{Racall that } \cov(X_{t},X_{t+k})
= \expect(X_{t}-\mu_{x_t})(X_{t+k}- \mu_{x_{t+k}})=\expect(X_{t}(X_{t+k}
-\mu_{x_{t+k}})-\mu_{x_t}(X_{t+k}-\mu_{x_{t+k}}))
=\expect(X_{t}X_{t+k}-\mu_{x_{t+k}}X_{t}-\mu_{x_t}X_{t+k}+\mu_{x_t}\mu_{x_{t+k}})
=\expect(X_{t}X_{t+k})-\mu_{x_{t+k}}\expect(X_{t})-\mu_{X}\expect(X_{t+k})+\mu_{x_t}\mu_{x_{t+k}}
=\expect(X_{t}X_{t+k})-\mu_{x_{t+k}}\mu_{x_t}- \mu_{X}\mu_{x_{t+k}}+\mu_{x_t}\mu_{x_{t+k}}
=\expect(X_{t}X_{t+k})-2\mu_{x_t}\mu_{x_{t+k}}+\mu_{x_t}\mu_{x_{t+k}} \cov(X_{t},X_{t+k})
= \expect(X_{t}-\mu_{x_t})(X_{t+k}- \mu_{x_{t+k}})=\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}
\text{Recall also that } \var(X_{t})=\expect(X_{t}-\expect(X_{t})^2)
=\expect(X_{t}-\expect(X_{t}))(X_{t}-\expect(X_{t})) =\expect(X_{t}^2-2X\expect(X_{t})
+(\expect(X_{t})^2)) =\expect(X_{t}^2)-\expect(2X_{t}\expect(X_{t}))+(\expect(X_{t}))^2
=\expect(X_{t}^2)-2\expect(X_{t})\expect(X_{t})+(\expect(X_{t}))^2
=\expect(X_{t}^2)-2(\expect(X_{t}))^2+(\expect(X_{t}))^2 =\expect(X_{t}^2)-(\expect(X_{t}))^2 \var(X_{t})
=\sigma^2 \text{Similarly} \var(X_{t+k})=\expect(X_{t+k}-\expect(X_{t+k})^2)
=\expect(X_{t+k}-\expect(X_{t+k}))(X_{t+k}-\expect(X_{t+k})) =\expect(X_{t+k}^2-2X
\expect(X_{t+k})+(\expect(X_{t+k})^2)) =\expect(X_{t+k}^2)-\expect(2X_{t+k}\expect(X_{t+k}))+(\expect(X_{t+k}))^2
=\expect(X_{t+k}^2)-2\expect(X_{t+k})\expect(X_{t+k})+(\expect(X_{t+k}))^2 =\expect(X_{t+k}^2)
-2(\expect(X_{t+k}))^2+(\expect(X_{t+k}))^2 =\expect(X_{t+k}^2)-(\expect(X_{t+k}))^2 \var(X_{t+k})
=\sigma^2 \text{From }\rho_{t,t+k} =\dfrac{\cov(X_{t},X_{t+k})} {\sqrt{\var(X_{t})\cdot\var(X_{t+k})}} \rho_{t,t+k}
=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2\cdot\sigma_{x_{t+k}}^2}}
=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sqrt{\sigma_{x_{t}}^2}\cdot\sqrt{\sigma_{x_{t+k}}^2}}
=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}} {\sigma_{x_{t}}\cdot\sigma_{x_{t+k}}}
=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}} {\sigma_{x}\cdot\sigma_x}
=\dfrac{\expect(X_{t}X_{t+k})-\mu_{x_t}\mu_{x_{t+k}}}{\sigma_x^2}
\end{autobreak}\label{eq3.5}
\end{align}

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