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在多列环境中使用对齐

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在多列环境中使用对齐

即使我按照社区的建议操作，代码仍然不起作用。这是怎么回事？

\subsubsection{Score function vs. Influence function}
\begin{multicols}{2}
    \paragraph{Score Function}
    Let $l(\theta|y)$ be the log-likelihood for $\theta \in \R$ and $U_\theta(y) = \dfrac{\partial}{\partial \theta} l(\theta|y)$ be the score function, we have
    \noindent   
    \begin{align*}
        \EE{U_\theta(Y)} &= 0 \\
        \var{\widehat{\theta}} &\simeq \frac{1}{n\var{U_\theta(Y)}} = \\ 
        &= \frac{1}{n \EE{U_\theta^2(Y)} - \EE{U_\theta(Y)}^2} = \\
        &= \frac{1}{n \EE{U_\theta^2(Y)}\\
    \end{align*}
    where $\EE{U_\theta^2(Y)}$ is the \textbf{Fisher Information}.
    
    \columnbreak
    
    \paragraph{Influence Function}
    In nonparametric estimation we have for the influence function that 
    \noindent   
    \begin{align*}
        \EE{L_F(Y)} &= 0 \\
        \var{\widehat{\theta}} &\simeq \frac{\var{L_F(Y)}}{n} = \\
        &= \frac{\EE{L_F^2(Y)} - \EE{L_F(Y)}^2}{n} = \\
        &= \frac{\EE{L_F^2(Y)}}{n}\\
    \end{align*}
\end{multicols}

答案1

恐怕我不知道\EE应该如何定义。不过，我确实猜测应该如何定义\R和\var。以下代码可以编译；但是，为了使其可用于您的目的，您需要提供的定义\EE。

在此处输入图片描述

\documentclass{article}
\usepackage{amsmath,amssymb,multicol}
\providecommand\R{\mathbb{R}}
\providecommand\EE[1]{#1} % ???
\DeclareMathOperator{\var}{Var} % variance operator, right?


\begin{document}
\setcounter{section}{5}   % just for this example
\setcounter{subsection}{3}

\subsubsection{Score function vs.\ influence function}

\begin{multicols}{2}

\paragraph{Score Function}
Let $l(\theta\mid y)$ be the log-likelihood for $\theta \in \R$ and $U_\theta(y) = \frac{\partial}{\partial \theta} l(\theta\mid y)$ be the score function. We have  
    \begin{align*}
        \EE{U_\theta(Y)} &= 0 \\
       \var\hat{\theta} &\simeq \frac{1}{n\var{U_\theta(Y)}} = \\ 
       &= \frac{1}{n \EE{U_\theta^2(Y)} - \EE{U_\theta(Y)}^2} = \\
        &= \frac{1}{n \EE{U_\theta^2(Y)}}
    \end{align*}
where $\EE{U_\theta^2(Y)}$ is the \textbf{Fisher Information}.
    
\columnbreak
    
\paragraph{Influence Function}
In nonparametric estimation we have for the influence function that   
    \begin{align*}
        \EE{L_F(Y)} &= 0 \\
        \var\hat{\theta} &\simeq \frac{\var L_F(Y)}{n} = \\
        &= \frac{\EE{L_F^2(Y)} - \EE{L_F(Y)}^2}{n} = \\
        &= \frac{\EE{L_F^2(Y)}}{n}
    \end{align*}

\end{multicols}

\end{document}

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