修改latex中block中公式与block标题间隔过大的问题

修改block中公式与block间隔过大的问题

如图的block中公式出现了空白:代码见下方

\begin{proof}[证明]\begin{align*}&Z=\alpha +\beta _XX+\beta _YY+\varepsilon \rightarrow XZ=\alpha X+\beta _XX^2+\beta _YXY+\varepsilon X&\\&E\left( Z \right) =\alpha +\beta _XE\left( X \right) +\beta _YE\left( Y \right) &\\&E\left( X \right) E\left( Z \right) =\alpha E\left( X \right) +\beta _XE^2\left( X \right) +\beta _YE\left( Y \right) E\left( X \right) &\\&E\left( XZ \right) =aE\left( X \right) +\beta _XE\left( X^2 \right) +\beta _YE\left( XY \right) &\\&\sigma _{xz}=\beta _X\sigma _{y}^{2}+\beta _Y\sigma _{xy} \qquad \sigma _{yz}=\beta _X\sigma _{xy}+\beta _Y\sigma _{y}^{2}&\\&\beta _Y=\frac{\sigma _{x}^{2}\sigma _{yz}-\sigma _{xz}\sigma _{xy}}{\sigma _{y}^{2}\sigma _{x}^{2}-\sigma _{xy}^{2}}\qquad\beta _X=\frac{\sigma _{y}^{2}\sigma _{xz}-\sigma _{yz}\sigma _{xy}}{\sigma _{y}^{2}\sigma _{x}^{2}-\sigma _{xy}^{2}}\end{align*}
\end{proof}

修改方式就是在block的下面，align环境的上面加一个

\setlength{\abovedisplayskip}{-15pt}

其中的-15pt可以自己调节,修改结果如下:

\begin{proof}[证明]\setlength{\abovedisplayskip}{-15pt}\begin{align*}&Z=\alpha +\beta _XX+\beta _YY+\varepsilon \rightarrow XZ=\alpha X+\beta _XX^2+\beta _YXY+\varepsilon X&\\&E\left( Z \right) =\alpha +\beta _XE\left( X \right) +\beta _YE\left( Y \right) &\\&E\left( X \right) E\left( Z \right) =\alpha E\left( X \right) +\beta _XE^2\left( X \right) +\beta _YE\left( Y \right) E\left( X \right) &\\&E\left( XZ \right) =aE\left( X \right) +\beta _XE\left( X^2 \right) +\beta _YE\left( XY \right) &\\&\sigma _{xz}=\beta _X\sigma _{y}^{2}+\beta _Y\sigma _{xy} \qquad \sigma _{yz}=\beta _X\sigma _{xy}+\beta _Y\sigma _{y}^{2}&\\&\beta _Y=\frac{\sigma _{x}^{2}\sigma _{yz}-\sigma _{xz}\sigma _{xy}}{\sigma _{y}^{2}\sigma _{x}^{2}-\sigma _{xy}^{2}}\qquad\beta _X=\frac{\sigma _{y}^{2}\sigma _{xz}-\sigma _{yz}\sigma _{xy}}{\sigma _{y}^{2}\sigma _{x}^{2}-\sigma _{xy}^{2}}\end{align*}
\end{proof}

修改latex公式与正文上下间距的问题

只需要添加如下代码即可

\setlength\abovedisplayskip{3pt}%shrink space
\setlength\belowdisplayskip{3pt}

假设所有的外生变量服从均值为0、方差为1(即$\sigma^2=1$的正态分布)\begin{itemize}\item <0-> 假定观察到Y=2,确定X的最优估计\setlength\abovedisplayskip{3pt}%shrink space\setlength\belowdisplayskip{3pt}\begin{align*}\E(X|Y=2)&=\E(3Y-U_Y|Y=2)\\&=6 \end{align*}

修改效果如下: