对两个方程进行括号

Question

有没有办法将第 4 和第 5 个等式末尾的两个括号连接在一起，以便它跨越这两个等式并且只有一个文本？

\begin{align}
f(\beta_i^{r,c},\alpha_{r,c} \mid n_{r,i},t_{c,i},(\lambda_1,\lambda_2) \text{ or } \lambda) \propto \\
\times \prod_{i=1}^{P} \prod_{c=1}^C ({\theta_{c,i}})^{t_{c,i}} &\;\;\bigg] \textbf{Individualdaten}\\
\times \prod_{i=1}^{P} \prod_{r=1}^R \frac{1}{B(\alpha)} \prod_{c=1}^C (\beta_i^{r,c})^{\alpha^{r,c}-1} &\;\;\bigg] \textbf{Priori Distribution}\\
\times \prod_{r=1}^{R} \prod_{c=1}^C \frac{1}{\lambda_2 ^{\lambda_1} \Gamma(\lambda_1)} \alpha_{r,c}^{\lambda_1-1} exp(-\alpha_{r,c} \lambda_2) &\;\;\bigg] \textbf{Priori Distribution}\\
\text{or}\\
\times \prod_{r=1}^{R} \prod_{c=1}^C \lambda exp(-\lambda\alpha_{r,c} &\;\;\bigg] \textbf{Priori Distribution}\\
\text{with } B(\alpha)=\frac{\prod_{c'=1}^C \Gamma(\alpha_{rc'})}{\Gamma(\Sigma_{c'=1}^C\alpha_{r,c'})}
\end{align}

Answer 1

是的。

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{tikzmark}
\begin{document}
  \begin{align}
f(\beta_i^{r,c},\alpha_{r,c} \mid n_{r,i},t_{c,i},(\lambda_1,\lambda_2) \text{ or } \lambda) \propto \\
\times \prod_{i=1}^{P} \prod_{c=1}^C ({\theta_{c,i}})^{t_{c,i}} &\;\;\bigg] \textbf{Individualdaten}\\
\times \prod_{i=1}^{P} \prod_{r=1}^R \frac{1}{B(\alpha)} \prod_{c=1}^C (\beta_i^{r,c})^{\alpha^{r,c}-1} &\;\;\bigg] \textbf{Priori Distribution}\\
\times \prod_{r=1}^{R} \prod_{c=1}^C \frac{1}{\lambda_2 ^{\lambda_1} \Gamma(\lambda_1)} \alpha_{r,c}^{\lambda_1-1} \exp(-\alpha_{r,c} \lambda_2) &\;\;
\tikzmarknode{brkt}{\bigg]}
\tikzmarknode{above}{\textbf{Priori Distribution}}\\
\text{or}\\
\tikzmarknode{l1}{\times \prod_{r=1}^{R} \prod_{c=1}^C \lambda
\exp(-\lambda\alpha_{r,c}}
&\\
\tikzmarknode{l2}{\text{with } B(\alpha)=\frac{\prod_{c'=1}^C
\Gamma(\alpha_{rc'})}{\Gamma(\Sigma_{c'=1}^C\alpha_{r,c'})}}
\end{align}
\begin{tikzpicture}[overlay,remember picture]
\draw[line width=0.6pt] (brkt.west |- l1.north) -| (brkt.north |- l2.south)
node[pos=0.75,right]{\textbf{Priori Distribution}}
-- (brkt.west |- l2.south);
\end{tikzpicture}
\end{document}