\documentclass[12pt, a4paper]{article}
\usepackage{epsfig}
\usepackage{subfigure}
\usepackage{amscd}
\usepackage{amssymb}
\usepackage{amsbsy}
\usepackage{amsthm}
\usepackage{graphicx}
\usepackage{natbib}
\usepackage{amsbsy}
\usepackage{amsmath}
\usepackage{rotating}
\usepackage{soul}
\usepackage{natbib}
\usepackage{frcursive}
\usepackage{xcolor}
\usepackage{float}
\usepackage{blkarray}
\usepackage[font={color=gray,bf},figurename=Fig.,labelfont={it}]{caption}
\usepackage{vmargin}
\begin{document}
\begin{gather}
\ Pr(X_{i,j} = x, Y_{i,j} = y ) =
\tau_{\lambda_1, \lambda_2}(x,y)
\frac{\lambda_1^x exp(-\lambda_1)}{x!}\frac{\lambda_2^y exp(-\lambda_2)}{y!}
\end{gather}
where
\begin{subequation}
\tau_{\lambda_1, \lambda_2}(x,y) = \left\{\begin{array}{ c l }
1-\lambda_1\lambda_2\rho & \quad \textrm{if } x = y = 0 \\
1+\lambda_1\rho & \quad \textrm{if } x = 0, y = 1 \\
1+\lambda_2\rho & \quad \textrm{if } x=1, y=0 \\
1 - \rho & \quad \textrm{if } x=1, y=1 \\
1 & \quad \textrm{otherwise }
\end{array}
\end{subequation}
\end{document}
我怎样才能使我的第二个方程居中?
答案1
正如他们所说,你做错了。编译当前的示例代码时,你会看到很多错误。这是一个更新版本,显示了更符合预期的输出:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\Pr(X_{i,j} = x, Y_{i,j} = y ) =
\tau_{\lambda_1, \lambda_2}(x,y)
\frac{\lambda_1^x \exp(-\lambda_1)}{x!}
\frac{\lambda_2^y \exp(-\lambda_2)}{y!}
\end{equation}
where
\begin{equation}
\renewcommand{\arraystretch}{1.2}
\tau_{\lambda_1, \lambda_2}(x,y) =
\left\{\begin{array}{ c @{\quad} l }
1 - \lambda_1\lambda_2\rho & \text{if } x = y = 0 \\
1 + \lambda_1\rho & \text{if } x = 0, y = 1 \\
1 + \lambda_2\rho & \text{if } x = 1, y = 0 \\
1 - \rho & \text{if } x = 1, y = 1 \\
1 & \text{otherwise}
\end{array}
\right.
\end{equation}
\end{document}
对于第二个方程,您可以考虑使用cases
环境而不是手动构建的环境array
。另外,如果您希望第二个方程中的每个元素都编号,则需要numcases
(从numcases
包裹)。
答案2
正如@Werner在他的回答中提到的,您可以考虑使用cases
在包中定义amsmath
或甚至更好的dcases
在包中定义mathtools
。 后者使cases
内容Idisplay
模式:
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\Pr(X_{i,j} = x, Y_{i,j} = y ) =
\tau_{\lambda_1, \lambda_2}(x,y)
\frac{\lambda_1^x \exp(-\lambda_1)}{x!}
\frac{\lambda_2^y \exp(-\lambda_2)}{y!}
\end{equation}
where
\[
\tau_{\lambda_1, \lambda_2}(x,y)
= \begin{dcases}
1 - \lambda_1\lambda_2\rho & \text{if } x = y = 0 \\
1 + \lambda_1\rho & \text{if } x = 0, y = 1 \\
1 + \lambda_2\rho & \text{if } x = 1, y = 0 \\
1 - \rho & \text{if } x = 1, y = 1 \\
1 & \text{otherwise}
\end{dcases}
\]
\end{document}