latex案例：编排数学公式
latex源码1：
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\title{formula}
\author{howard}
\begin{document}
\maketitle
Bayes Formula: $P(H|X)=\frac{P(X|H)P(H)}{P(X)}$
\vskip\baselineskip
$P(X|H)=0.85, P(H)=0.001, P(X)=0.4$
\vskip\baselineskip
$P(H|X)=\frac{P(X|H)P(H)}{P(X)}=\frac{0.85*0.001}{0.4}=0.000215=0.0215\%$
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$P(X|H)\frac{P(H)}{P(X)}=P(H|X)$
\vskip\baselineskip
$\frac{P(H_1)}{P(H_2)}\cdot\frac{P(X|H_1)}{P(X|H_2)}=\frac{P(H_1|X)}{P(H_2|X)}$
\vskip\baselineskip
$\frac{P(H_1|X)}{P(H_2|X)}=\frac{P(X|H_1)\frac{P(H_1)}{P(X)}}{P(X|H_2)\frac{P(H_2)}{P(X)}}=\frac{P(X|H_1)P(H_1)}{P(X|H_2)P(H_2)}=\frac{P(X|H_1)}{P(X|H_2)}\cdot\frac{P(H_1}{P(H_2)}$
\vskip\baselineskip
$\because\frac{P(H_1|X)}{P(H_2|X)}=\frac{P(H_1)}{P(H_2)}\cdot\frac{P(X|H_1)}{P(X|H_2)}=\frac{75\%}{25\%}\cdot\frac{90\%}{30\%}=9:1$
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\hspace{0.4cm}$P(H_1|X)+P(H_2|X)=1$
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$\therefore P(H1|X)=\frac{9}{9+1}=0.9=90\%$
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Naive Bayes Formula:
$$P(X|C_i)=\sum_{k=1}^{n} p(X_k|C_i)$$
\vskip\baselineskip
$P(X|C_i)=\sum_{k=1}^{n} p(X_k|C_i)$
\end{document}
运行结果：
latex源码2：
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\title{formula}
\author{howard}
\begin{document}
\maketitle
\begin{align}
2x&+3y+5z=6 \\
-5x&+4y-z=-2 \\
3x&-2y+3z=7
\end{align}
\begin{align*}
P(HD=Yes)&=\sum_\alpha \sum_\beta P(HD=Yes | E=\alpha,D=\beta)P(E=\alpha,D=\beta)\\
&=\sum_\alpha \sum_\beta P(HD=Yes | E=\alpha,D=\beta)P(E=\alpha)P(D=\beta)\\
&=0.25\times0.7\times0.25+0.45\times0.7\times0.75+0.55\times0.3\times0.25+0.75\times0.3\times0.75\\
&=0.49\\
\end{align*}
\begin{align*}
P(BP=Yes)&=\sum_\gamma P(BP=Yes|HD=\gamma)P(HD=\gamma)\\
&=0.85\times0.49+0.2\times0.51=0.5185\\
\end{align*}
\begin{align*}
&P(HD=Yes|BP=Yes,D=Yes,E=Yes)\\
&=[\frac{P(BP=Yes|HD=Yes,D=Yes,E=Yes)}{P(BP=Yes|D=Yes,E=Yes)}] \times P(HD=Yes|D=Yes,E=Yes)\\
&=\frac{P(BP=Yes|HD=Yes)P(HD=Yes|D=Yes,E=Yes)}{\sum_\gamma P(BP=Yes|HD=\gamma)P(HD=\gamma|D=Yes,E=Yes)}\\
&=\frac{0.85\times0.25}{0.85\times0.25+0.2\times0.75}\\
&=0.5862\\
\end{align*}
\begin{align*}
P(H|X)&=P(H|X_1)\cdot P(H|X_2)\cdot P(H|X_3)\\
&=\frac{P(X_1|H)\cdot P(H)}{P(X_1)}\cdot \frac{P(X_2|H)\cdot P(H)}{P(X_2)}\cdot \frac{P(X_2|H)\cdot P(H)}{P(X_3)}\\
&=\frac{\frac{20}{20}\times0.2}{0.25}\times\frac{\frac{18}{20}\times0.2}{0.18}\times\frac{\frac{15}{20}\times0.2}{0.2} \\
&=\frac{4}{5}\times1\times\frac{3}{4}=\frac{3}{5}=0.6
\end{align*}
\begin{align*}
E(X_k)=a_k(\theta_1,\theta_2,...,\theta_m),k=1,2,...,m
\end{align*}
$$A_k=\frac{1}{n}\sum^n_{i=1}X^k_i,k=1,2,..,m$$
$$a_k(\theta_1,\theta_2,...,\theta_m)=E(X_k)=\frac{1}{n}\sum^n_{i=1}X^k_i,k=1,2,..,m$$
\begin{align*}
F(x;\mu,\sigma)=\dfrac{1}{\sigma\sqrt{2\pi}}\int^x_{-\infty}e^{-\frac{(t-\mu)^2}{2\sigma^2}}dt
\end{align*}
\begin{align*}
f(x)=\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(t-\mu)^2}{2\sigma^2}}
\end{align*}
\begin{align*}
&h(x)=f(x)g(x) \\
&h^{'}(x)=f^{'}(x)g(x)+f(x)g^{'}(x)
\end{align*}
$$L(\mu,\sigma^2;x)=\prod^n_{i=1}f(x_i;\mu,\sigma^2)=(2\pi\sigma^2)^{-\frac{n}{2}}e^{-\frac{1}{2\sigma^2}\sum^n_{i=1}(x_i-\mu)^2}$$
\[
\left[
\begin{array}{lcr}
4 & 1 & 0 \\
-1 & 1 & 3 \\
2 & 0 & 1 \\
1 & 3 & 4
\end{array}
\right]
\]
\[
y=\left\lbrace
\begin{aligned}
x+2 \hspace{0.5cm} x<2\\
x^2 \hspace{0.5cm} x\geq2\\
\end{aligned}
\right.
\]
\end{document}
运行结果：