-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathimportantProperty.tex
More file actions
53 lines (47 loc) · 1.98 KB
/
importantProperty.tex
File metadata and controls
53 lines (47 loc) · 1.98 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
\documentclass[11pt,a4paper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsfonts,amsmath,amssymb,suetterl}
\usepackage{lmodern}
\usepackage[T1]{fontenc}
\usepackage{fancyhdr}
\usepackage{float}
\usepackage[utf8]{inputenc}
\usepackage{fontawesome}
\DeclareUnicodeCharacter{2212}{-}
\usepackage{mathrsfs}
\usepackage[nodisplayskipstretch]{setspace}
\setstretch{1.5}
\fancyfoot[C]{\thepage}
\renewcommand{\footrulewidth}{0pt}
\parindent 0ex
\setlength{\parskip}{1em}
\begin{document}
\section*{Page Break Equation}
\begingroup
\allowdisplaybreaks
\begin{align*}
\mathbb{P}(Y(t) = k)
&= \sum_{i=0}^\infty \mathbb{P}(X(t) = k+i)\mathbb{P}(Y(t) = k\, | \, X(t)=k+i)\\
&= \sum_{i=0}^\infty \frac{\lambda^{k+i}}{(k+i)!}e^{-\lambda}\binom{k+i}{k}p^k(1-p)^i\\
&= \sum_{i=0}^\infty \frac{\lambda^k \lambda^i}{k!i!}e^{-\lambda}p^k(1-p)^i\\
&= \frac{(p\lambda)^k}{k!}e^{-\lambda} \sum_{i=0}^\infty \frac{((1-p)\lambda)^i}{i!}\\
&= \frac{(p\lambda)^k}{k!}e^{-\lambda}e^{(1-p)\lambda}\\
&= \frac{(p\lambda)^k}{k!}e^{-p\lambda}.\\
\end{align*}
\endgroup
%
\section*{Breaking Equation in aligned line}
\begin{align*}
f_{T_n}(t)
&= -\left(\lambda + \lambda^2t+\cdots+\frac{\lambda^{n-2}t^{n-3}}{(n-3)!}+\frac{\lambda^{n-1}t^{n-2}}{(n-2)!}\right)e^{-\lambda t}\notag \\
& \quad \quad + \left(1 + \lambda t + \frac{(\lambda^2t)}{2!}+\cdots+\frac{(\lambda t)^{n-2}}{(n-2)!}+\frac{(\lambda t)^{n-1}}{(n-1)!}\right)\lambda e^{-\lambda t}\\
&= \frac{\lambda^nt^{n-1}}{(n-1)!}e^{-\lambda t}
\end{align*}
%
\section*{tfrac,frac,dfrac}
\begin{align*}
\mathbb{P}(A(\tfrac{3}{2})= 3, B(\tfrac{3}{2}) = 3\, | \, A(\tfrac{3}{4}) = 2, B(\tfrac{3}{4})=1)\\
\mathbb{P}(A(\dfrac{3}{2})= 3, B(\dfrac{3}{2}) = 3\, | \, A(\dfrac{3}{4}) = 2, B(\dfrac{3}{4})=1)\\
\mathbb{P}(A(\frac{3}{2})= 3, B(\frac{3}{2}) = 3\, | \, A(\frac{3}{4}) = 2, B(\frac{3}{4})=1)
\end{align*}
\end{document}