0

I want to write following on my latex. Please help me.

enter image description here

\documentclass[12pt,oneside,a4paper]{report}
\pagestyle{plain}
\usepackage{enumerate}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{setspace}
\usepackage{bm}
\usepackage{mathdesign}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{glossaries}
\usepackage{enumitem}
\makeglossaries
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{defn}{Definition}[section]
\newtheorem{remk}{Remark}[section]
\newtheorem{assump}{Assumptions}[section]
\numberwithin{equation}{section}
\setlength{\topmargin}{-.7in}
\setlength{\textheight}{9.5in}
\setlength{\oddsidemargin}{-.3in}
\setlength{\textwidth}{7in}
\renewcommand{\qedsymbol}{$\blacksquare$}
\begin{document}\begin{theorem}
If the $(k\times 1)$ vector $\bm{x}$ is distributed normally with mean vector $\bm{\mu}$ and variance-covariance matrix $\bm{I}_k$, then
\begin{enumerate}[(i).]
\item $\mathbb{E}[\phi(\bm{x}'\bm{x})\bm{x}]=\bm{\mu}\mathbb{E}[\phi(\chi^{2}_{(k+2,\Delta)})]$,
\item $\mathbb{E}[\phi(\bm{x}'\bm{x})\bm{x}\bm{x}']=\bm{I}_{k}\mathbb{E}[\phi(\chi^{2}_{(k+2,\Delta)})]+\bm{\mu}\bm{\mu}'\mathbb{E}[\phi(\chi^{2}_{(k+4,\Delta)})]$
\end{enumerate}
where $\Delta=\frac{1}{2}\bm{\mu}'\bm{\mu}$ is the noncentrality parameter.
\end{theorem}
 \begin{assump}[Regularity Conditions]
\begin{enumerate}[start=0,label={(\bfseries R\arabic*):}]
\begin{enumerate}
\item The pdfs are distinct.
\item The pdfs have common support for all $\theta$.
\end{enumerate}
\end{assump}
\end{document}
  • Welcome to TeX.SX! Please help us to help you and add a minimal working example (MWE) that illustrates your problem. It will be much easier for us to reproduce your situation and find out what the issue is when we see compilable code, starting with \documentclass{...} and ending with \end{document}. – user31729 Feb 22 '15 at 15:25
  • I have created an Assumptions environment using \newtheorem but I am unable to start from the next line and to create a custom items numbers starting from zero. – Kashif Ali Feb 22 '15 at 15:26
4

Without further information this is a starter...(I've not done the theorem environment around the enumeration)

The enumitem package is the key for easy customization of itemize/enumerate lists. It provides the means for use a starter value (start=0) and the label=.... option.

enumitem can be used together or as enumerate using the shortlabels option.

\documentclass{article}
\usepackage{enumerate}
\usepackage[shortlabels]{enumitem}

\begin{document}

% Traditional style
\begin{enumerate}[a)] 
\item This is
\item an easy
\item customization of standard enumerate
\end{enumerate}


\begin{enumerate}[start=0,label={(\bfseries R\arabic*):}]
\item This is
\item an easy
\item customization of enumerate
\end{enumerate}
\end{document} 

enter image description here

  • The problem is that I have used enumerate package in my remaining document and this is just a little part in middle of the document. So, When I tried to implement your code. It gives me an error on previous enumerate commands. Any help. – Kashif Ali Feb 22 '15 at 15:35
  • 1
    @KashifAli: That's why I asked about the MWE ;-) Now this leaves my work useless :-( – user31729 Feb 22 '15 at 15:36
  • I have edited my question and given some MWE. – Kashif Ali Feb 22 '15 at 15:38
  • @KashifAli That is not an MWE. Please read the link Christian Hupfer provided. An MWE is a complete small document. – cfr Feb 22 '15 at 15:40
  • 2
    @KashifAli enumitem can emulate enumerate, can't it? So can't you just switch? – cfr Feb 22 '15 at 15:41
1

As there remains some problems with the placement of labels, I propose a variant that looks more like the image in the O.P.'s post. I replaced amsthm with ntheorem, which has a break theorem style and writes the optional argument of theorems in boldface if \theoremname is boldface.

Loading mathdesign did nothing, as no font was chosen as an option. I added charter.

\documentclass[12pt,oneside,a4paper]{report}
\pagestyle{plain}
\usepackage{mathtools}
\usepackage{graphicx}
\usepackage{setspace}
\usepackage{bm}
\usepackage[charter]{mathdesign}
\usepackage[shortlabels]{enumitem}

%\usepackage{amssymb}
%\usepackage{amsthm}
%\newtheorem{theorem}{Theorem}[section]
%\newtheorem{corollary}{Corollary}[section]
%\newtheorem{lemma}{Lemma}[section]
%\newtheorem{defn}{Definition}[section]
%\newtheorem{remk}{Remark}[section]
%\newtheorem{assump}{Assumptions}[section]
%\renewcommand{\qedsymbol}{$\blacksquare$}

  \usepackage[thmmarks, amsmath, thref]{ntheorem}

 \theoremstyle{plain}
 \theoremheaderfont{\upshape\bfseries}
 \theorembodyfont{\itshape}
 \theoremseparator{.}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{defn}{Definition}[section]
\newtheorem{remk}{Remark}[section]

\theoremstyle{break}
 \theoremseparator{.\medskip}
\newtheorem{assump}{Assumptions}[section]

 \theoremstyle{nonumberplain}
 \theoremheaderfont{\itshape}
 \theorembodyfont{\upshape}
 \theoremsymbol{\ensuremath{\blacksquare}}
 \newtheorem{proof}{Proof}

\numberwithin{equation}{section}
\setlength{\topmargin}{-.7in}
\setlength{\textheight}{9.5in}
\setlength{\oddsidemargin}{-.3in}
\setlength{\textwidth}{7in}

\begin{document}
\setcounter{chapter}{6}\setcounter{section}{1}
\begin{theorem}
  If the $(k \times 1)$ vector $\bm{x}$ is distributed normally with mean vector $\bm{\mu}$ and variance-covariance matrix $\bm{I}_k$, then
  \begin{enumerate}[(i).]
    \item $\mathbb{E}[\phi(\bm{x}'\bm{x})\bm{x}]=\bm{\mu}\mathbb{E}[\phi(\chi^{2}_{(k+2,\Delta)})],$

    \item $\mathbb{E}[\phi(\bm{x}'\bm{x})\bm{x}\bm{x}'] = \bm{I}_{k}\mathbb{E}[\phi(\chi^{2}_{(k+2,\Delta)})] + \bm{\mu}\bm{\mu}'\mathbb{E}[\phi(\chi^{2}_{(k+4,\Delta)})],$
  \end{enumerate}
  where $\Delta=\frac{1}{2}\bm{\mu}'\bm{\mu}$ is the noncentrality parameter.
\end{theorem}

\begin{assump}[Regularity Conditions]
  \begin{enumerate}[start=0,label={\upshape(\bfseries R\arabic*):},wide = 0pt, leftmargin = 3em]
    \item The pdfs are distinct; i.e., $ \theta \neq \theta' \Rightarrow f(x_i; \theta) \neq f(x_i; \theta')$. Text text text text text text text text text text text text text text text text text.
    \item The pdfs have common support for all $\theta$.
    \item The point $ \theta_0$ is an interior point of $ \Omega $.
  \end{enumerate}
\end{assump}
\end{document}

enter image description here

  • Thanks @Bernard for such an elegant code relating to the solution of my problem. Thanks a lot. – Kashif Ali Feb 23 '15 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.