Automatically generate new commands with symbol names

Question

As an extension of my earlier question Automatically generate new commands, I want to generate commands for Greek random variables as well, that is:

\usepackage{bm}
\newcommand{\myrv}[1]{\bm{#1}}
\newcommand{\rvepsilon}{\myrv{\epsilon}}
\newcommand{\rvnu}{\myrv{\nu}}

Unfortunately, the tricks in the answers to the earlier question don't work in this case. For example, when I tried

\documentclass{article}

\usepackage{bm}

% Notation for Greek random variables
\newcommand{\myrv}[1]{\bm{#1}}

\makeatletter
\newcommand*\defrvar[1]{
  \expandafter\newcommand\csname rv#1\endcsname[1][]{\myrv{#1}}}

\newcommand*\defrvars[1]{
  \@for\@i:=#1\do{\expandafter\defrvar\expandafter{\@i}}}
\makeatother

\defrvars{epsilon,nu}

\begin{document}
$\rvepsilon = 1$
\end{document}

The output is "epsilon = 1" instead of the desired output "ε = 1".

Is there a way to define the commands automatically by looping over a list, say {alpha,beta,gamma,...}?

Answer 1

You're making a couple of mistakes:

\documentclass{article}

\usepackage{bm}

% Notation for Greek random variables
\newcommand{\myrv}[1]{\bm{#1}}

\makeatletter
\newcommand*\defrvar[1]{%
  \expandafter\newcommand\csname rv#1\endcsname{\myrv{\csname #1\endcsname}}}

\newcommand*\defrvars[1]{%
  \@for\@i:=#1\do{\expandafter\defrvar\expandafter{\@i}}%
}
\makeatother

\defrvars{epsilon,nu}

\begin{document}
$\rvepsilon = 1$
\end{document}

1. The \rvepsilon command shouldn't have an optional argument
2. You missed \csname...\endcsname, so \defrvar{epsilon} defined \rvepsilon to give \myrv{epsilon}; now it gives \myrv{\csname epsilon\endcsname}

The unavoidable expl3 version:

\documentclass{article}

\usepackage{bm}
\usepackage{xparse}

% Notation for Greek random variables
\NewDocumentCommand{\myrv}{m}{\bm{#1}}

\ExplSyntaxOn
\NewDocumentCommand{\defrvars}{ m }
 {
  \clist_map_inline:nn { #1 }
   {
    \cs_new:cpx { rv##1 } { \myrv { \exp_not:c { ##1 } } }
   }
 }
\ExplSyntaxOff

\defrvars{epsilon,nu}

\begin{document}
$\rvepsilon = 1$
\end{document}