Operator with both arguments and limits

Question

I'd like to define a command \E that can be used precisely as follows: it should translate

\E{f(x)}
\E_x{f(x)}

into

\mathbb{E}(f(x))
\mathbb{E}_x(f(x))

respectively. Specifically, the goal is to render them as

respectively.

But notice that _x neither has the syntax of an optional nor of a mandatory argument.

Is such a thing possible? If so, how can I do it?

Answer 1

Yes, it's possible, by coupling this with \DeclarePairedDelimiter.

New answer (January 2021)

This also allows upper limits.

\documentclass{article}
\usepackage{mathtools,amssymb}

\NewDocumentCommand{\E}{e{^_}}{%
  \operatorname{\mathbb{E}}%
  \IfValueT{#1}{^{#1}}%
  \IfValueT{#2}{_{#2}}%
  \parens
}
\DeclarePairedDelimiter{\parens}{(}{)}

\begin{document}

$\E{f(x)}$

$\E[\big]{f(x)}$

$\E[\Bigg]{f(x)}$

$\E*{\dfrac{1}{2}}$

$\E_x{f(x)}$

$\E_x[\big]{f(x)}$

$\E_x[\Bigg]{f(x)}$

$\E_x*{\dfrac{1}{2}}$

$\E^x{f(x)}$ $\E_x^y[\big]{f(x)}$

\end{document}

If only lower limits are needed, one can simplify the definition into

\NewDocumentCommand{\E}{e{_}}{%
  \operatorname{\mathbb{E}}%
  \IfValueT{#1}{_{#1}}%
  \parens
}
\DeclarePairedDelimiter{\parens}{(}{)}

Old answer

\documentclass{article}
\usepackage{mathtools,amssymb}

\makeatletter
\DeclareRobustCommand{\E}{\operatorname{\mathbb{E}}\@ifnextchar_{\m@Es}{\m@Epd}}
\newcommand{\m@Es}[2]{_{#2}\m@Epd}
\DeclarePairedDelimiter{\m@Epd}{(}{)}
\makeatother

\begin{document}
$\E{f(x)}$

$\E[\big]{f(x)}$

$\E[\Bigg]{f(x)}$

$\E*{\dfrac{1}{2}}$

$\E_x{f(x)}$

$\E_x[\big]{f(x)}$

$\E_x[\Bigg]{f(x)}$

$\E_x*{\dfrac{1}{2}}$
\end{document}