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}

\E{f(x)}
or\E_x{f(x)}
? – Werner Jan 29 '15 at 23:13_x
? – Sigur Jan 29 '15 at 23:13[x]
which is the next-best alternative I can think of, but at least for learning I'd like to know if this is possible. – user541686 Jan 29 '15 at 23:14\DeclareMathOperator{\E}{\mathbb{E}}
and type\E(f(x))
or\E_{x}(f(x))
; there's no advantage in having braces instead of parentheses. – egreg Jan 29 '15 at 23:16