Expectation of min of two random variable Looking for your kind help to solve the following expectation problem. 
Let assume, 
$C_{u} = min(C_{a},C_{b})$
where 
$C_{a}$ & $C_{b}$ are random variables. Is the following is true?
$E[C_{u}]$ = $E[min(C_{a},C_{b})]$ = $min(E[C_{a}],E[C_{b}])$ 
If yes, then how? 
BR, 
 A: Let $C_a$ and $C_b$ be independent random variables that take value $0$ and $1$ each with probability $\frac{1}{2}$. Then $\min(C_a,C_b)=0$ with probability $\frac{3}{4}$ and $\min(C_a,C_b)=1$ with probability $\frac{1}{4}$, giving expectation $\frac{1}{4}$.
But $\min(E(C_a,C_b)=\frac{1}{2}$.
There are many other counterexamples.
A: It is not true.  Say you toss two coins.  Let $X_i$ be the number of "heads" you get from the $i$th coin. Then
$$
(X_1,X_2)=\begin{cases} (1,1) \\  (1,0) \\ (0,1) \\ (0,0) \end{cases}
$$
each with probability $1/4$.  In three cases, the minimum is $0$ and in one case it is  $1$.
Notice that $\min\{\operatorname{E}(X_1),\operatorname{E}(X_2)\}=\min\{1/2,1/2\}=1/2$, but
$$\operatorname{E}(\min\{X_1,X_2\}) = \frac 1 4\cdot0 + \frac14\cdot0 + \frac14\cdot0 + \frac14\cdot1 = \frac14 \ne \frac 1 2.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\,{\rm E}\bracks{\min\pars{x,y}} & =
{\rm E}\bracks{\frac{x + y - \verts{x - y}}{2}}
=
\\[5mm] & = \frac{%
\,{\rm E}\bracks{x} + \,{\rm E}\bracks{y} - \,{\rm E}\bracks{\verts{x - y}}}{2}
\\[8mm] & =\frac{%
\,{\rm E}\bracks{x} + \,{\rm E}\bracks{y}
-\verts{\,{\rm E}\bracks{x} - \,{\rm E}\bracks{y}}}{2}
\\[2mm] & +\frac{\verts{\,{\rm E}\bracks{x} - \,{\rm E}\bracks{y}}
-\,{\rm E}\bracks{\verts{x - y}}}{2}
\\[8mm] &=
\min\pars{\,{\rm E}\bracks{x},\,{\rm E}\bracks{x}}
\\[2mm] & +\frac{\verts{\,{\rm E}\bracks{x} - \,{\rm E}\bracks{y}}
-\,{\rm E}\bracks{\verts{x - y}}}{2}
\end{align}

$$
\,{\rm E}\bracks{\min\pars{x,y}}
-\min\pars{\,{\rm E}\bracks{x},\,{\rm E}\bracks{x}}
=\dsc{\frac{\verts{\,{\rm E}\bracks{x} - \,{\rm E}\bracks{y}}
-\,{\rm E}\bracks{\verts{x - y}}}{2}}
$$
The $\dsc{\mbox{right hand side}}$ doesn't, in general, vanishes out.
