Expected minimum of a linear function on the unit cube Let $c\in\mathbb{R}^n$ and let $X_1,X_2$ be two independent uniform samples on the unit cube in $\mathbb{R}^n$.  Is there anything at all (in an analytic sense) we can say about the expectation $E\min\{c^T X_1 , c^T X_2\}$? How about if I have $k>2$ independent samples?
 A: Let us provide an explicit (albeit complicated) expression for $EM_k$, where 
\begin{equation*}
 M_k:=\min_{1\le i\le k}c^TX_i
\end{equation*}
and $k$ is any natural number. Without loss of generality, each of the coordinates $c_j$ of the vector $c=(c_1,\dots,c_n)$ is nonzero; otherwise, one can reduce the dimension $n$. Note that almost surely 
\begin{equation*}
 s+M_k\ge0,\quad\text{where}\quad s:=\sum_j \max(0,-c_j).  
\end{equation*}
So, 
\begin{multline*}
 EM_k=-s+E(s+M_k)=-s+\int_0^\infty P(s+M_k>x)\,dx \\ 
 =-s+\int_{-s}^\infty P(M_k>u)\,du=-s+\int_{-s}^\infty P(c^T X_1>u)^k\,du. 
\end{multline*}
In turn, by Theorem 1 (used here with $w=-c$ and $z=-u$), 
\begin{equation*}
P(c^T X_1>u)=\frac{(-1)^n}{n!\prod_1^n c_j}\,\sum_{J\subseteq[n]}(-1)^{|J|}(c^T\,1_J-u)_+^n.  
\end{equation*}
So, $P(c^T X_1>u)^k$ is a linear combination of products of the form 
\begin{equation*}
 \prod_{r=1}^k(c^T\,1_{J_r}-u)_+^n=I\{u<\min_{1\le r\le k} c^T\,1_{J_r}\}\prod_{r=1}^k(c^T\,1_{J_r}-u)^n,
\end{equation*}
which are piecewise polynomial. So, 
\begin{multline}
 EM_k
 =-s+
 \Big(\frac{(-1)^n}{n!\prod_1^n c_j}\Big)^k \\
 \times \sum_{J_1,\dots,J_k\subseteq[n]} (-1)^{\sum_{r=1}^k|J_r|}
\int_{-s}^{\min_{1\le r\le k}  c^T\,1_{J_r}} du\,\prod_{r=1}^k(c^T\,1_{J_r}-u)^n. \tag{1}
\end{multline}
The integrands in (1) are certain polynomials, and so, the integrals in (1) can be explicitly expressed. 
A: For an upper bound, you have $E\min(X,Y)\le \min(E X,E Y)$, and so
$$ E\min(c^TX_1,c^TX_2)\le c^T\cdot(1/2,\ldots,1/2)=\frac12\sum_{i=1}^n c_i.$$
For the lower bound, notice first that $\min(a,b)=(a+b-|a-b|)/2$, whence
$$ E\min(c^TX_1,c^TX_2) = 
\frac12\sum_{i=1}^n c_i
-\frac12E|c^TX_1-c^TX_2|.
$$
It remains to upper-bound the latter term. Hölder's inequality comes to mind: we can bound $|c^T(X_1-X_2)|$ by
$||c||_2||X_1-X_2||_2$, 
or by 
$
||c||_1||X_1-X_2||_\infty
$, or, say, by
$
||c||_\infty||X_1-X_2||_1
$.
Let's see where the first bound leads.
We have 
$$E||X_1-X_2||_2^2=\sum_{i=1}^nE(X_1(i)-X_2(i))^2
=\frac{1}{6}n,
$$
the latter is a routine calculation, see https://en.wikipedia.org/wiki/Triangular_distribution .
Now $E||X_1-X_2||_2 = E\sqrt{||X_1-X_2||_2^2||}
\le\sqrt{E||X_1-X_2||_2^2}=\sqrt{n/6}
$.
This yields a lower bound of
$$
\frac12\sum_{i=1}^n c_i
-\frac{||c||_2}2\sqrt{\frac{n}6}
\le E\min(c^TX_1,c^TX_2).
$$
You'll get other estimates via the other applications of Hölder, which will be better or worse depending on $||c||_p$, for $p\in[1,\infty]$.
