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?



Expected minimum of a linear function on the unit cube
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](https://arxiv.org/abs/math/0607715) (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. 


