# 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?

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.
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. H?lder'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 H?lder, which will be better or worse depending on $$||c||_p$$, for $$p\in[1,\infty]$$.