# Inequality between the expected values of the minimum of $k$ i.i.d. uniform random variables distributed on a set and its subset

Let $$A \subseteq [0,1]$$ be a Borel set and let $$B$$ be another Borel set such that $$B \subseteq A$$ and $$B = [0,a]$$ for some $$a \in [0,1]$$.

Let $$x_1, x_2, \ldots, x_k$$ be $$k$$ i.i.d. random variables distributed uniformly on $$A$$, and let $$y_1, y_2, \ldots, y_k$$ be some other $$k$$ i.i.d. random variables distributed uniformly on $$B$$. Finally, let $$X = \min\{ x_1, x_2, \ldots, x_k \}$$ and $$Y = \min\{ y_1, y_2, \ldots, y_k \}$$

Intuitively, it seems obvious that $$\mathbb{E}[Y] \leq \mathbb{E}[X]$$.

What would be a formal and easy reasoning of this inequality?

• What if $A=[0,1]$ and $B=\{0\}\cup [1/2,1]$?
– user140541
Aug 5, 2020 at 15:09
• @d.k.o. right, thanks for the observation! I will add an extra restriction that $B=[0,a]$ for some $a \in [0,1]$
– lisk
Aug 5, 2020 at 17:19
• As edited, the statement is correct. I don't know how formal or easy you consider the following, but I would argue this via coupling $x_i$ s to $y_i$s by sampling $x_i$ iid first, and setting $y_i = x_i$ if $x_i \in [0, a]$ and uniformly in $[0,a]$ otherwise. Note that $x_i = y_i$ in the first case and $y_i < x_i$ otherwise. Then, take expectations.
– E-A
Aug 5, 2020 at 18:26

Since the support of $$x_1$$ is a superset of $$B=[0,a]$$, $$\mathsf{P}(X\ge t)=[\mathsf{P}(x_1\ge t)]^k\ge [\mathsf{P}(y_1\ge t)]^k =\mathsf{P}(Y\ge t),$$ and $$\mathsf{E}X=\int_0^{\infty}\mathsf{P}(X\ge t)\, dt\ge \int_0^{\infty}\mathsf{P}(Y\ge t)\, dt=\mathsf{E}Y.$$