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?

  • 3
    $\begingroup$ What if $A=[0,1]$ and $B=\{0\}\cup [1/2,1]$? $\endgroup$
    – user140541
    Aug 5, 2020 at 15:09
  • $\begingroup$ @d.k.o. right, thanks for the observation! I will add an extra restriction that $B=[0,a]$ for some $a \in [0,1]$ $\endgroup$
    – lisk
    Aug 5, 2020 at 17:19
  • 1
    $\begingroup$ 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. $\endgroup$
    – E-A
    Aug 5, 2020 at 18:26

1 Answer 1


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 $$ \mathsf{E}X=\int_0^{\infty}\mathsf{P}(X\ge t)\, dt\ge \int_0^{\infty}\mathsf{P}(Y\ge t)\, dt=\mathsf{E}Y. $$

  • $\begingroup$ indeed, easy. Thanks! $\endgroup$
    – lisk
    Aug 6, 2020 at 4:57

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