If $Y_i = 1 - X_i$, why is $Y_{(1)} = X_{(n)}$ if $X \sim U(0, 1)$?

Question

I saw in my notes that if $Y_i = 1 - X_i$, then $Y_{(1)} = X_{(n)}$ if $X_i \sim U(0, 1)$?

Note that $X_{(i)}$ is the i-th order statistic so $X_{(n)} = \max\{X_1, \ldots, X_n\}$.

Isn't $Y_{(1)} = \min\{1 - X_1, \ldots, 1 - X_n\}$? If so, wouldn't $Y_{(1)} = X_{(1)}$?

It's possible something may have gotten lost in translation in my notes, but I wanted to make sure I'm not misunderstanding something here.

Answer 1

• Multiplying the $X_i$ by $-1$ reverses the order: $a <b \iff -a > -b$
• Adding $1$ does not change the order: $-a > -b \iff 1-a > 1-b$
• So the highest $X_i$ corresponds to the lowest $-X_i$ and the lowest $Y_i=1-X_i$
• So $$Y_{(1)} =\min\{Y_1, Y_2, \ldots, Y_n\} \\ = \min\{1-X_1, 1-X_2, \ldots , 1-X_n\}\\ = 1+\min\{-X_1, -X_2, \ldots, -X_n\} \\ = 1-\max\{X_1, X_2, \ldots ,X_n\} \\ =1 -X_{(n)}$$