Why is the event $\{X_{(j)} \le x_i\}$ is equivalent to the event $\{Y_i \ge j\}$?

Question

From Statistical Inference by Casella and Berger:

Let $X_1, \dots X_n$ be a random sample from a discrete distribution with $f_X(x_i) = p_i$, where $x_1 \lt x_2 \lt \dots$ are the possible values of $X$ in ascending order. Let $X_{(1)}, \dots, X_{(n)}$ denote the order statistics from the sample. Define $Y_i$ as the number of $X_j$ that are less than or equal to $x_i$. Let $P_0 = 0, P_1 = p_1, \dots, P_i = p_1 + p_2 + \dots + p_i$.

If $\{X_j \le x_i\}$ is a "success" and $\{X_j \gt x_i\}$ is a "failure", then $Y_i$ is binomial with parameters $(n, P_i)$.

Then the event $\{X_{(j)} \le x_i\}$ is equivalent to the event $\{Y_i \ge j\}$

Can someone explain why these two are equivalent?

$\{X_{(j)} \le x_i\} = \{s \in \text{dom}(X_{(j)}) : X_{(j)}(s) \le x_i\}$

$\{Y_i \ge j\} = \{s' \in \text{dom}(Y_i) : Y_i(s') \ge j\}$

I'm having trouble understanding how these random variable functions show this equivalence.

Answer 1

The following statements are equivalent for every $\omega\in\Omega$:

• $\omega\in\{X_{(j)}\leq x_i\}$
• $X_{(j)}(\omega)\leq x_i$
• $|\{k\in\{1,\dots,n\}\mid X_k(\omega)\leq x_i\}|\geq j$
• $Y_i(\omega)\geq j$
• $\omega\in\{Y_i\geq j\}$

Looking at the first and the last bullet we conclude that: $$\{X_{(j)}\leq x_i\}=\{Y_i\geq j\}$$

Answer 2

If $X_{(j)}\le x_i$, then clearly all the other smaller order statistics $X_{(1)},\ldots, X_{(j-1)}$ are also less than or equal to $x_i$ (since they are less than or equal to $X_{(j)}$). So at least $j$ of the order statistics are less than or equal to $x_i$, or in other words, $Y_i \ge j$.

Conversely, if $Y_i \ge j$, then by definition at least $j$ of the order statistics are less than or equal to $x_i$. This implies that $X_{(j)} \le x_i$. (Otherwise, $X_{(j)}$ is greater than $x_i$, and then so are all the greater order statistics, so at most $j-1$ order statistics are less than or equal to $x_i$, a contradiction.)