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

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.

Formula for the volume of $\{x_i \in [-N,N]:\sum_{i=1}^n x_i = 0\}$

I am not an expert in convex geometry but if we define $ a_i \sim \mathcal{U}([-N,N])$ where $ [-N,N] \subset \mathbb{R}$ and $ S_n = \sum_{i=1}^n a_i$ I’d like to know whether for arbitrary $ N \in \mathbb{R}_+$ :

  1. The following limit always holds true:

\begin{equation} \lim_{n \to \infty} P(S_n =0)=0 \tag{*} \end{equation}

  1. There exists a simple formula for the volume:

\begin{equation} Vol(\{x_i \in [-N,N]:\sum_{i=1}^n x_i = 0\}) \end{equation}

So far I have managed to address the discrete case by modelling it as a random walk on $ \mathbb{Z}$ . Basically, I managed to show that if we assume $ a_i \sim \mathcal{U}([-N,N])$ where $ [-N,N] \subset \mathbb{Z}$ then:

\begin{equation} \lim_{n \to \infty} P(S_n =0)=0 \tag{1} \end{equation}

\begin{equation} \lim_{n \to \infty} P(|S_n| \leq N )=0 \tag{2} \end{equation}

\begin{equation} \lim_{n \to \infty} P(|S_n| > N )=1 \tag{3} \end{equation}

A description of my analysis is on my blog, Kepler Lounge. That said, I highly doubt that I’m the first person to make this discovery.

I suspect that the result I’m looking for is known to experts in convex analysis but I don’t know which references I should consult.

If $(X_{i}, \varphi_{ij})$ is a inverse system of nonempty sets and surjective maps, then the inverse limit is nonempty (proof verification)

I had a troubles with this problem, so I thought it would be good to write my solution because I’m not sure of all the details.

Since I already accepted the answer, I can not open a bounty to check my solution, so I opened another question. If this is a problem, more experienced users can alert me and I will delete the question.


I tried to use Max and Arturo’s hints.

Solution.

Let $ (X,\varphi_{i})$ the inverse limit of $ (X_{i},\varphi_{ij})$ , that is, $ $ X = \left\{x \in \prod_{i}X_{i} \mid \varphi_{ij}\circ\pi_{j}(x) = \pi_{i}(x)\;\forall i,j; j \geq i \right\}$ $ and $ \varphi_{i} = \pi_{i}|_{X}$ where $ \varphi_{i}: X \to X_{i}$ .

We can consider $ I = \mathbb{N}$

Lema 1. If $ I$ is a countable direct set, then there is a cofinal countable subset totally ordered in $ I$ . Moreover, we can consider $ \mathbb{N} \subset I$ .

Proof. Since $ I$ is countable, we can write $ I = \{a_{j} \mid j \in \mathbb{N}\}$ . Let $ (x_{n})$ be a sequence defined by:

(i) $ x_{0} = a_{0}$

(ii) Defined $ x_{n}$ , take $ i = \min\{j \in \mathbb{N} \mid a_{j} \geq x_{n}; a_{j} \geq a_{n}\}$ and define $ x_{n+1} = a_{i}$ .

Claim: $ (x_{n})$ is increasing.

We have that $ x_{n+1} = a_{i}$ where $ i$ satisfies $ a_{i} \geq x_{n}$ , that is, $ x_{n+1} \geq x_{n}$ .

Claim: for each $ a \in I$ , there is $ x_{i} \geq a$ .

Given $ a_{k} \in I$ , note that $ x_{k+1} = a_{i}$ where $ i$ is smallest natural that, in particular, satisfies $ a_{i} \geq a_{k}$ , that is, $ x_{k+1} \geq a_{k}$ .

We know that $ \tilde{I} = \{x_{1} \leq \cdots \leq x_{n} \leq \cdots\}$ is totally ordered. Since $ \tilde{I}$ is a countable subset totally ordered, $ \mathbb{N}$ immerserd in $ I$ , that is, there is no problem to assume $ \tilde{I} = \mathbb{N}$ . $ \Box$

Lema 2. Let $ I$ be a direct set and $ J \subset I$ cofinal. If $ (X_{i},\varphi_{ij})$ is an inverse system indexed by $ I$ , then $ $ \varprojlim_{i \in I}X_{i} \simeq \varprojlim_{j \in J}X_{j}.$ $

Proof. We have that $ $ \varprojlim_{i \in I}X_{i} = \left\{x \in \prod_{i}X_{i} \mid \varphi_{ik}\circ\pi_{k}(x) = \pi_{i}(x)\;\forall i,k; k \geq i \right\}$ $ and $ $ \varprojlim_{j \in J}X_{j} = \left\{x \in \prod_{j}X_{j} \mid \varphi_{jl}\circ\pi_{l}(x) = \pi_{j}(x)\;\forall j,l; l \geq j \right\}.$ $ Thus, we get a projective map $ $ \begin{eqnarray*} \pi: \prod_{i}X_{i} & \to & \prod_{j}X_{j}\ x_{i} & \mapsto & x_{j}, \end{eqnarray*}$ $ and, since $ \pi\left(\varprojlim_{i \in I}X_{i}\right) \subset \varprojlim_{j \in J}X_{j}$ , the function $ $ \psi = \pi|_{X}: \varprojlim_{i \in I}X_{i} \to \varprojlim_{j \in J}X_{j}$ $ where $ X = \varprojlim_{i \in I}X_{i}$ is well-defined.

Now, let $ (x_{j}) \in \varprojlim_{j \in J}X_{j}$ . If $ i \in I\setminus J$ and $ J$ is cofinal, there is $ j \in J$ with $ j \geq i$ . Define $ x_{i} = \varphi_{ij}(x_{j})$ .

Claim. the equallity does not depends of $ j$ .

Let $ \tilde{j} \in J$ such that $ \tilde{j} \geq i$ . Since $ J \subset I$ and $ I$ is a direct set, there is $ k \in J$ such that $ k \geq j$ and $ k \geq \tilde{j}$ . Thus, $ \varphi_{jk}(x_{k}) = x_{j}$ and $ \varphi_{\tilde{j}k}(x_{k}) = x_{\tilde{j}}$ . Therefore, $ $ \begin{eqnarray*} \varphi_{ij}(x_{j}) & = & \varphi_{ij}(\varphi_{jk}(x_{k}))\ & = & (\varphi_{ij}\circ\varphi_{jk})(x_{k})\ & = & \varphi_{ik}(x_{k})\ & = & (\varphi_{i\tilde{j}}\circ\varphi_{\tilde{j}k})(x_{k})\ & = & \varphi_{i\tilde{j}}(\varphi_{\tilde{j}k}(x_{k}))\ & = & \varphi_{i\tilde{j}}(x_{\tilde{j}}). \end{eqnarray*}$ $

Given $ (x_{j}) \in \varprojlim_{j \in J}X_{j}$ we get $ (x_{i}) \in \prod_{i}X_{i}$ such that $ \pi(x_{i}) = x_{j}$ . Let $ i,k \in I$ with $ k \geq i$ . We know that there is $ j \in J$ with $ j \geq k$ and so, $ j \geq i$ . We have that $ \varphi_{kj}(x_{j}) = x_{k}$ and $ \varphi_{ij}(x_{j}) = x_{i}$ . Thus, $ $ \varphi_{ik}(\pi_{k}(x_{i})) = \varphi_{ik}(x_{k}) = \varphi_{ij}(x_{j}) = x_{i} = \pi_{i}(x_{i}),$ $ that is, $ (x_{i}) \in \varprojlim_{i \in I}X_{i}$ . Therefore, this map is the inverse map for $ \psi$ . So, $ \psi$ is a canonical bijection and so, the inverse limits are isomorphic. $ \Box$

By lema 1 we can assume $ \mathbb{N} \subset I$ is a cofinal subset of $ I$ and, by lema 2, we can suppose $ I = \mathbb{N}$ .

Now, take $ x_{0} \in E_{0}$ and, since $ X_{i} \neq \emptyset$ and $ \varphi_{ij}$ is surjective, define indutivelly $ x_{n} \in E_{n}$ so that $ x_{n} \in \varphi^{-1}_{n-1,n}(x_{n-1})$ for $ n \geq 1$ .

Claim. $ x_{m} = \varphi(m,n)(x_{n})$ for $ m \leq n$ .

If $ n-m = 0$ , then $ \varphi_{nn}(x_{n}) = x_{n}$ by hypothesis. If $ n-m=1$ , then by definition of $ x_{n}$ , $ \varphi_{n-1,n}(x_{n}) = x_{n-1}$ . If the result is true for $ n-m = k$ , then $ $ \varphi_{n-k-1,n}(x_{n}) = \varphi_{n-k-1,n-k}\circ\varphi_{n-k,n}(x_{n}) = \varphi_{n-k-1,n-k}(x_{n-k}) = x_{n-k-1}.$ $ Now, whereas $ \varphi_{i} = \pi_{i}|_{X}: X \to X_{i}$ , we have that $ $ \varphi_{mn}\circ\pi_{n}(x_{n}) = \varphi_{mn}(x_{n}) = x_{m} = \pi_{m}(x_{n}),$ $ therefore, $ (x_{n}) \in X$ .