What is the probability that $X_i$ is the $k^{th}$ order statistic in consecutive trials?

Consider $$n$$ r.vs $${X_1, X_2,…,X_n}$$. Each is i.i.d drawn from some distribution $$f(.)$$. What is the probability that $$X_i$$ is the $$k^{th}$$ order statistic in any two consecutive trials?

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:

$$$$\lim_{n \to \infty} P(S_n =0)=0 \tag{*}$$$$

1. There exists a simple formula for the volume:

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

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:

$$$$\lim_{n \to \infty} P(S_n =0)=0 \tag{1}$$$$

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

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

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, $$(x_{n}) \in X$$.