[proofplan] Continuity of the common distribution function implies that each $X_i$ has no atoms, so pairwise ties occur with probability zero. On the event of no ties, the sample values determine exactly one strict ordering of the indices. Independence and identical distribution make the joint law invariant under coordinate permutations, so every strict ordering event has the same probability. Since the $n!$ ordering events are disjoint and cover the no-tie event, each has probability $1/n!$, and translating orderings into rank vectors gives the result. [/proofplan]

[step:Show that ties have probability zero] Let $(\Omega,\mathcal{F},\mathbb{P})$ denote the underlying probability space on which the real-valued random variables $X_i: \Omega \to \mathbb{R}$ are defined. For each $m \in \mathbb{N}$, let $\mathcal{B}(\mathbb{R}^m)$ denote the Borel $\sigma$-algebra on $\mathbb{R}^m$. Let $\mu$ denote the common law of $X_1,\dots,X_n$, so $\mu(A) = \mathbb{P}(X_i \in A)$ for every $A \in \mathcal{B}(\mathbb{R})$ and every $i \in \{1,\dots,n\}$. Since $F$ is continuous, $\mu$ has no atoms: for every $a \in \mathbb{R}$, \begin{align*} \mu(\{a\}) = F(a) - \lim_{t \uparrow a} F(t) = 0. \end{align*} For distinct indices $i,j \in \{1,\dots,n\}$, independence gives that the joint law of $(X_i,X_j)$ is $\mu \otimes \mu$. Let $\Delta_{\mathbb{R}} := \{(x,y) \in \mathbb{R}^2 : x = y\}$ denote the diagonal. The diagonal $\Delta_{\mathbb{R}}$ is closed in $\mathbb{R}^2$, hence Borel. Applying Tonelli's theorem to the non-negative Borel function $\mathbb{1}_{\Delta_{\mathbb{R}}}: \mathbb{R}^2 \to \{0,1\}$ gives \begin{align*} (\mu \otimes \mu)(\Delta_{\mathbb{R}}) = \int_{\mathbb{R}} \mu(\{x\}) \, d\mu(x). \end{align*} Therefore \begin{align*} \mathbb{P}(X_i = X_j) = (\mu \otimes \mu)(\Delta_{\mathbb{R}}) = 0. \end{align*} Define the no-tie event \begin{align*} E := \bigcap_{1 \le i < j \le n} \{X_i \ne X_j\}. \end{align*} By the union bound over the finitely many unordered pairs, \begin{align*} \mathbb{P}(E^c) \le \sum_{1 \le i < j \le n} \mathbb{P}(X_i = X_j) = 0. \end{align*} Hence $\mathbb{P}(E) = 1$. [/step]

[step:Partition the no-tie event by strict orderings] Let $S_n$ denote the set of permutations $\pi: \{1,\dots,n\} \to \{1,\dots,n\}$. For each $\pi \in S_n$, define the ordering event \begin{align*} A_\pi := \{\omega \in \Omega : X_{\pi(1)}(\omega) < X_{\pi(2)}(\omega) < \cdots < X_{\pi(n)}(\omega)\}. \end{align*} If $\pi,\sigma \in S_n$ with $\pi \ne \sigma$, then $A_\pi \cap A_\sigma = \varnothing$, because two different strict total orderings of the same $n$ distinct [real numbers](/page/Real%20Numbers) cannot both hold. Conversely, if $\omega \in E$, then the values $X_1(\omega),\dots,X_n(\omega)$ are pairwise distinct, so there is a unique permutation $\pi \in S_n$ such that \begin{align*} X_{\pi(1)}(\omega) < X_{\pi(2)}(\omega) < \cdots < X_{\pi(n)}(\omega). \end{align*} Therefore \begin{align*} E = \bigcup_{\pi \in S_n} A_\pi \end{align*} as a disjoint union. [/step]

[step:Use exchangeability to give every ordering event the same probability]Fix $\pi \in S_n$. Since $X_1,\dots,X_n$ are independent and identically distributed with common law $\mu$, define the random vector $X: \Omega \to \mathbb{R}^n$ by $X(\omega) = (X_1(\omega),\dots,X_n(\omega))$. The joint law of $X$ is the product measure $\mu^{\otimes n}$. Define the coordinate permutation map $T_\pi: \mathbb{R}^n \to \mathbb{R}^n$ by $T_\pi(x_1,\dots,x_n) = (x_{\pi(1)},\dots,x_{\pi(n)})$. The product measure $\mu^{\otimes n}$ is invariant under $T_\pi$. Indeed, on every measurable rectangle $B_1 \times \cdots \times B_n$ with $B_k \in \mathcal{B}(\mathbb{R})$, the preimage under $T_\pi$ is $B_{\pi^{-1}(1)} \times \cdots \times B_{\pi^{-1}(n)}$, and both rectangles have $\mu^{\otimes n}$-measure $\prod_{k=1}^n \mu(B_k)$. Since these rectangles form a $\pi$-system generating $\mathcal{B}(\mathbb{R}^n)$, the uniqueness theorem for finite measures gives $\mu^{\otimes n}(T_\pi^{-1}(B)) = \mu^{\otimes n}(B)$ for every $B \in \mathcal{B}(\mathbb{R}^n)$. Let \begin{align*} C := \{(x_1,\dots,x_n) \in \mathbb{R}^n : x_1 < x_2 < \cdots < x_n\}. \end{align*} The set $C$ is open in $\mathbb{R}^n$, because it is the finite intersection of the open inverse images of $(0,\infty)$ under the continuous maps $x \mapsto x_{k+1} - x_k$ for $k \in \{1,\dots,n-1\}$; hence $C \in \mathcal{B}(\mathbb{R}^n)$. Then $A_\pi = X^{-1}(T_\pi^{-1}(C))$. By invariance of $\mu^{\otimes n}$ under $T_\pi$, \begin{align*} \mathbb{P}(A_\pi) = \mu^{\otimes n}(T_\pi^{-1}(C)) = \mu^{\otimes n}(C) = \mathbb{P}(A_{\operatorname{id}}), \end{align*} where $\operatorname{id}: \{1,\dots,n\} \to \{1,\dots,n\}$ is the identity permutation. Thus all events $A_\pi$ have the same probability.[/step]

[guided]The point of this step is to turn the phrase “the sample is exchangeable” into an exact measure statement. Define the random vector $X: \Omega \to \mathbb{R}^n$ by $X(\omega) = (X_1(\omega),\dots,X_n(\omega))$. Because the random variables are independent and identically distributed with common law $\mu$, the law of $X$ is the product measure $\mu^{\otimes n}$ on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$. Now fix a permutation $\pi \in S_n$ and define the coordinate permutation map $T_\pi: \mathbb{R}^n \to \mathbb{R}^n$ by $T_\pi(x_1,\dots,x_n) = (x_{\pi(1)},\dots,x_{\pi(n)})$. This map only relabels coordinates. Since every coordinate has the same marginal law $\mu$, the product measure is unchanged by this relabelling. Concretely, for measurable rectangles $B_1 \times \cdots \times B_n$ with $B_i \in \mathcal{B}(\mathbb{R})$, \begin{align*} \mu^{\otimes n}\bigl(T_\pi^{-1}(B_1 \times \cdots \times B_n)\bigr) = \mu^{\otimes n}(B_{\pi^{-1}(1)} \times \cdots \times B_{\pi^{-1}(n)}) = \prod_{k=1}^n \mu(B_k) = \mu^{\otimes n}(B_1 \times \cdots \times B_n). \end{align*} Since measurable rectangles form a $\pi$-system generating $\mathcal{B}(\mathbb{R}^n)$, the uniqueness theorem for finite measures implies that the same equality holds for every Borel set in $\mathbb{R}^n$. Define the increasing cone \begin{align*} C := \{(x_1,\dots,x_n) \in \mathbb{R}^n : x_1 < x_2 < \cdots < x_n\}. \end{align*} This set is Borel: for each $k \in \{1,\dots,n-1\}$, the map $x \mapsto x_{k+1} - x_k$ from $\mathbb{R}^n$ to $\mathbb{R}$ is continuous, so the condition $x_k < x_{k+1}$ defines the open inverse image of $(0,\infty)$, and $C$ is the finite intersection of these open sets. The event $A_\pi$ is exactly the preimage of $T_\pi^{-1}(C)$ under $X$, because $X(\omega) \in T_\pi^{-1}(C)$ means \begin{align*} T_\pi(X_1(\omega),\dots,X_n(\omega)) \in C, \end{align*} which is precisely \begin{align*} X_{\pi(1)}(\omega) < X_{\pi(2)}(\omega) < \cdots < X_{\pi(n)}(\omega). \end{align*} Therefore, using the law of $X$ and the coordinate-permutation invariance of $\mu^{\otimes n}$, \begin{align*} \mathbb{P}(A_\pi) = \mu^{\otimes n}(T_\pi^{-1}(C)) = \mu^{\otimes n}(C) = \mathbb{P}(A_{\operatorname{id}}). \end{align*} Thus every strict ordering of the observations has the same probability.[/guided]

[step:Compute the common probability of an ordering] Since the events $(A_\pi)_{\pi \in S_n}$ are pairwise disjoint and their union is $E$, and since $\mathbb{P}(E) = 1$, we have \begin{align*} 1 = \mathbb{P}(E) = \sum_{\pi \in S_n} \mathbb{P}(A_\pi). \end{align*} The set $S_n$ has $n!$ elements, and the previous step shows that all summands are equal. Therefore, for every $\pi \in S_n$, \begin{align*} \mathbb{P}(A_\pi) = \frac{1}{n!}. \end{align*} [/step]

[step:Translate strict orderings into rank vectors] Let $r = (r_1,\dots,r_n)$ be a permutation of $\{1,\dots,n\}$. Define $\pi_r \in S_n$ to be the unique permutation satisfying \begin{align*} r_{\pi_r(k)} = k \end{align*} for every $k \in \{1,\dots,n\}$. On the no-tie event $E$, the equality $(R_1,\dots,R_n) = r$ holds exactly when \begin{align*} X_{\pi_r(1)} < X_{\pi_r(2)} < \cdots < X_{\pi_r(n)}. \end{align*} Hence \begin{align*} \{R_1 = r_1,\dots,R_n = r_n\} \cap E = A_{\pi_r}. \end{align*} Since $\mathbb{P}(E) = 1$, adding or removing $E^c$ does not change probability. Therefore \begin{align*} \mathbb{P}(R_1 = r_1,\dots,R_n = r_n) = \mathbb{P}(A_{\pi_r}) = \frac{1}{n!}. \end{align*} Since this holds for every permutation $r$ of $\{1,\dots,n\}$, the rank vector $(R_1,\dots,R_n)$ is uniformly distributed over the $n!$ permutations of $\{1,\dots,n\}$. [/step]