Proof of $P(R_{i} = j) = 1/n$ for ranked statistics I need some help to understand a proof in a book about nonparametric statistics. Topic is the distribution of ranked statistics.
Let $X_{1},....,X_{n}$ be a sample of continous iid random variables. Let $R_{i} = r_{i}$ be the rank of the i-th random variable in the sample. $P(R_{i} = j) = \frac{1}{n}, j \in \{1,..,n\}$ is to be proven. Proof:
The author defines these 3 events:
$ A = (R_{i} = k)$
$B = (R_{1} = r_{1},...,R_{k-1}=r_{k-1},R_{k+1}=r_{k+1},...,R_{n}=r_{n}), r_{j} \neq k, j \neq i$
$C = (R_{1} = r_{1},...,R_{i-1} = r_{i-1}, R_{i} = k, R_{i+1} = r_{i+1},...,R_{n}=r_{n})$
$(r_{1},....,r_{n})$ is a permutation of (1,...,n). Then the proof goes on to say that $C = A \cap B$ therefor:
$P(A) = \frac{P(C)}{P(B|A)} = \frac{\frac{1}{n!}}{\frac{1}{(n-1)!}} = \frac{1}{n} $
I understand the last section, but I don´t understand what the event B is and what is k in this proof? So I don´t understand why C is the intersection of A and B. I worked through a concrete example, but it seems non sensical to me.
 A: It appears that the proof in your post uses the joint distribution of the rank vector $(R_1,\ldots,R_n)$ and the conditional distribution of $(R_1,\ldots,R_{i-1},R_{i+1},\ldots,R_n)$ given $R_i$ to arrive at the marginal distribution of $R_i$. But the result follows more directly.
By definition, $R_i=j$ means $X_i=X_{(j)}$ (the $j$th order statistic) for every $j=1,2,\ldots,n$. Since $X_1,\ldots,X_n$ are i.i.d continuous random variables, the probability that any one of them equals $X_{(j)}$ must be $\frac{(n-1)!}{n!}=\frac1n$.
Alternatively, rank of $X_i$ is the number of observations in the sample less than or equal to $X_i$, i.e.
$$R_i=\sum_{j=1}^n I(X_j\le X_i)=1+\sum_{j(\ne i)=1}^n I(X_j\le X_i)\,,$$
where $I$ is an indicator function.
If $T=\sum_{j(\ne i)=1}^n I(X_j\le X_i)$, then the conditional distribution of $T$ given $X_i$ is $\text{Binomial}(n-1,F(X_i))$ where $F$ is the common distribution function of the $X_i$'s. Using this, it would follow that the marginal distribution of $R_i$ is uniform on $\{1,2,\ldots,n\}$.
