Calculating probability related to maximum of random variables Let $X_1, X_2, \cdots, X_n$ be non-negative continuous iid random variables. The goal is to find the probability:
\begin{align*}
\Pr(\max_{k+1 \leq i \leq j } X_i < \max_{1 \leq i \leq k }X_i)
\end{align*}
and I solve it by the following method:
\begin{eqnarray}
\Pr(\max_{k+1 \leq i \leq j} X_i \leq \max_{1 \leq i \leq k} X_i)=\int_0^\infty \Pr(\max_{k+1 \leq i \leq j} X_i \leq \omega) G_{\max_{1 \leq i \leq k} X_i} (d\omega)=\cdots=&\frac{k}{j}.
\end{eqnarray}
I guess that this problem can be solved by permutations methods. I was wondering if anyone could explain it.
 A: It's simpler than that.
Note that we don't care about the $X_i$ for which $j < i \leq n$, since they aren't part of the problem statement.  In what follows, we'll simply assume the indices are in $1, \dots, j$.
First, observe that either $\max_{k+1 \leq i \leq j } X_i$ or $\max_{1 \leq i \leq k }X_i$, whichever is larger, is $\max_i X_i$.  Restating the question taking this into account leads to:
Let $X_1, X_2, \cdots, X_j$ be non-negative continuous iid random variables. The goal is to find the probability that $\arg\max_i X_i \leq k$.
At this point, it should be clear how to proceed.  Since each of the $j$ observations is equally likely to be the maximum due to the continuity and the i.i.d. properties, the probability that the maximum is in the first $k$ observations is $k/j$.
A: First, the condition "non-negative" is redundant, as the result still holds without this condition (only relative rank ordering matters).
A permutation argument goes as follows: Since $X_1, \ldots, X_n$ are i.i.d., the distribution of $(X_1, \ldots, X_n)$ given the order statistic $(X_{(1)}, \ldots, X_{(n)})$ is
\begin{align}
P[X_1 = x_{(i_1)}, \ldots, X_n = x_{(i_n)} | X_{(1)} = x_{(1)}, 
\ldots, X_{(n)} = x_{(n)}] = \frac{1}{n!},
\end{align}
where $(i_1, \ldots, i_n)$ is a permutation of $(1, \ldots, n)$ (for a rigorous proof to this very intuitive result, see Example 2.4.1 of Testing Statistical Hypotheses by E. L. Lehmann and J. P. Romano). Accordingly, conditioning on the order statistic $(X_{(1)}, \ldots, X_{(n)}) = (x_{(1)}, \ldots, x_{(n)})$, determining the probability of the event $\max\limits_{k + 1 \leq i \leq j}X_i < \max\limits_{1 \leq i \leq k}X_i$ becomes a classical probability problem, meaning that it is sufficient to count the number of all possible permutations and the number of permutations such that $\max\limits_{k + 1 \leq i \leq j}X_i < \max\limits_{1 \leq i \leq k}X_i$ respectively. To this end, call the fixed $n$ distinct values $x_{(1)} < \cdots < x_{(n)}$ "$n$ positions". Obviously, the number of all possible permutations is $n!$, which is the number of ways of arranging $\{X_1, \ldots, X_n\}$ to the $n$ positions.  To count the number of permutations that meet the event constraint, we take three steps:  in the first step, we pick up $j$ positions from the $n$ positions to allocate $\{X_1, \ldots, X_j\}$, which has $\binom{n}{j}$ of ways. In the second step, we select one item from $\{X_1, \ldots, X_k\}$ to put it to the largest position of the $j$ positions picked up in the first step and arrange the remaining $j - 1$ items in $\{X_1, \ldots, X_j\}$ to the remaining $j - 1$ positions. In this way, the constraint $\max\limits_{k + 1 \leq i \leq j}X_i < \max\limits_{1 \leq i \leq k}X_i$ is met. There are $\binom{k}{1}\times (j - 1)! = k(j - 1)!$ ways of completing this step.  In the third step, we arrange the remaining $n - j$ items $\{X_{j + 1}, \ldots, X_n\}$ to the remaining $n - j$ positions, which has $(n - j)!$ of ways.  It then follows by the fundamental rule of counting that the number of permutations satisfying $\max\limits_{k + 1 \leq i \leq j}X_i < \max\limits_{1 \leq i \leq k}X_i$ is $\binom{n}{j} \times k(j - 1)! \times (n - j)! = n!\frac{k}{j}$, which results in
\begin{align}
P\left[\left.\max_{k + 1 \leq i \leq j}X_i < \max_{1 \leq i \leq k}X_i \right\vert X_{(1)} = x_{(1)}, \ldots, X_{(n)} = x_{(n)}\right] = 
\frac{n!\frac{k}{j}}{n!} = \frac{k}{j}. \tag{1}
\end{align}
(Note, your problem actually made no use of random variables $X_{j + 1}, \ldots, X_n$, so the above argument can be further simplified by only taking $X_1, \ldots, X_j$ into account: the probability $\frac{k}{j}$ is just $\frac{k(j - 1)!}{j!}$, in which the numerator is the number of ways of selecting one item from $\{X_1, \ldots, X_k\}$ and placing it to the largest position then arranging the remaining $j - 1$ items in $\{X_1, \ldots, X_j\}$, and the denominator is the number of ways of arranging $j$ items $X_1, \ldots, X_j$.)
Denote by $f(x_{(1)}, \ldots, x_{(n)})$ the joint density  of $(X_{(1)}, \ldots, X_{(n)})$ (which is is equal to $n!f(x_{(1)})\cdots f(x_{(n)}), x_{(1)} < \cdots < x_{(n)}$, where $f$ is the common density of $X_1, \ldots, X_n$. However, this expression is not needed to reach the final answer). It then follows by $(1)$ that
\begin{align}
 & P\left[\max_{k + 1 \leq i \leq j}X_i < \max_{1 \leq i \leq k}X_i\right] \\
=& \int_{x_{(1)} < \cdots < x_{(n)}}P\left[\left.\max_{k + 1 \leq i \leq j}X_i < \max_{1 \leq i \leq k}X_i \right\vert X_{(1)} = x_{(1)}, \ldots, X_{(n)} = x_{(n)}\right]
f(x_{(1)}, \ldots, x_{(n)})dx_{(1)}\cdots dx_{(n)} \\
=& \int_{x_{(1)} < \cdots < x_{(n)}}\frac{k}{j}f(x_{(1)}, \ldots, x_{(n)})dx_{(1)}\cdots dx_{(n)} \\
=& \frac{k}{j}. 
\end{align}
As you can see, this permutation argument (if made rigorous) is actually much more verbose than the standard direct calculation:
\begin{align}
 & P\left[\max_{k + 1 \leq i \leq j}X_i < \max_{1 \leq i \leq k}X_i\right] \\
=& \int_{-\infty}^\infty P\left[\max_{k + 1 \leq i \leq j}X_i < x\right]f_{\max_{1 \leq i \leq k}X_i}(x)dx \\
=& \int_{-\infty}^\infty F(x)^{j - k}kF^{k - 1}(x)f(x)dx \\
=& k\int_{-\infty}^\infty F^{j - 1}(x)dF(x) \\
=& k\int_0^1 u^{j - 1}du \\
=& \frac{k}{j}.  
\end{align}
