Expected value of number of values greater than max of uniform variables

Question

Let $X_i$ be $i\in \{1,\ldots, n-k\}$ uniform variables over $[0,1]$

Let $Y_j$ be $j\in \{1,\ldots, k\}$ uniform variables over $[0,1]$

I would like to know $\mathbb{E}[|\{i|X_i > \max\limits_j(Y_j)\}|]$

I think the underlying probability follows a binomial distribution where the probability is being greater than the expected value of $\max(Y_j)$, which would give $\mathbb{E} = \frac{n-k}{k+1}$, but i don't know how to prove it.

Answer 1

I am assuming that the $X_i$'s and $Y_j$'s are independent.

Let $Z = \max\{ Y_1, Y_2, \dots, Y_k\}$ and $N$ denote the random number of $X_i$'s that are greater than $Z$. Note that $N$ can be written as \begin{align*} N = \sum_{i = 1}^{n-k} \mathbb{1}\{X_i > Z\}, \end{align*} where $\mathbb{1}\{A\}$ denotes the indicator of the event $A$. Thus, \begin{align*} \mathbb{E}[N|Z] = \sum_{i = 1}^{n-k} \mathbb{E}[\mathbb{1}\{X_i > Z\}] = (n-k)(1 - Z). \end{align*} Using the tower rule, we know, \begin{align*} \mathbb{E}[N] = \mathbb{E}[\mathbb{E}[N|Z]] = \mathbb{E}[(n-k)(1-Z)] = (n-k) \left( 1- \frac{k}{k+1} \right) = \frac{n-k}{k+1}. \end{align*}

Answer 2

Let $Y:=\max_j Y_j$, then for $t\in(0,1)$ we have $$ \mathbb P(Y\leqslant t) = \mathbb P\left(\bigcap_{j=1}^k Y_j\leqslant t\right) = \prod_{j=1}^k \mathbb P(Y_j\leqslant t) = \mathbb P(Y_1\leqslant t)^k = \frac k{1+k}. $$ We then compute $$\mathbb P(X_1>Y)= \int_0^1\int_0^t kt^{k-1}\ \mathsf ds\ \mathsf dt = \frac k{1+k}.$$ From here it is clear that the probablity that $m$ of the $X_i$'s exceed $Y$ is $$ \sum_{m=0}^{n-k}m\binom{n-k}m \left(\frac k{1+k}\right)^m\left(\frac 1{1+k}\right)^{n-k-m}= \frac{k(n-k)}{k+1},\quad 0\leqslant m\leqslant n-k. $$