I have two sets of variables as: $x=(x_1, \ldots, x_n)$ and $y=(y_1, \ldots, y_m)$. Both members of $x$ and $y$ are independently and uniformly distributed between $[0,1]$.
I want to ask about the probability of the event that there at lease one member of $x$ has greater value than members of $y$.
My intuition is that if $n =1$ and $m = 1$, the questioned probability equals to $1/2$. Hence given the value of $n$ and $m$, the probability can be determined as
\begin{align} \Pr\{\exists~ (x_{i} > y_{j})\} & = 1 - \Pr\{\forall(x_i < y_j)\} \\ & =1 - \frac{1}{2^{n \times m}} \end{align} where $2^{n \times m}$ is the number of pairs between $x$ and $y$.
Please give me some comments or hints about this problem. I am very appreciated it. Thank you very much.
Let's start step by step. The first step of finding the probability of the complement was the correct way to begin. If $A$ is the event that there exists $(i,j)$ such that $x_i > y_j$ then we are interested in finding $\Pr(A) = 1 - P(A^c)$ where, $A^c$ is the complement of $A$. The event $A^c$ can then be described as $\forall i \in \{1,2, \dots, n\}$, $\exists j \in \{1,2, \dots, m\}$ such that $x_i < y_j$, which is saying that no member of $x$ has a value greater than all the elements in $y$. This is equivalent to saying $x_i < z$ for all $i$ where $z = \max\{y_1, y_2, \dots, y_m\}$.
If $f_Z$ is the density function of $Z = \max\{Y_1, Y_2, \dots, Y_m\}$, then our required probability is \begin{align} \Pr(A^c) & = \int_{0}^1 \Pr( \forall i \ (x_i < z)) f_Z(z) dz \\ & = \int_{0}^1 \prod_{i = 1}^n \Pr(x_i < z) f_Z(z) dz \\ & = \int_{0}^1 (z^n) (m z^{m-1}) dz \\ & = \frac{m}{m + n} \\ \end{align}
In the second step, I used the independence of $x_i$'s and in the third I used that they are uniformly distributed.
Thus the required probability $\displaystyle \Pr(A) = \frac{n}{m + n}$
I directly used the known formula for $f_Z(z)$ here. We can also derive it very simply. Note that \begin{align} \Pr(Z < z) & = \Pr( \max\{Y_1, Y_2, \dots, Y_m \} < z ) \\ & = \Pr( \forall j \in \{1,2,\dots, m\} \ (y_i < z)) \\ & = \prod_{j = 1}^m \Pr(y_j < z) \\ & = z^m \end{align} I again used the independence and the uniform distribution of $y_j$'s. Therefore, we have $F_Z(z) = z^m$. On differentiating it, we obtain the required pdf.
There is another simple way of understanding the solution. Event $A$ is another way of stating that the maximum of the set $\{x_1, \dots, x_n, y_1, \dots, y_m\}$ is actually an element from $\{x_1, \dots, x_n\}$. Since all of them are independent and uniformly distributed, the probability that $x_i$ is the maximum is $\dfrac{1}{n+m}$. Then $\displaystyle \Pr(A) = \sum_{i = 1}^n \Pr(\max = x_i) = \sum_{i = 1}^n \frac{1}{m + n} = \frac{n}{m + n}$.