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Suppose I am a student in a class of $m$ students and I rank $i$-th among them, what is the probability that I do better than a student that ranks $j$-th in a class of $n$ different students? I modelled this problem assigning to each student his "incompetency", measured by a real number (so a student is better than another student if in this scale he scores less) and assuming that these scores are extracted for each student in i.i.d. manner from a fixed continuous distribution.

Formally, suppose $X_1,X_2, X_3,\dots,Y_1,Y_2,Y_3,\dots$ are i.i.d. real random variables with common distribution $\mu$, that we assume continuous, i.e. $\forall x \in \mathbb{R}, \mu(\{x\}) = 0$. If $m\in \mathbb{N}$, let $X^1_m,X^2_m,\dots,X^m_m$ denote respectively the reordering from $X_1,\dots,X_m$ such that $X^1_m$ is the smallest in the set $\{X_1,\dots,X_m\}$, $X^2_m$ is the second smallest in the set $\{X_1,\dots,X_m\}$, and so on (and so $X^1_m \le X^2_m \le \dots \le X^m_m$ and ties appear with zero probability, since $\mu$ is continuous). If $n \in \mathbb{N}$, define analogously $Y_n^1,\dots,Y^n_n$.

Given $m,n \in \mathbb{N}, i \in \{1,\dots,m\}, j \in \{1,\dots,n\}$, what it the probability of the event $$\{X^i_m \le Y^j_n\}?$$

Actually, I have in mind a simple strategy using brute force (i.e. disintegration, independence and integration by parts) to get to the result, but the calculations are a bit cumbersome... has anyone any idea how to get to the result in a simpler manner?

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1 Answer 1

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It seems to me that your problem is equivalent to this. Have a urn with $m$ white balls and $n$ black balls. You draw $i+j-1$ balls from it. These are your top $i+j-1$ students. We don't need the exact ranking amongst these $i+j-1$ and neither amongst the remaining balls. We just need to know that every drawn ball is better than any non-drawn ball. We pick $i+j-1$ balls because:

  • we cannot have $i$ white balls and $j$ black balls at the same time;
  • however, there must be either at least $i$ white balls or at least $j$ black balls (mutually exclusive).

So, if happens that you have at least $i$ white balls, it means that the $i$-th student of the first group is better than the $j$-th student of the second group. The other way around if you have $j$ or more black balls.

In the end, you just need to calculate the probability of having at least $i$ white balls amongst the balls drawn (hint: hypergeometric).

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  • $\begingroup$ Nice solution! This means the answer does not depend on $\mu$, does it? $\endgroup$ Aug 1, 2020 at 9:06
  • $\begingroup$ @FormulaWriter Correct, it doesn't. $\endgroup$
    – nicola
    Aug 3, 2020 at 11:57

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