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Suppose $X_1,X_2,\ldots,X_n $ are $n$ i.i.d. random variables with a continuous distribution $F(x)$ and density function $f(x)$. What is the probability distribution that any given $X_i$ is among the top $k$ largest of the $n$ $X$'s?

For example, there are $10$ individuals. Each of them draw a random number from a distribution $F(x)$ with density function $f(x)$. The draws are i.i.d. What is the probability that individual $i$'s draw will be among the top $3$ largest numbers (i.e. either top, second or third)?

Many thanks.

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    $\begingroup$ Well... the probability is k/n, by mere symmetry. $\endgroup$ – Did Aug 4 '14 at 18:20
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$X_1,\ldots,X_n$ are the $n$ i.i.d. random variables.

Let $X_{(1)}<\cdots< X_{(n)}$ be the order statistics, i.e. the same random variables sorted. (By continuity of the c.d.f., we need not write "$\le$".)

Then $X_i=X_{(j)}$. Given $i$, what is the distribution of $j\text{ ?}$

Let $Y_1,\ldots,Y_n$ be $X_{\sigma(1)},\ldots,X_{\sigma(n)}$, where $\sigma$ is some permutation.

Lemma: The joint distribution of $Y_1,\ldots,Y_n$ is the same as the joint distribution of $X_1,\ldots,X_n$.

Hence $\Pr(Y_i = Y_{(j)})=\Pr(X_i=X_{(j)})$. But $\Pr(Y_i=Y_{(j)})$ is the probability that $X_{\sigma(i)}$ is in the $j$th position when sorted. Thus the probability that $X_{\sigma(i)}$ is in the $j$th position is the same as the probability that $X_i$ is in the $j$th position.

This works regardless of which permutation $\sigma$ is, so it shows that every index has the same probability of being in the $j$th position. This is true for every value of $j$.

Hence the rank of $X_i$ in the sorting is uninformly distributed on the set $\{1,\ldots,n\}$. It probability of being in any particular subset is the size of that subset divided by $n$.

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  • $\begingroup$ Nice, but I really do not see the need of all this. Every individual evidently has here the same chance to end up in the top $k$ of $n$ (as Did implicitly remarks). $\endgroup$ – drhab Aug 4 '14 at 19:10

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