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Let $X_1, X_2, \ldots, X_n$ be $n$ random variables uniformly distributed in $[0, 1].$ I need to find out the expected value of $k$-th minimum. I tried finding out the cumulative distribution function but couldn't.

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    $\begingroup$ en.wikipedia.org/wiki/Order_statistic $\endgroup$
    – user762914
    Jun 14, 2020 at 18:24
  • $\begingroup$ Almost an hour ago I posted an answer, but now in recent minutes I've updated it so that it can be seen that the distribution of the $k$th order statistic is a beta distribution. $\qquad$ $\endgroup$ Jun 14, 2020 at 19:40
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    $\begingroup$ Please share your work on the problem. $\endgroup$ Jun 14, 2020 at 19:46

2 Answers 2

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You have $X_1,\ldots,X_n$ i.i.d. in $[0,1].$ (Your posting doesn't mention independence. I would have included that explicitly, just as it is included here in the letters "i.i.d.". Many people posting here omit to include that.) And you have $X_{(1)} < \cdots < X_{(n)},$ the order statistics, which are those same random variables sorted into increasing order.

\begin{align} & \Pr( X_{(k)}\le x) \\[8pt] = {} & \Pr\big( \text{at least $k$ of the $n$ observations are} \le x \big) \\[8pt] = {} & \Pr\left(\begin{array}{l} \text{at least $k$ successes in $n$ independent trials} \\ \text{with probability $x$ of success on each trial} \end{array} \right) \\[8pt] = {} & \sum_{\ell\,=\,k}^n \binom n \ell x^\ell (1-x)^{n-\ell}. \end{align} That is the cumulative probability distribution function of the order statistic $X_{(k)}.$

But if you differentiate it to get the probability density function, all but one of the terms cancel and you get a very simple expression.

Here is the density function: \begin{align} & \frac d {dx} \sum_{\ell\,=\,k}^n \binom n \ell x^\ell (1-x)^{n-\ell} \\[8pt] = {} & \sum_{\ell\,=\,k}^n \binom n \ell \Big( \ell x^{\ell-1} (1-x)^{n-\ell} - x^\ell (n-\ell)(1-x)^{n-\ell-1} \Big) \\ & \qquad\qquad \text{by the product rule} \\[10pt] = {} & \sum_{\ell\,=\,k}^n \binom n \ell \ell x^{\ell-1} (1-x)^{n-\ell} - \sum_{\ell\,=\,k}^n \binom n \ell x^\ell (n-\ell)(1-x)^{n-\ell-1} \\[8pt] = {} & \sum_{\ell\,=\,k}^n n \binom{n-1}{\ell-1} x^{\ell-1} (1-x)^{(n-1)-(\ell-1)} - \sum_{\ell\,=\,k}^{n-1} n \binom{n-1}\ell x^\ell (1-x)^{(n-1)-\ell} \\ & \qquad\qquad \text{The upper bound of summation changed from} \\ & \qquad\qquad \text{$n$ to $n-1$ because the term involving $n$ is zero.} \\[8pt] = {} & \underbrace{ \sum_{j\,=\,k-1}^{n-1} n \binom{n-1} j x^j (1-x)^{(n-1)-j}}_\text{substitution: $j\,=\,\ell\,-\,1$} - \sum_{\ell\,=\,k}^{n-1} n \binom{n-1}\ell x^\ell (1-x)^{(n-1)-\ell} \end{align} Now these two sums are exactly the same except that the first one has a term where the index is $k-1$. Therefore, all of the others terms cancel out when the two expressions are subtracted. The result is $$ n\binom{n-1}{k-1} x^{k-1} (1-x)^{(n-1)- (k-1)}. $$ This is a beta density.

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You can use "the circle trick". Hence you get $k/(n+1)$: the circle is divided into $n+1$ parts and you need the first $k$ parts out of them.

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