# Expected value of $k$-th minimum among $n$ random variables uniformly distributed in $[0, 1]$.

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.

• en.wikipedia.org/wiki/Order_statistic
– user762914
Jun 14, 2020 at 18:24
• 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. $\qquad$ Jun 14, 2020 at 19:40
• Please share your work on the problem. Jun 14, 2020 at 19:46

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 $$X_{(k)}.$$
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.
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.