\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
\(\Prob (Y_k) \leq y\) is the probability that there is at least \(k\) numbers that are less than equal to \(y\).
If there are \(k \leq m \leq n\) numbers less than or equal to \(y\), then there must be \(n - m\) numbers greater than or equal to \(y\). The probability of the first thing happening for each number is \(y\), and for the second thing happening for each number is \(1 - y\). We also have to choose \(m\) numbers from the \(n\) to make them less than or equal to \(y\). Therefore, \[ \Prob (Y_k \leq y) = \sum _{m = k}^{n} \binom {n}{m} y^m (1 - y)^{n - m}. \]
We have \[ m \binom {n}{m} = m \cdot \frac {n!}{m! (n - m)!} = \frac {n!}{(m - 1)! (n - m)!} = n \cdot \frac {(n - 1)!}{(m - 1)! (n - m)!} = n \binom {n - 1}{m - 1}. \]
We have \[ (n - m) \binom {n}{m} = (n - m) \cdot \frac {n!}{m! (n - m)!} = \frac {n!}{m! (n - m - 1)!} = n \cdot \frac {(n - 1)!}{m!(n - m - 1)!} = n \binom {n - 1}{m}. \]
The cumulative distribution function \(F_{Y_k}\) is \[ F_{Y_k}(y) = \sum _{m = k}^{n} \binom {n}{m} y^m (1 - y)^{n - m}. \]
Therefore, the probability density function \(f_{Y_k}\) is \begin {align*} f_{Y_k}(y) & = F_{Y_k}'(y) \\ & = \sum _{m = k}^{n} \binom {n}{m} \left [m y^{m - 1} (1 - y)^{n - m} - (n - m) y^{m} (1 - y)^{n - m - 1}\right ] \\ & = \sum _{m = k}^{n} y^{m - 1} (1 - y)^{n - m - 1} \left [m \binom {n}{m} (1 - y) - (n - m) \binom {n}{m} y\right ] \\ & = n \left [\sum _{m = k}^{n} \binom {n - 1}{m - 1} y^{m - 1} (1 - y)^{n - m} - \sum _{m = k}^{n - 1} \binom {n - 1}{m} y^{m} (1 - y)^{n - m - 1} \right ] \\ & = n \left [\sum _{m = k}^{n} \binom {n - 1}{m - 1} y^{m - 1} (1 - y)^{n - m} - \sum _{m = k + 1}^{n} \binom {n - 1}{m - 1} y^{m - 1} (1 - y)^{n - m} \right ] \\ & = n \binom {n - 1}{k - 1} y^{k - 1} (1 - y)^{n - k}. \end {align*}
Since \(Y_k \in [0, 1]\), we must have \[ \int _{0}^{1} f_{Y_k} (y) \Diff y = 1, \] and hence \[ n \binom {n - 1}{k - 1} \int _{0}^{1} y^{k - 1} (1 - y)^{n - k} \Diff y = 1, \] and therefore we have \[ \int _{0}^{1} y^{k - 1} (1 - y)^{n - k} \Diff y = \frac {1}{n \binom {n - 1}{k - 1}}. \]