Question

Let \(Y_{1}

Short answer

The \(k\)th order statistic \(Y_{k}\) from a random sample of size \(n\) from a uniform distribution (0,1) follows a Beta distribution with parameters \(\alpha = k\) and \(\beta = n - k + 1\).

Step-by-step solution

Probabilistic representation

Let's suppose we have a random Variables \(Y_1, Y_2, ..., Y_n\) such that \(Y_1<Y_2<...<Y_n\) are uniform on the interval (0,1). Let's consider \(Y_{k}\), the \(k\)th smallest of these variables, and calculate its probability density function (pdf). We can express this as \(P(Y_k \le y) = P(\text{at least } k \text{ of the } Y_i's \text{ are less than or equal to } y)\). Because the \(Y_i\)s are independent and identically distributed, we use combinatorics and the binomial distribution to express the probability.

Use of combinatorial calculation

The probability that exactly \(j\) of the \(n\) variables are less than or equal to \(y\) (where \(y\) is between 0 and 1), is given by the binomial distribution with success probability \(y\), and can be expressed as \({n \choose j}y^j(1-y)^{n-j}\). Summing over \(j\) from \(k\) to \(n\) yields \(P(Y_k \le y) = \sum_{j=k}^{n} {n \choose j} y^j (1-y)^{n-j}\). This is a cumulative distribution function (CDF).

Deriving pdf from cdf

To get the pdf from the cdf, we take the derivative of \(P(Y_k \le y)\) with respect to \(y\). Differentiating the sum term by term, we get \(f(y) = {n \choose k}k y^{k-1}(1-y)^{n-k}\).

Comparison with Beta distribution

By comparison, this function \(f(y)\) is the pdf of a Beta distribution with parameters \(\alpha = k\) and \(\beta = n - k + 1\) because the Beta distribution has pdf \(f(x; \alpha, \beta) = C x^{\alpha - 1} (1 - x)^{\beta - 1}\), where \(C\) is the normalizing constant. Thus the pdf for the \(k\)th order statistic \(Y_{k}\) is a Beta(\(\alpha=k, \beta=n-k+1\)) distribution.