Calculation of pdf

Question

A random variable $X$ is uniform distributed on $[0, \alpha]$ with unknown $\alpha >0$. There are made $n \in \mathbb N$ independent measurements. The results $(X_1,\dots ,X_n)$ are ordered from the smallest to the greatest, yet. I want to calculate the density of $X_i$. I am not able to solve this, in the solution they say it is given by $\frac{\alpha^{-n}x^{i-1}(\alpha -x)^{n-1}\Gamma (n+1)}{\Gamma (i) \Gamma (n-i+1)}1_{0 \le x \le \alpha}$. Help is much appreciated.

Answer 1

Let $X_{(i)}$ be the $i$th smallest statistic ($i$th order statistic) then,

$$P(X_{(i)} \leq x) = \sum_{k=i}^{n}\binom{n}{k}P(X\leq x)^{k} ((1-P(X\leq x))^{n-k} = \sum_{k=i}^{n}\binom{n}{k}\left(\frac{x}{\alpha}\right)^{k}\left(1-\frac{x}{\alpha}\right)^{n-k}$$

Intuition behind above equation: Let $Y = \sum_{j=1}^{n}\mathbb{I}(X_j \leq x)$. Note that $Y$ is a binomial random variable with success probability $P(X\leq x)$ and number of trials $n$.

It is easy to verify that $P(X_{(i)} \leq x) = P(Y \geq i)$.

Finally,

$$\begin{align}f_{X_{(i)}}(x) &= \frac{\partial P(X_{(i)} \leq x)}{\partial x} \\ &= \frac{1}{\alpha^n}\sum_{k=i}^{n}\binom{n}{k} (kx^{k-1}(\alpha-x)^{n-k} - x^{k}(n-k)(\alpha-x)^{n-k-1})\\ &= \frac{1}{\alpha^n}\sum_{k=i}^{n}\binom{n}{k} x^{k-1}(\alpha-x)^{n-k-1}(k(\alpha-x) - x(n-k))\\ &= \frac{1}{\alpha^n}\sum_{k=i}^{n}\left(\binom{n}{k}k x^{k-1}(\alpha-x)^{n-k} - \binom{n}{k}(n-k) x^{k}(\alpha-x)^{n-k-1}\right)\\ &= \frac{1}{\alpha^n}\left(\sum_{k=i}^{n}\binom{n-1}{k-1}n x^{k-1}(\alpha-x)^{n-k} - \sum_{j=i+1}^{n}\binom{n-1}{j-1}n x^{j-1}(\alpha-x)^{n-j}\right)\\ &= \frac{n!}{(i-1)!(n-i)!}\frac{1}{\alpha}\left(\frac{x}{\alpha}\right)^{i-1}\left(1-\frac{x}{\alpha}\right)^{n-i}\end{align}$$

You can also derive the pdf of $i$th order statistic of a general continuous distribution too using the same procedure which will lead to the formula mentioned in post by @GrahamKemp.

Answer 2

A random variable $X$ is uniform distributed on $[0, \alpha]$ with unknown $\alpha >0$. There are made $n \in \mathbb N$ independent measurements. The results $(X_1,\dots ,X_n)$ are ordered from the smallest to the greatest, yet. I want to calculate the density of $X_i$. I am not able to solve this, in the solution they say it is given by $\frac{\alpha^{-n}x^{i-1}(\alpha -x)^{n-1}\Gamma (n+1)}{\Gamma (i) \Gamma (n-i+1)}1_{0 \le x \le \alpha}$.

Help is much appreciated.

You want to calculate the probability density for some sample being $x$, in an arrangement with $i-1$ smaller and $n-i$ larger samples; where $x$ exists in the support of $[0..\alpha]$ and $X$ is conditionally uniform over that support for a given parameter $\alpha$.

So that is: $\quad f_{X_i}(x;\alpha) ~{= \dfrac{n!}{(i-1)!~1!~(n-i)!}\cdot f_{X}(x;\alpha)\cdot{F_{X}(x;\alpha)}^{i-1}\cdot{\big(1-F_{X}(x;\alpha)\big)}^{n-i} \\ \vdots\\ = \dfrac{\Gamma(n+1)~x^{i-1}~(\alpha-x)^{n-i}}{\Gamma(i)~\Gamma(n-i+1)~\alpha^n}\cdot\mathbf 1_{(0\leqslant x\leqslant \alpha)}}$

Now do you see how this order statistic's density function is obtained?