Question

Derive the distribution of the range of a sample of size $2$ from a distribution having density function $f\left(x\right)=2x,0<x<1.$

Short answer

Distribution of the range :$f_R\left(a\right)=\frac{4}{3}{a}^3-4a+\frac{8}{3}$

Step-by-step solution

:  Probability density function :

The probability density function is defined as the integral of the variable density density over a certain range. It is represented by the letter f(x).

Explanation :

Density function, $f\left(x\right)=2x$

Such that $0<x<1$.

The number of units in a sample size, $n=2$.

According to the statement,

$f\left(x\right)=2x$for $0<x<1$

Which yields

$F\left(x\right)=x^2$for the same value of x.

Then, $X_1,X_2$be the random variable.

Keep in mind that the range is a random variable that denotes the distance between the sample's maximum and least value.

So because sample has two variables, then

$R=\left|X_1-X_2\right|$

Take any $a\in (0,1)$.

Then

role="math" localid="1647252046377" $$\begin{gathered} P(R\le a)=1-P(R>a) \\ =1-2{\int }_a^1\left({\int }_0^{x_1-a}2x_1.2x_{2}d{x}_2\right) \\ =1-2{\int }_a^{1}2x_1{(x_1-a)}^{2}d{x}_1 \\ =1-2\left(-\frac{a^4-6a^2+8a-3}{6}\right) \\ =1-\frac{1}{3}(a^4-6a^2+8a-3) \end{gathered}$$

Hence by differentiating,

We have

$f_R\left(a\right)=\frac{4}{3}{a}^3-4a+\frac{8}{3}$