1.5 Distribution of the Range
The range \(R = X_{(n)} - X_{(r)} > 0\) \[\left (X_{(n)}\, , \, X_{(1)}\right )\longrightarrow \, \left (R\, , \, U\right )\] where \(R = X_{(n)} - X_{(1)}, \hspace {0.2cm} U = X_{(1)}\). Hence \(X_{(1)} = U, \, \, X_{(n)} = R + U\) \[f_{1, n}(x,y) = n\, (n - 1)\, p(x) \, p(y)\, \left (P(y) - P(x)\right )^{n - 2}.\] \[J = \begin {vmatrix} \dfrac {\partial X_{(n)}}{\partial R} & \dfrac {\partial X_{(n)}}{\partial U}\\\\ \dfrac {\partial X_{(1)}}{\partial R} & \dfrac {\partial X_{(1)}}{\partial U}\\ \end {vmatrix} = \begin {vmatrix} 1 & 1 \\ 0 & 1\\ \end {vmatrix} = 1.\] \begin {align*} f_{R,U}(r,u) & = n\, (n - 1)\, p(u)\, p(r + u)\, \left [P(r + u) - P(u)\right ]^{n - 2}\, \begin {vmatrix} J\\ \end {vmatrix}\\ & = n\, (n - 1)\, p(u)\, p(r + u)\, \left [P(r + u) - P(u)\right ]^{n - 2}. \end {align*}
Therefore \[f_R(r) = n\, (n - 1)\, \int _{u}p(u)\, p(r + u)\, \left [P(r + u) - P(u)\right ]^{n - 2}\, du\hspace {0.5cm}\cdots \hspace {0.3cm} 1.5.1\]
Example 1.5.1. Find the pdf of the range given a random of size \(n\) from exponential with mean \(\frac {1}{\lambda }\)
Solution. \begin {align*} f_R(r) & = n\, (n - 1)\, \int _{0}^{\infty }\lambda \, e^{-\lambda \, u}\, \lambda \, e^{-\lambda (u + r)}\, \left (1 - e^{-\lambda (u + r)} - 1 + e^{-\lambda u}\right )^{n - 2}\, du\\ & = \lambda ^2 n (n-1) e^{-\lambda r}\int _0^{\infty }e^{-2\lambda u}\left (e^{-\lambda u}\right )^{n - 2}\, \left (1 - e^{-\lambda r}\right )^{n - 2}\, du\\ & = \lambda ^2\, n\, (n - 1)\, e^{-\lambda r}\, \left (1 - e^{-\lambda r}\right )^{n - 2}\int ^{\infty }_0e^{-\lambda n u}\, du\\ & = n\, (n - 1)\, \lambda ^2\, e^{-\lambda r}\, \left (1 - e^{-\lambda r}\right )^{n - 2}\Big [\frac {e^{-\lambda n u}}{n\lambda }\Big |_0^{\infty }\\ & = (n - 1)\, \lambda \, e^{-\lambda r}\, \left (1 - e^{-\lambda r}\right )^{n - 2}, \hspace {0.3cm} r >0. \end {align*}
\(F_r(r) = \left (1 - e^{-\lambda r}\right )^{n - 1}, \hspace {0.3cm} r \in (-\infty , \infty )\) (cdf or exponential) □
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