1.4 Distribution of the Range of a Random Sample
Let \(X_1,X_2,\dots ..., X_n\) be a random sample from some distribution \(f_X(.)\). Let \(X_{(j)}\), be the \(j^{\text {th}}\) order statistic, \(j=1,2,\dots , n\), then \(X_{(n)}\) and \(X_{(1)}\) are the \(n^{\text {th}}\)
and \(1^{\text {st}}\) order statistics respectively.
Find the expression for the distribution of the range \(R\), i.e \(R=X_{(n)}-X_{(1)}\).
\[f_{X_{(i)},X_{(j)}}(x,y) =\frac {n!}{(i-1)!(j-i-1)!(n-j)!}f(x)f(y)\Big (F_X(x)\Big )^{i-1}\Big (F_X(y)-F_X(x)\Big )^{j-i-1}\] \[\Big (1-F_X(y)\Big )^{n-j}\]
\[f_{X_{(1)},X_{(n)}}(x,y)=\frac {n!}{(n-2)!}f(x)f(y)\Big (F_X(y)-F_X(x)\Big )^{n-2}\]
\[\Big (X_{(1),X_{(n)}}\Big )\longrightarrow \Big (R,U\Big )\]
Let \(R=X_{(n)}-X_{(1)}=g_1(X_{(1)},X_{(n)})\), and \(U=X_{(1)}=g_2(X_{(1)},X_{(2)})\)
\[\implies \hspace {0.5cm} X_{(1)}=U,\hspace {0.6cm} X_{(n)}=R+U\]
\begin {align*} J=\text {Jacobian} &= \begin {vmatrix} \dfrac {\partial X_{(1)}}{\partial R}&\dfrac {\partial X_{(1)}}{\partial U}\\\\ \dfrac {\partial X_{(n)}}{\partial R}&\dfrac {\partial X_{(n)}}{\partial U}\\ \end {vmatrix}\\\\ &= \begin {vmatrix} 0&1\\1&1\\ \end {vmatrix}\\ &=-1 \end {align*}
\begin {align*} f_{U,R}(u,r) &=f_{(X_{(1)},X_{(n)})}\Big (g^{-1}_1(X_{(1)},X_{(n)}), g^{-1}_2(X_{(1)},X_{(n)})\Big )\Big |J\Big |\\ &=\frac {n!}{(n-2)!}f(u)f(r+u)\Big (F_X(r+u)-F_X(u)\Big )^{n-2} \end {align*}
\[\therefore \hspace {0.5cm} f_{U,R}(u,r)=n(n-1)f(u)f(r+u)\Big (F_X(r+u)-F_X(u)\Big )^{n-2}\]
\[\implies \hspace {0.5cm} f_R(r)=\int _un(n-1)f(u)f(r+u)\Big (F_X(r+u)-F_X(u)\Big )^{n-2}du\]
Example 1.4.1. Let \(X_1,X_2,.......,X_n\) be a random sample from exponential with mean \(1/\lambda \) find the pdf for \(R=X_{(n)}-X_{(1)}\). \[f_X(x)=\lambda e^{-\lambda x},\hspace {0.5cm} F_X(x)=1-e^{-\lambda x}\] \begin {align*} f_R(r) &=\int ^{\infty }_0n(n-1)\lambda e^{-\lambda u}\lambda e^{-\lambda (u+r)}\Big (1-e^{-\lambda (u+r)}-1+e^{-\lambda u}\Big )^{n-2}du\\\\ &=n(n-1)\lambda ^2e^{-\lambda r}\int ^{\infty }_0e^{-2\lambda u}\Big (e^{-\lambda u}-e^{-\lambda r}\,e^{-\lambda u}\Big )^{n-2}du\\\\ &=n(n-1)\lambda ^2e^{-\lambda r}\Big (1-e^{-\lambda r}\Big )^{n-2}\int ^{\infty }_0e^{-\lambda nu}du\\\\ &=(n-1)\lambda e^{-\lambda r}\Big (1-e^{-\lambda r}\Big )^{n-2},\hspace {0.5cm} r>0.\\\\ \therefore \hspace {0.5cm} f_R(r) &=(n-1)\lambda e^{-\lambda r}\Big (1-e^{-\lambda r}\Big )^{n-2},\hspace {0.5cm} r>0.\\ \end {align*}
If \(X_1\thicksim g(\alpha _1,\lambda )\), \(X_2\thicksim g(\alpha _2,\lambda )\) then \(Z=X_1+X_2\thicksim g(\alpha _1+\alpha _2,\lambda )\) provided \(X_1\) and \(X_2\) are independent.
If \(X_j\thicksim g(\alpha _j,\lambda )\) then \(Z=\sum \limits ^k_{j=1}X_j\thicksim g\Big (\sum \limits ^k_{j=1}\alpha _j,\lambda \Big )\) provided \(X_j's\) are independent.
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