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Let $X_{1}$,$X_{2}$ and $X_{3}$ be a random sample from exponential distribution with density function - $$f(x) = \lambda e^{-\lambda x}$$ $x>0$ and $\lambda>0$
Show that $X_{(1)}$,$X_{(2)}$-$X_{(1)}$ and $X_{(3)}$-$X_{(2)}$ are independent,where $X_{(i)}$ , i=1,2,3 are order statistics.

AshishSinha5
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1 Answers1

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The pdf of $\mathbf{X}=(X_{(1)}, X_{(2)}, X_{(3)})$ is $f_{\mathbf{X}}(x_1, x_2, x_3)=6\lambda^3 e^{-\lambda (x_1+x_2+x_3)}I_{(0<x_1<x_2<x_3)}$.

Setting $\mathbf{Y}=(Y_1, Y_2, Y_3)=(X_{(1)}, X_{(2)}-X_{(1)}, X_{(3)}-X_{(2)})$, we have $\det(\displaystyle\frac{\partial \mathbf{x}}{\partial \mathbf{y}})=1$. Also, $0<x_1<x_2<x_3$ is equivalent to $y_1, y_2, y_3>0$. Thus the pdf of $\mathbf{Y}$ is

$\begin{align*} f_{\mathbf{Y}}(y_1, y_2, y_3) &=6\lambda^3 e^{-\lambda (3y_1+2y_2+y_1)}I_{(y_1, y_2, y_3>0)}\\ &=(3\lambda e^{-3\lambda y_1}I_{(y_1>0)})(2\lambda e^{-2\lambda y_2}I_{(y_2>0)})(\lambda e^{-\lambda y_3}I_{(y_3>0)})\\ &=f_{Y_1}(y_1)f_{Y_2}(y_2)f_{Y_3}(y_3) \end{align*}$

where $f_{Y_i}$ is a pdf of $Y_i, i=1,2,3$. Thus $Y_1, Y_2, Y_3$ are independent.

shrimpabcdefg
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