vn1 j where s1 v1 s1 v2 s2 v2 s1 vn1 j

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Unformatted text preview: t; sn+1 < ∞ s1 < s2 < . . . < sn+1 = sn+1 . ⇒ s1 = v1 vn+1 s2 . . . = v2 vn+1 sn = vn vn+1 sn+1 = vn+1 fV (v1 , v2 , . . . , vn+1 ) = fS (v1 vn+1 , v2 vn+1 , . . . , vn+1 )|J | where ∂ s1 ∂v1 ··· ∂s1 ∂v2 ∂s2 ∂v2 ··· ∂s1 ∂vn+1 |J | = ∂s2 ∂v1 ∂s2 ∂vn+1 ··· . . . ∂sn+1 ∂v1 ∂sn+1 ∂v2 . . . ∂sn+1 ∂vn+1 vn+1 = 0 0 ··· 0 0 . . . vn+1 0 ··· 0 . . . v1 v2 v3 · · · 1 63 n = vn+1 n ⇒ fV (v1 , v2 , . . . , vn+1 ) = exp (−vn+1 ) vn+1 0 < vn+1 < ∞ v1 < v2 < . . . < vn+1 ∞ f V ( v1 , v 2 , . . . , v n ) = 0 n exp (−vn+1 ) vn+1 dvn+1 Successive integration by parts gives: fV ( v1 , v 2 , . . . , v n ) = n ! for v1 < v2 < . . . vn . We show that this is precisely the joint pdf of n ordered uniforms. Given X1 , X2 , . . . , Xn ∼ fX (x) then the joint pdf of the order statistics is fX(1) ,...,X(n) (x(1) , x(2) , . . . , x(n) ) = n!fX (x(1) )fX (x(2) ) . . . fX (x(n) ) Intuition: there are n! ways of gettin...
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