Hence result by recursion b exponential spacings let

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Unformatted text preview: tatistics of a sample of size n − 1 uniform over (0, U(n) ). Hence result by recursion. (b) Exponential spacings Let X1 , . . . , Xn+1 ∼ Exp(1), with pdf e−x , x > 0, independently. Define, Sk = X1 + . . . + Xk , k = 1, 2, . . . , n + 1. CLAIM U(k) = Sk , Sn+1 k = 1, . . . , n ⇒ U(1) < . . . < U(n) are ordered U (0, 1)’s. Proof n+1 fX (x1 , . . . , xn+1 ) = n+1 e −xi = exp − i=1 xi xi ≥ 0 i=1 s1 = x1 s2 . . . = x1 + x 2 sn+1 = x1 + x2 + . . . + xn+1 ⇒ x1 = s 1 x2 = s2 − s1 x3 . . . = s3 − s2 xn+1 = sn+1 − sn fS (s1 , s2 , . . . , sn+1 ) = fX (s1 , s2 − s1 , . . . , sn+1 − sn )|J | 62 where ∂ x1 ∂s1 ∂x2 ∂s1 ∂x1 ∂s2 ∂x2 ∂s2 ··· ∂x1 ∂sn+1 ∂x2 ∂sn+1 ··· |J | = . . . 1 −1 0 . . . . . . 0 ∂xn+1 ∂sn+1 ··· 0 −1 · · · 1 0 = ∂xn+1 ∂s1 ∂xn+1 ∂s2 ··· 0 . . . 0 0 ··· =1 1 n ⇒ fS (s1 , s2 , . . . , sn+1 ) = exp −s1 − (si+1 − si ) i=1 = exp(−sn+1 ) Now, let v1 = s1 , v2 sn+1 = s2 , . . . vn sn+1 = sn , vn+1 sn+1 0 &l...
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