# 06_05 - fX(1,X(n(x1 xn dx1 dx n P x1 dx1 < X(1 < x1 dx1 2 2...

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p. 6-41 Q : Examine whether X (1) , …, X ( n ) are independent using the Theorem in LNp.6-21. f X (1) ,...,X ( n ) ( x 1 ,...,x n ) dx 1 ··· dx n P x 1 dx 1 2 <X (1) <x 1 + dx 1 2 ,..., x n dx n 2 ( n ) n + dx n 2 = ( i 1 ,...,i n ): permutations of (1 ,...,n ) P x i 1 dx i 1 2 1 i 1 + dx i 1 2 x i n dx i n 2 n i n + dx i n 2 = ( i 1 ,...,i n ): permutations of (1 ,...,n ) f ( x 1 ) ×···× f ( x n ) dx 1 dx n = n ! × f ( x 1 ) f ( x n ) dx 1 dx n . Theorem. If X 1 , …, X n are i.i.d. with cdf F and pdf f , then 1.The pdf of the k th order statistic X ( k ) is 2.The cdf of X ( k ) is f X ( k ) ( x )= n 1 ,k 1 ,n k f ( x ) F ( x ) k 1 [1 F ( x )] n k . Proof . F X ( k ) ( x n m = k n m [ F ( x )] m F ( x )] n m . X (1) X (2) p. 6-42 Theorem. If X 1 , …, X n are i.i.d. with cdf F and pdf f , then 1.The joint pdf of X (1) and X ( n ) is for s t , and 0 otherwise. 2.The pdf of the range R = X ( n ) X (1) is for r 0, and 0 otherwise. f X (1) ,X ( n ) ( s, t n ( n 1) f ( s ) f ( t )[ F ( t ) F ( s )] n 2 , f R ( r −∞ f X (1) ,X ( n ) ( u, u + r ) du, X ( k )

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p. 6-43 Theorem. If X 1 , …, X n are i.i.d. with cdf F and pdf f , then 1.The joint pdf of X ( i ) and X ( j ) , where 1 i < j n , is f X ( i ) ,X ( j ) ( s, t ) = n ! ( i 1)!( j i 1)!( n j )! f ( s ) f ( t ) × [ F ( s )] i 1 [ F ( t ) F ( s )] j i 1 [1 F ( t )] n j , for s t , and 0 otherwise. Reading : textbook, Sec 6.3, 6.6, 6.7 f S j ( s )= −∞ f X ( j 1) ,X ( j ) ( u, u + s ) du, 2.The pdf of the j th spacing S j = X ( j ) X ( j 1) is for s 0, and zero otherwise.
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06_05 - fX(1,X(n(x1 xn dx1 dx n P x1 dx1 < X(1 < x1 dx1 2 2...

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