HW9-sol - NTHU MATH 2810, 2011 Solution to Homework 9 14....

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Unformatted text preview: NTHU MATH 2810, 2011 Solution to Homework 9 14. X: Y: Let D | X FD ( d ) P( D P (Y 2 L2 2 L2 Y| d) X P (| X Y dxdy 0 y L d y d L L ~ Unif (0, L ) ~ Unif (0, L ) Y | d) P( X Y X d) d) L m in( y d , L ) dxdy 0 y L d y dxdy d L d 2 L 38. 1 2 Y 3 4 5 (a) P( X (b) j, Y i) d L 0 X 1 2 3 4 5 1 1 1 1 1 1 1 1 1 P(X=1,Y=1)= P(2,1)= P(3,1)= P(4,1)= P(5,1)= 5 5 2 5 3 5 4 5 5 P(2,2)= 1 1 1 1 1 1 1 1 P(3,2)= P(4,2)= P(5,2)= 5 2 5 3 5 4 5 5 P(3,3)= 1 1 1 1 1 1 P(4,3)= P(5,3)= 5 4 5 3 5 5 P(4,4)= 1 1 1 1 P(5,4)= 5 4 5 5 P(5,5)= 1 1 5 5 1 1 5 j j 1, ,5 i 1, ,j P( X j |Y i) P( X j, Y i) P(Y i ) 1 1 5 j 5 1 k i 5k 1 j 5 k i 1 k 5 j i Jointly made by NTHU MATH 2810, 2011 Solution to Homework 9 (c ) P( X 5, Y 5) 1 1 P( X 5) 25 5 5) P( X 5) P(Y 5) P(Y 5) 1 25 P( X 41. (a ) f X |Y ( x | y ) 5, Y Hence, X and Y are not independent. f X ,Y ( x, y ) f Y ( y) 0 xe xe xe xe 0 x ( y 1) ( y 1) 2 xe x ( y 1) x ( y 1) x 0 dx xe xy f Y | X ( y | x) f X ,Y ( x, y ) f X ( x) x ( y 1) y 0 x ( y 1) dy (b) FZ ( z ) P( Z z x z) x ( y 1) P ( XY dydx 0 z) (1 e z ) e x dx 1 e z xe 0 0 f Z ( z) e z z 0 58. X 1 ~ Exp( ) X 2 ~ Exp( ) f X 1 , X 2 ( x1 , x 2 ) Y1 Y2 X1 e X1 X1 ln Y2 0 |J| | 1 hY1 ,Y2 ( y1 , y 2 ) 2 e ( x1 x2 ) x1 0 x2 0 X2 X2 Y1 ln Y2 1 1 y2 | 1 y2 y2 2 f ( ln y 2 , y1 ln y 2 ) | J | y2 e y1 y1 ln y 2 y2 1 Jointly made by NTHU MATH 2810, 2011 Solution to Homework 9 Theoretical Exercises 14. X ~ Geo( p ) Y ~ Geo( p ) (a) X 1 (b) P( X i| X Y n) P ( X i, Y n i ) P ( X Y n) p (1 p ) i 1 p (1 p ) n C1n 1 p 2 (1 p) n 2 i 1 Y n P( X 2 i| X Y n) n 1 C1n 1 1 n 1 n 1 1 n 1 22. W ~ (t , ) fW | X 1 x1 , , X n xn (w | X 1 x1 , , Xn xn ) f ( x1 , w t f ( w, x1 , , xn ) f ( x1 , , xn ) n , x n | W w ) f ( w) f ( x1 , , xn ) wt 1 w e i 1 wxi e 1 f ( x1 , n (t ) t ( , xn ) (t ) f ( x1 , w , xn ) w t n 1 xi ) w i 1 e (1) (1) n W | X1 x1 , , Xn x n ~ (t n, i 1 xi ) Jointly made by NTHU MATH 2810, 2011 Solution to Homework 9 31. X1, fX (k By Lecture notes p.6 43 , X n ~ Unif (0,1) 1) , X ( k ) n! s k 2 (1 w) n k (k 2)!(n k )! For 1 k n 1 X ( 0 ) 0 , X ( n 1) 1 ( s, w) Let U W fU (u ) X ( k ) U X (k and 1) W 0 s S s u 1 S f X 1 u ( k 1) , X ( k ) ( s, u s ) ds Let s (1 u ) x ds (1 u )dx 0 1 n! s k 2 (1 u s ) n k ds ( k 2)!(n k )! n! (1 u ) k 2 x k 2 (1 u ) n k (1 x) n k (1 u )dx ( k 2)!(n k )! 0 1 1 0 1 n! (1 u ) n (k 2)!(n k )! n! (1 u ) n (k 2)!(n k )! n (1 u ) n P( X ( k ) X (k 1) 1 x k 2 (1 x) n k dx (k 2)!(n k )! (n 1)! 1 1 x k 2 (1 x) n k dx 0 Beta(k 1, n k 1) 0 u 1 t) P(U t) t In fact, U ~ Beta(1, n) n (1 u ) n 1 du (1 t ) n 34. Let Y1 1 n X (1) Yn Yn X (n) y1 ) n 2 f Y ,Y ( y1 , y n ) M Y1 n(n 1) ( y n R Yn - Y1 2 2M R 2M R Y1 Yn and | J | 1 2 2 2m r 2 m r hR , M ( r , m) f ( , ) | J | n ( n 1) r n 2 2 2 0 r 1 r 2 m 1 r 2 Jointly made by ...
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