STAT 514 - HW 3 Solution Key

STAT 514 - HW 3 Solution Key - 5.21 Let X1,v,X” be iid...

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Unformatted text preview: 5.21 Let X1,v--,X” be iid random vnfiabies and m be the median, for two r.v.’s, P(rnax(Xl,X2)> m): 1—P{max(X,,X2):g m)=1—[P(Xi g m)]2 : 1 {if = 133(max(Xl,---,Xn)L‘»m):1—P(nnax(XE,‘--,X”)gm)=1u[}‘-’(Xi gmfl” =1W[1Y‘ 5.23 P(Z s z) = 2:112 s 21X : x)P(X = x): Z{1—P(Z > 4X m x)}P(X : x) X no I 1*: an x =1— Emil—Poazmulm (W) anew—e ”Eb/.20”) emixz, x! 8—1 x=(} x! Iv: Thus, fz(z)m§m—1 for 0<z<1. e_ 5.24 LEE YT'WXB), ZflX(n),th€B _. n! :2 2wyn$2_n(n-1) w n_2 fy.z(yaz}m(n_2)[[6][ 49] _. 6” (z y) for 0<y<z<6 Let U=Y/Z, V=Z then Y=UV, Z=Vand Mm». Thus. mum): fy,z(W=V)* v e ”(2%“...wa - ”(2.3% W for 0<u<1, 0<v<6. The joint pdf of U and V is factored by functions of u and v, so U and V are independent. 1 1+6}: E 65:3 6 . _ 19.1 EXlzLx 2 dx=3x2+~g— a? then 6M0M=3X ml - -~ 6 . _ . . EHWM ==E(3X):3EX1==33:9 so, 91140.»; 23X IS unblased. a 1+QXI 1 62:41 1 E 82 3 2 EXl—Ifxz dx2wmx3+ :M", VOF’X :EXZ—(EX)2:~——-z “'6 I 2 6 8 m1 3 1 l I 3 9 9 : Then, VaréMOM =Varejf):~~Sv«:I/£1r1'.}i’1 2 3‘6- ——>0as Ira->00 n n Therefore, by Theorem 10.1.3 in CasellarflBerger éMOM = 35:: is consistent. 10.2 By Slutsky’s Theorem U" : aWn +in W9, then by Theorem 5.5.13 in Caseiiar/Berger U H = aWn + b” ——”—~>6. ...
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