solutions8 - cS r` 3 xW 3 3z b" v sqp nm y p ut irGWoel...

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Unformatted text preview: cS r` 3 xW 3 3z b " v sqp nm y p ut irGWoel ` jh y 6 w 1@ w i2gf h v 6 vv @ RC 0 @ A9 5 v 87 4 8 @6 A9 5 87 4 6 8 and d k W ef d` 1 6 @ y 6 x vv y w 6 v q@ D 0 1@ x w u tW 3 W 3 ` 3 YT` 3 " " b ` 3 W 3 " and u W " W` " ` b ES ts&qrYqXVpg ES ` " W 21ih" b 4 S gfW Xed ` Q b ` " W cS aYXVU& . 3 3 " 3 3 3 3 Chapter 10, Problem 1. and hence we solve @ A9 5 87 4 8 A6 9 @ 5 87 4 6 8 " )('&% $#! and 2210" which gives @ I 5 A9 6 P B S T 8 Q R8 FD GEC B !3 B H3 This is maximized by y b cS y w " 1@ w b ` " W ~} | cS 1 !A| { z v y 6 w yx` i1@ w wvW 3 3 y . Now Let where Chapter 10, Problem 2. Homework 8: Partial Solutions and hence we solve 1 . The mle is . The likelihood is 2 R 0 6 Y R 6 TY R h @ Q TY Q R h 3R Q , the maximum data point. Note y 6w R 3 IGE C v Q34 PHFD3z 4 5 B b v " k v @ " 6 3 534 Q 4 confidence interval is 5 4 5 B b v " k v @ " 6 G @ A9G 7 5 & ( @ 8 & ( 6 4 z3 2 v 3 " k v 4 @ 1v)( 0 & where we get is and . The mle is . The & " k v @ 3 " 3 3z " #!v S Q @ g 3 A9 6 I @ 5 Q 8 z 8 '% " & $ z '& S z Q @ A9 6 I @ 5 3 3 z k v @ 8 " 8 z h h z h v k 4 5 4 xW u ` U z I 8 8 With . The MSE is uu Ecu k v 5 b 4 S (fW 1 5 b 4 S gfW 1 " k es" ` ` Q @ 5 z Q The nonparametric plug-in estimator is gradient of than the nonparametric plug-in. An approximate The estimates standard error is The asymptotic standard error of this is is Chapter 10, Problem 3. Chapter 10, Problem 4. The mle is R S T3R that . Hence, . Solving for . By simulation, the mle has MSE about .015, substantially smaller 2 T`XU R Y W Q U 3R V 4 v 5 B C ! Y5 b Q & Q . Then, . The estimated stan. An approximate @ $& 3 2 5 B S ! 27 5 B S 3R Y7GP 3R " & P 3R Q34 534 2 5 5 U U R Y6x R " & 5 R & R & 4 x R & Q Q h & 4 U 4 2 R & 4 U R P 4 U R W Q U RQ h 3R & 3 Q Q Q W R Q @ Q k W @ 4 Q @ Q 43 Q2 Q v 4 2 0 wdA@ 1 1 Chapter 10, Problem 6. (a) (b) Let . is . The mle is v Q % k xA$ # " 6 4 % !l 1 1 Q @ )0(UA&' | % A9 6 I @ 5 3 8 5 !l 8 I @Q A9 6 I @ 8 5 8 3 8 & 8 6 Y Q R 4 k dard error of tor is 95 per cent confidence interval is Thus, and 5 as (c) Chapter 10, Problem 5. has mean 2 numbers. . . Now, . The likelihood is . The mle is obtained by setting . Consistency follows from the weak law of large 3 so so the method of moments estima, the log-likelihood is yielding @ 5 v % 54 " 4 )r 3 @ 3 @ A9G & ( @ 8 & ( I @ 3 3 @ 3 @ 3 Q 2 Q 3 v 2 4 4 Q 0 v)( & @ d T@ & Q y @ v 6 w y k @ U ! U T@ A@ k 6 w Q @ @ 5 5 d ! y @ k v @ w k Q Q y @ @ w Q Q 4 4 @ v 6 6 @ q@ Q Q @ 3 2 3 3 Q and and the gradient of . is is 2 Q ~ 2 & 4 3 2 Q 5 S @ 4iU 2 3 & k ~ 4 k ~ 4 k h 4 kQ W Q & & 4 Q h Q Q $ @ R 25 R Y5 v ' R & 4 r R & 2 4 2 Q Q D 2 1 l rp @ 3 q Q 4 3 U @ 4 so (d) Note that 2 2 5S 2 4 2 Q of By the delta method, the estimated standard error of U @ ! Since The matrix is still consistent. , we have Chapter 10, Problem 7. (a) (d) The bootstrap code is: & (c) (b) The likelihood is (e) By the law of large numbers, is . The ARE is of second derivatives is converges in probability to so the mle is inconsistent. On the other hand, converges in probability to 4 # $ 3 " 2 , the Fisher information matrix is . For an arbitrary distribution and . The true value . So B <- 10000 tau.boot <- rep(0,B) for(i in 1:B){ xx1 <- rbinom(1,n1,p1.hat) xx2 <- rbinom(1,n2,p2.hat) tau.boot[i] <- (xx1/n1)-(xx2/n2) } 5 ...
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This note was uploaded on 05/25/2008 for the course STAT 36-625 taught by Professor Larrywasserman during the Fall '02 term at Carnegie Mellon.

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