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- New York
- Cornell University
- Engineering
- Engineering 2700
- Richard
- Solutions08.pdf

Anonymous

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Homework 8 1. a. implies that ?2)( 2 =XE ?= ? ? ? ? ? ? ? ? 2 2 X E . Consider n X i 2 ? 2 ? =? . Then () ( ) ? ? ? ? ==== ? ? ? ? ? ? ? ? = ??? n n nn XE n X EE ii 2 2 2 2 22 ? 22 , implying that is an unbiased estimator for ? ? ? . b. , so 1058.1490 2 = ? i x 505.74 20 1058.1490 ? ==? 3. a. We wish to take the derivative of , set it equal to zero and solve for p. () ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?xn x pp x n 1ln ()( )( ) p xn p x pxnpx x n dp d ? ? ?= ? ? ? ? ? ? ??++ ? ? ? ? ? ? ? ? 1 1lnlnln ; setting this equal to zero and solving for p yields n x p =? . For n = 20 and x = 3, 15. 20 3 ? ==p b. () () ()pnp n XE nn X EpE ===? ? ? ? ? ? = 11 ? ; thus is an unbiased estimator of p. p? c. ()4437.15.1 5 =? 4. a. () 2 1 1 2 1 1)( 1 0 + ?= + + =+= ? ?? ? ? ? dxxxXE , so the moment estimator is the solution to ? ? 2 ? 1 1 + ?= ? X , yielding 2 1 1 ? ? ? = X ? . Since .325 ? ,80. =?== ?x b. ()()( ) ? ?? n n n xxxxxf ...1;,..., 211 += , so the log likelihood is . Taking () ( ? ++ i xn ln1ln ?? ) ?d d and equating to 0 yields ? ?= + )ln( 1 i x n ? , so 1 )ln( ? ??= ? i X n ? . Taking ( ) i xln for each given yields ultimately . i x 12.3 ? =? 5. a. [] []() [ ] n i nn n x xxx x x x ? ? ? ? ? ? 2/exp ...2/exp...2/exp 2 1 22 1 1 ?? =? ? ? ? ? ? ?? ? ? ? ? ? ? . The natural log of the likelihood function is ()() ? ? 2 ln...ln 2 i ni x nxx ? ?? . Taking the derivative wrt ? and equating to 0 gives 0 2 2 2 = ? +? ?? i xn , so 2 2 i x n ? =? and n x i 2 2 ? =? . The mle is therefore n X i 2 ? 2 ? =? , which is identical to the unbiased estimator suggested in Exercise 15. b. For x > 0 the cdf of X if ( ) ( )xXPxF ?=?; is equal to ? ? ? ? ? ?? ? ?2 exp1 2 x . Equating this to .5 and solving for x gives the median in terms of ? : ? ? ? ? ? ?? = ?2 exp5. 2 x implies that () ?2 5.ln 2 x? = , so ?? 38630.1 ~ ==x . The mle of ? ~ is therefore () 2 1 ? 38630.1 ? . 7. a. () ()5.59,1.5718.13.58 25 396.1 3.58 =±=± b. () ()9.58,7.5759.3.58 100 396.1 3.58 =±=± c. () ()1.59,5.5777.3.58 100 358.2 3.58 =±=± d. 82% confidence 09.18.82.1 2 =?=?=?? ? ?? , so 34.1 09. 2 == zz ? and the interval is () ()7.58,9.57 100 334.1 3.58 =± . e. () 62.239 1 358.22 2 = ? ? ? ? ? ? =n so n = 240. 8. a. ()() =±=± 9.328439 25 100645.1 8439 (8406.1, 8471.9). b. 04.08.92.1 2 =?=?=? ? ?? so 75.1 04. 2 == zz ? 10. d.f. = n ? 1 = 7, so the critical value for a 95% C.I. is 365.2 7,025. =t . The interval is () ()8.32,6.276.22.30 8 1.3 365.22.30 =±= ? ? ? ? ? ? ? ? ± .

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