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Exam05solution - Midterm Exam 1 Math 756/856 March 9 2005...

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Unformatted text preview: Midterm Exam 1 Math 756/856, March 9, 2005 Nameflfimfl ID Score Since I haven’t provided you the Z— table t- table and x2 -tabte, you can. use 20,, to m and xi‘n in your solution. 2a means that P(Z > 20,) — a, the 1000: upper percentage point of standard normal distribution ta,“ means that P(t (n) > to, W) — cc, and xan means that P(X2(n) > xi“) = a, where n is the degrees of freedom 1. (40pts) Let X1, X2, - - 3X“ be an i.i.d. sample from a Normal distribution with mean it and known variance 0 = 1, i.e., X1, X2, ---,Xn ~ N01,, 1), with the density function 1 _(L—_Efi :1; = —e 2 , n ) T,” (a) (8pts) Find the method of moments estimate of is, denoted with Li. (b) (8pts) Find the maximum likelihood estimate of 4“: denoted with fit. (c) (8pts) What is the distribution of {1? Based on this distribution, construct a 95% confidence interval for n. (d) (8pts) Is it an unbiased estimator? Why? What is the variance of fit? (e) (8pts) What is the Fisher information I (p)? Is there any other unbiased estimate of n have Smaller variance than ,1}? Why? (a) XwA/ffl,!) Mug gxs/gg fiMere/a =X ._. (.5) Mad-- —- EM; ,u)__ fi/fi) 6 mzjj M“) WM“) ‘ MM?” " ego ya)“ .-.= —§,{§m) ~é§m w) / a a 51/66 0'1 3%: gugij/ 2% M6):- [UV m! 1415421):er Maw - ' ”0% .5wa gang Ma WWWM MW 2. (40pts) Let X1, X2, - - - ,Xfl be an i.i.d. sample from a distribution with probability density distribution fix] 6) = 6:1:‘9’1, a; 21 6 > 1. (a) (Spts) Find the method of moments estimate of 6, denoted with 6. (b) (8pts) Find the maximum likelihood estimate for 6, denoted with 6. (c) (Spts) Based on the factorization theorem, find a sufficient statistic for 6. (d) (8pts) What is the Fisher information 1(6), what is the asymptotic variance of the m.].e, 6. A (e) (Spts) Based on the asymptotic normality of rule, 6, construct a 95% confidence interval for 6. (a) Xxx/flac/MwEQ" 79!. W m 96 5? [500 146 49,3205 ”19% 4% (9%.. 4% 3&1? t? f" X fiat! Xfifié 655:?- (5) We?) WM): £97237)“ [6) 6M) 72 6!? w Wm. M) -~ -——- ”56“” r 11...,— J/ =3 WWI-=07? a 1%“; 7a fire—fir fif— C a , ( ) flxbg-‘Mw’ 9) :‘QZZWJ/ “fl mm gym X”) y: -fl/ W mevfi W): 4% m. M / MM, 77%):Z7flé 425 a (mg/424M Wfiw (d) 750%) =-- M“. WW) == [579-— —- (w) W 89 W 119% *5 i35331;gififW =- -EH-» '7‘; A . f 2 3. (20pts) Let X1, X2, ' ' - ,Xn be an i.i.d. sample from following distribution with the density function f(:z:|,\) : AEJIB—M, 0 < a; < oo, A > U. 5pts) Argue that Y : 2232131} is a suflicient statistic for A, why? (a K t (c t l C 1)) [5p 5) Find the maximum likelihood estimate for A, denoted with i. ) (5p 8) Compute E(1i,) based on the distribution of Y. (d) ( Spts) From part (c), find the constant C such that C;\ is an unbiased estimator of A. ‘- 4t 9, 1/0“ a”! 4: 0&5)“ - .- - ’ rH (6L) {haXzJ'fiXn'flQf- ll fl, Xth if], Egg-X46 xiv! vital-W). wigs) Mg) £15”. ...
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