tutorial9 solution

# tutorial9 solution - THE UNIVERSITY OF HONG KONG...

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Unformatted text preview: / THE UNIVERSITY OF HONG KONG DEPARTDJENT OF STATISTICS AND ACTUABIAL SCIENCE STAT1301 PROBABILITY AND STATISTICS I EXAJXIPLE CLASS 9 EX.1 Let X N Poisson(a), Y N Poisson([3) be two independent Poisson random variables. Find (a) the momentgenerating function of X; the pmf and the moment generating function of X + Y; (c) the moment generating function of X — Y. 2 edceé—O . L) Kw GEFWXH; %) =~ men m: a) r- WM, Li“) gig? (3'P(7=E"X)=o,7’f%7z) : eiwﬁ) 2 t" 9( 7‘ g (3’20 Wag/z ﬂXﬁ/ﬂﬂf) { acetuepm -/) 2 8 2 Q [3 X1; 7 Nﬂwm (0466) 77% 14510010! )HT: ijéfef) EX.2 The inside diameter of a cylindrical tube is a random variable with a mean of 3 cm and a standard deviation of 0.02 cm, the thickness of the tube is a random variable with a mean of 0.3 cm and a standard deviation 0.005 cm, and the two random variables are independent. 1 (a) Find the moment generating function of X N N a, b); (b) Hence, Show that if X N N (a, bl and Y N N (c, d%, Where X and Y are indepen— dent, then X + Y ~ N(a + 0,122 + d2); (c) Find the mean and the standard deviation of the outside diameter of the tube; . (d) Using the Chebyshev’s inequality, ﬁnd the lower bound for the probability that Wis Within 0.05 cm from the mean; (e) Evaluate the probability in part (d) by assuming that the inside diameter and the thickness of the tube are independently distributed as normal. , ﬂ) X/v/W'Aly) ’ngz d) (5; @zéghv’CS-ZWL/y, a: w Wm , : [[umk-Kms) amgaﬁig 73% % KM ? [7’ ’1 n 2((t/2E6': k” L e“ - (X’Af :: __ W) __ : ’Agle-Eéwz’ Ezﬂx / ~ 04? 2 °“ I LEEQW 25" 96C .. A “Kiwi if; 41+ aj/MX 2; ‘ 2‘ .04 0i,“ . r t 3 6W E a UL 2579 6594/ 2; L @ DIVA/(3,602 ) Z q 5% i M x wig @2559 ~ Ema W) L , ( MC X11 Y, £.'N:_Iﬂ+2'{:v/\/G’6J9-ova§) Hw‘rCf) » HIM/{4r} u Zena; 86f“; ﬁle/1cm pé/Nrﬂk 005’) .. (a+C)t+C£ﬁ42fL » ZAP/1 535, r “6 L “WC/VkrﬁtiWJ " X'FY’VWMC/élﬂj ' : Q§C2.2}61)*/ C} but 0 lg 24%, mama. 2 <— magma Tée ﬂlLM§;%ﬂL 2 ’ J. made, Mama; H: 9+2? 0 97% iffECT) :ECOJaZECT): 3 72(03):}.63 a,“ 23/: WT) = M» (1)) +4 VMT)=&02L+4CU.005’) L: omer o" Jami = 0- @2234 CM EX.3 . (3.) Given Z N nb(r,p) with mgf (ﬂ) , Show that if X,Y X +Y N nb(2,p). (b) Hence, ﬁnd the conditional pmf of X, given X + Y = n when X and Y are in— dependent and identically distributed geometric random variables with paxameter p. Nﬁ‘xwwﬁw’ HWW’W’WWHZM £23 geom(p), then~ - X: m) 5 0% an 666“) : efﬂélﬁépjx'fp (@0663 9%“): if; ewﬁmffﬂ 2 a”? ﬁCeY/«Wv eff/77K / 6 < 71—- V c 70 A; m 31% jWCDJ , If < we”; f, fo‘r 66): HX[{2€/{7[fj L :— \$079)] b) W(X:’X [X‘H/zm] P(X'f‘]’:n ) t <,X :rwf) +'z’=nj : 7/2/14) MUM) z W 9~ "7:; <6 gimmc MW [ 62.1% nglp2(/7y)/I~L t ' 1,2: / QZ/12/"~/”-I 3 Hem Wm” 25 dzsﬁgw m. ﬂsoﬁ m%['//2r--.rﬂ~/,7, EXA ' Suppose the distribution of Y, conditional on X = m, is N (:3, 11:2) and that the“ marginal distribution of X is uniform(0,l). Find (a) 130/); . (b) Var(Y); and (C) Cov(X, Y). M Em: Eager/>0} (1‘ WM “VA/(£070) ' :CX) QTZXW: 9( “'21 (:‘Xw/ﬂf (0f!) 5/ VH7): VarZEKY/XJJ 7L Eli/445720] ’ I 2 Wm («6X7 6606221, 7% : ‘ +4- : 3%!- [15 “3) IM‘ a) @Mvéggnmg ﬁg [XECYWJ Cir—(XL) ( ‘5.”- 3 av (Xﬁr): EMU’EOOECT) .2: .L- f L 3%?) m r .9.” {7/ ...
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tutorial9 solution - THE UNIVERSITY OF HONG KONG...

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