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STAT1301 (T4 ans)

# STAT1301 (T4 ans) - THE UNIVERSITY OF HONG KONG DEPARTMENT...

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Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACT UARIAL SCIENCE STAT 1301 PROBABILITYAND STATISTICS I EXANEPLE CLASS 4 Moment The r-th moment about the origin of X: H; = Eix") The r—th moment about the expectation of X: ' M- = 306 " Fr)" Moment Generating Function . ‘3“ Given a random variable X, the moment generating function is deﬁned by “X (f) :1 g 6%) a» j at)? {[10 64‘ ‘A Mgor) = 5'02“] r and we have . 2 {£14696 gf "+16%??ij A 8 mm), ._. Who) = 3(an : {row thW+ 152% x Z! x at” go “A”? gngx/r h («Liam {flax/Mt; mt... V V09: dig/X0" fl’xﬂ '6 n— 1. For a random variableX ~Bz'nomz‘al (n, p) [643: E€eﬁ<ﬂ fCKQEJZﬁ As the moment generating function of X is :xca A v Matt) = (pet + 1 ~29)” Find E(X) and Var(X) by using the formula. EMF/75(0) ECX‘)514W/ﬂ@€‘~é/~ﬁ)”inﬂf Q/Wﬂdeo : A @7636 l’ﬂfﬂﬂeﬂko 2 22\$)ng8 ’ W09 ‘ , 2< ”AF Alias-2d xn Foam (A) “(WW W/zé—ieﬂx rims): g; @{X-i-fe" 2. Prove that if X has mgf Mxﬁ) and Y = a + EX, then Y has the mgf MY(t]=e“ luggameyp ”A ﬂ‘d‘f} : (£6ng ;: X43 7:": 6 C Eéeké'H7CX) :— jig:— U 46 {’X z: 6 i 5552 ’ 617 ix it €09= {12(0):}lafgfefw/g: z 6 ECG )26 [11\$ &6 _ ECK‘}:/1x((9€ 2916630: {ﬁledemﬁoézo ’ - 2: 2%) i» 5 3. Prove that if X and Y are independent random variables with mgf’s Mg and My and Z =X+ Y, 32:1? @092 Mz (t) = M3 (0M? (t) on the common interval Where both mgf’s exist. (Hint: if E(g(x))and E(h(x))exist and X and Y are independent R.V., then E(g(x)h(x))= E(g(x)) E(h(x))) my: ﬂat) L" W Em ewe/M mac 966659”? 75 565571 61/. f a” gen/0255mm, Wanner/1), mm ; éQD/xygg'y 15km xawmmmwy It (17 2mm 55 :mwmw Wanggw , rﬂmzceﬁn ~ 6, Markov Inequality; Let f (X) be a nonnegative function of the random variableX. If E[f(X)] exists, then, for every positive constant c, / E X Leif X 12 4230669. » Prom 2 c) s E“. ”- (ﬁt A a: {ac-{max} #129deer {7110 m” . C v“ . Chebxshev Inequality - Eéaﬂj: goo W5“) Let X be a random variable and c be a positive constant, then - PrGX~ul 2 ca) \$6715, 2 2%; £62350?) + xi: 63qu where p. = EU?) and 0’2 = Var(X). ‘0 O % 231% 2(qu 3“ , Hum-anew , meme, . 751190th a gimrcraemmg 1 z I / . 4. Let P(X = i): 6, 2': —4, — 3, — 2,--- ,2, 3, 4. Find PQXI 2 3) and check t e answer by usmg (I FWZCKEEL‘) Chebyshev’s inequality. FWD/3):éix‘”<O¢/70C”J)W0¢3)ffé<f¥) 4/0“”) Meg; fir g’éC/dw jngHT-O é g}; {2 gay/[€629] A {a E a §{[[4€)L{C”J)tr“f}‘{tgj;%§ 709,452 /’§f [Pt 5. Suppose that it is known the number of items produced in a factory during a week is a random variable with mean 50. Ifthe variance of a week’s production is known to equal to 25, what is the lower bound of the probability that this week’s production will be between 40 and 60? Kim / s6: 5’ WMU<X<W : FC/K’fO/ao) :- // [’0st [wag/me; 5“ ...
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STAT1301 (T4 ans) - THE UNIVERSITY OF HONG KONG DEPARTMENT...

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