Assignment 3_Solution

Assignment 3_Solution - The University of Hong Kong...

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Unformatted text preview: The University of Hong Kong Department of Statistics and Actuarial Science Stat 1301 Probability & Statistics Assignment 3 suggested solution Section A \$223qu :1 = 0.2 + :11 + 0.3 +f{4}l +02 + 6.1 :2» fig} = 01 b) x 1 2 3 4 5 6 [51.“ F(3.7)=F(3)=0.6 2. a) gag +x)"‘2dx = 1,—c(1 +5415: = 1 => c = 1 . .. A" —2~ ‘ v a” 1 __1 , b)PCXixIl—fo(1+u3 G-u=-'C1+u) |0=1~E_E=;~,x:1 c)P£1 «if. x a: 2) = Ea +5-22ch = ...{1 + ,0 ~21}? =§ 3. 8) fo) = PUX] S x) = PlC—x g‘X :1 x) = FCx) ~Fﬂ-x} b) F (1‘3 = PW 5’— x) = PlI—v'i a: X 5:. VI?) = FEE} — Fit-m 4 a) Find the roots of the polynomial .2: —~ x2 = 0 Possible values for c2 = 0 and ,8 = 1 b) f;c{x-x2)dx=1,(xg~§)l% = LC = 5 5. 6. See example class 7 7. a) Joint cdf F(x,y) b) Marginal density f(x), f(x=‘l)=0.2, f(x=2)=0.5, f(x=3)=0.3 Marginal density f(y), f(y=1)=0.3, f(y=2)=0.26, f(y=3)=0.25, f(y=4)=0.19 f{1,1} =: 0.05 ‘7‘: f(x = :0va = 1) = 0.05 X and Y are not independent. C) X 1 2 3 _ _ F Y) 0.3 0.56 0.81 8. See Example Class 11 1. B1 (a) (“) (If ) 1P(X = ac) = w (“Lb—m ,a; = 0,1,--- ,Inin{k,a}. ( k ) (b) Note that k and :1: are ﬁxed when a and b diverge to inﬁnity. IP’(X = m) a! b! __ m!(a-w)! (Ic--a:)!(b-k+:c)! '— (a+b)! k!(a+b-k)! k! (a—I—b—k)! a! b' m!(k—:c)! (a—f—b)! (a—x)!(b—k-+ 3:)! (k)a(a——1)---(a,-93+1)-b(b—l)~-(b——k+a:+1) x (a+b)(a+b—1)---(a+b—k+1) k a (1—1 a—x+1 :1: a+b a+b—l a+b~w+1 \._.__._.____.\,._._______/ .._____.-- a+b—m a+b—a:——1 a+b—k+1 s._______‘,—_—_/ ..) (mm—mm a: ll Comment: The asymptotic distribution of X is a binomial distri— bution as a and b diverge to inﬁnity. 2. B2 Let N (t),t 2 0 be a Poisson process of rate A > 0, T1 the time of the ﬁrst event, and for n > 1, Tn denote the elapsed time between the 3. (n -— 1)th and nth events. {Tmn = 1, 2, - - is called the sequence of interarrival times. The event {T1 > t} takes place if and only if no event of the Poisson process occur in the interval [0,t] and thus lP’(T1 > t) = lP’(N(t) = 0) = exp{—At}. Hence, T1 has an exponential distribution of rate A. Since lP’{T2 > tITl = s} = lP’(O event in(s, 8 +t]lT1 = s) = lP’(O event in(s, 8 + t]|T1 = .9) (independent increments) exp{—At} (stationary increments), we have IP’(T2 > t) = 1E[1P’{T2 > t|T1}] = exp{—At}. Therefore, T2 is also an exponential random variable of rate A and, fur- thermore, that T2 is independent of T1. Repeating the same argument yields Tn, n = 1, 2, - - - are iid exponential random variables of rate A. B3 (a) It is readily seen fX,y(x, y) 2 O, and +00 +00 / fX,Y(\$7 y)dwdy __ 2 _ 2 2 (a: 2P1!) +(1 My hwy 1:0 “:00 1 : ———————eX ~ /_oo Lo 27n/1 ~p2 pl 2(1-p2) +00 2 +00 __ 2 m «276 -90 1/27r(1-p2) 2(1—p2) +00 1 y2 = 1. Thus ny is the joint density function of (X, Y). 4. (b) The marginal density function of X is PYCU) = _+me,Y(\$7y)d\$ _ “puma -mw m _ Lo 27n/1—p2 p{ 2(1—p2) }d _ “Lex 223 1 +°°ex lac—WV m _ m 13{2} 27r(1~p2)/.oo p{ 2(1~p2)}d = ﬁexp{%—} 1 _ Thus Y N N(0,1). Symmetrically we have X N N(0, 1). B4 (a) For any subset A1, . - - ,An E R, IP{X1 6 A1,--- ,Xn 6 An} 2 Z P(X1=£E1,-" ,Xn=xn) sue/11 ZnEAn \$16A1 wnEAn i=1 931641 “’T‘EA" i=1 Thus X1, - ' - ,Xn are independent. (b) Not necessarily. Refer to Lecture Note Section 7.2.5, pp.44. Section B 5. Let X be the random variable of the serving time. Let S be the time that the customer at A have been served when Mrs. Lee arrives at counter B. Probability that counter A will be free after time t P[|f;t’c:s—:-t}nﬁﬁ>s}:l __' ' = PHI :4; 3 + :1}: > s) = Mij [1 — {Matti} — (1 —~ 8 '35} = 1. — 6'7“ = P(X «a t] =Probability of counter B will be free after time t. Due to the property that the exponential distribution is memory—less, the probability that counter A will be free is the same as the counter B will be free. 6a)Note that then umber of arrival between [0,t] follows Poisson (At) Let X=# of buses arrival between 10;] Let X1 = number of full bus bemreem It}, 1:], X 2 = number of non. - full bus between. [0, T] (1'er ” 22.! P(X = n} = 9"“ PUQ = m) = PCX = j, m buses are fail} = 2 PM = j} pmﬂw pﬂ’m Fm Fm PM 233°34er ‘4‘? ji ._m pm m’ (ﬁbril —p3)j _ T y _m = j: 9 (magwmgujmwy =35; gum): ‘9 A (1"?) J 2m '3 H-m 29mm —— Wife (m1 it?) mi L3 z=c§ :2 P ( ’17) 9.9155 (1,1,{1 _ e-Q’ﬁCI-Qﬁ‘ mi (1 fp mm, —- 19)) ___: _-______________________ g—h‘p mi {imp “1 mi g-lztp Similarly para z m} z 34st,”; (it-r?“ ‘ 8 m. b)P(X = X1 =m) = PUL' = m, all are full?) = “2:? p m ...
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This note was uploaded on 04/29/2010 for the course STAT 1301 taught by Professor Smslee during the Spring '08 term at HKU.

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Assignment 3_Solution - The University of Hong Kong...

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