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Unformatted text preview: STAT 3133 Quiz 2 (913 M'ihs)  Fan, 2093 Name: M
Instructor: Jiahua Chen ‘ ‘ 1.13.: ‘1 . Read and understand the questions carefully,
. Correct answers are usually very short.
. Not all problems are equally difﬁcult. CONNA “ll
1b 2 2 3b a 3 1. Let /\ represent ‘a'rene’wal event. an = P(/\ oCcurs at trial 72) and:
fn = P (A occurs at trial n for the ﬁrst time). Answer the following questions with a brief 5....“ AvaA+1Am CAyLGAlGa \J LULJ.  A? (a) [2] Suppose 11.4 > 0, us > 0, and more generally, an > 0 if and only if n = “1 + 6kg f*r
some non—negative interrers k. What is the period of A?
«He wad 6% w 74 in will; WW5 , in W“ i“ 3‘4 l “3 “"H‘W‘k’l
. I I I . . . , I I ,. V \
yr: ‘l’kl‘lékz = 2(2k.+;kz) , n?) Amlkfhuf’ 2 , 0M )1 can be to and ’+ (gallows; H.)
) CA n 3 Mn 793 72
(b) [2] Suppose u = 2:10 HR = 25. Is A recurrent or transient? ‘ ' S { N“ 4,“ WM}! 2
MN MW WWMMWW  o
.gwe ) as». x M +vans‘e’n}. ' (c) [2] Strappose f0 = 0 and fn = 3‘" for n 2 1. Is A recurrent or transient? — ﬂ . 1‘; s n s
Fig): 2.10.6" 7 2/3 4‘8" = Z_ : ‘§ __. ‘9
“:0 m. h=i ’_§3_ 38
I
uw=+~= ; =_3‘=_S. .9 = n):I——‘=—2<vo => .2
"FM “‘33 3 l 3 V‘ U 3 3 {757.5%}: 2. Let {X3310 be a discrete time stochastic process. Answer the following questions with a » brief explanation. (a) [2] Suppose the state space of {Xn}:°=o, .S' = {It/3 : k =_ 0, 1, . . Is it still possible that :meailmveﬁuMﬁikOJJlﬂé? r0 Wish We 1% 995/6 my :4 Mat . {M W sler oh 1‘54 99A Spy‘42 cam L2 WMgawoleJ A) an. ,h ) 10AM WWI (b) [2] Suppose P(X4 = 2/3IX3 = 1/3,X2 = 1/3) = 0.2 and P(X4 = 2/3X3 = 1/3,X1=
0) = 0.1. Is it still possible that {Xn}§:0 is a discrete time Markov chain? Ma, Lama 7+ dob m‘ 9min?“ m ALka propmo. mx 5) 04‘,“ N51 (1,3,4
abwmwmssk(x3),wmvm+kmwmaa MUM!)me Pom: 5;) W W55 #1 5M1 Xsd‘g).
/ \ n v ., M n... '5 :..,A. : ﬂu“ _'4.L WAN“ 1 TH .' "4.: .
kc; {4] Suppose A1 has UXyO‘ﬂcutlal dlbhrlbuhlull W'ihu mean J. m it can; pass; {X73310 is a discrete time Markov chain? wPle=/X) " a,“
\/M. 6) Jim getaw “9%). mnclm UA/r‘ﬂvyea 54 ovaJail: (9 14a 94:12 cw 5: Well laws/ale as»
@ 9WW ll“ WWW rarely we W M t1 h.
a...
Q
\../ 3. Assume {Xn};',°=0 is a Markov Chain with transition probability matrix F l U of) U (1.5]
n K n K
P = UU U OU O I
0 0 0.6 0.4
O 0 0.2 0.8 [3] (a) Find the two step transition probability matrix. a 9.: o 0.: 0 0‘: o 0.? “0.1: o 9.2: ML PW=PL= 0.5 o 0.5 0 o>’ 0 0'2: 0K 03 W5
oooléoﬁtf o 00‘ .‘f' 0
. .34
0 o 0.2. 03 0 O 0% 0
o o 0.2 0.3 0 o m? 0.72 [3] (b) Suppose that the probability function of X0 is given by the vector ,60 = (0.2, O, O, 0.8).
Find the probability function of X2.
0 ,7/5 D 0 ' ZS 0‘ V f1=ﬁ°¢l):[o.2,o,o,o.8][ a mg 03 0M o 0 NW 05L
0 w 0'72 ’ :[otox , (2)0196", Mrs] [6] (c) Classify the state space. For each class, determine Whether it is recurrent or tran
sient. Determine their periods. : ([759: lc [4 [awe sited“; and WW W213]
. _..?" I V ~14” 0 +3 T3wa 75,0153 lhecimtle wms ﬂue m. 0M3 m plus an 1% M4, we, '2‘ ‘ ‘ u "b edu"‘”?l th' MC duc'bi? I I  3 » l l (d) Wnat does It mean y m C1016 S 18 re 1 e nth/7% Wmmm’rwé ¥a avid/14
K! H ‘ n ‘ ‘Hﬁ , ' ‘ . y l “Mn N V: outqu 9m .+ can bi usual s we WM [4] (e) Find the long run proportions of times when the MC is in state 0, in state 2. (Do not blindly solve 1rP=7r). Sal91¢ O 'M iii 7%. frim‘W W3,» 1%»: 1% #1. #75 m,
nth/AC WM N01 is 5% sad: 0. H Wm N07 ham 9% I WW. (0 V, “HM gr,in chwwa 62:?" W] , «(2:11 Wm W=CW2,7151_
0.7, art“ _ J. M “me   ‘71— 3/ .
i 00.47112 +087i; =7r3 => 1T3" 2772 Ml 171W; \ I) % . [3] (f)Calcu1atehm"“’°°E[X“l A mum/mg it tires 442 MC u m 91942 2.
A5 n—>v0 , )(n shbbzes wtk 4k
WWH‘Q Ami/iv“ 1T:J:D 0 3L 31], 2+6, 1 4. Let «'Zn}ff’=O be a usual branching process With Z0 = 1 and Zn = 23;? X71415, for n > O
with family sizes XW being iid random variables. Assume X071 has binomial distribution with parameters 3, p. That is7 3 k)p’°(1 p)3"° Paco,1 = k) = < for k = 0,1,2,3. (a) [3] For What values of p the probability of extinction is 1? Willy? ~
Tim jawbdm'bfﬁ WWWW in i WM 1% main M43 929' [A Kari “M /, EEXM] = s) ( b) [4] When 3) = 0.5, compute the probability of (ultimate) extinction. ‘2‘: Wm : i (3 6.;7K 32' f— Ds‘WSS 1' 0 3n. OILS S$ +  ' {Satay—swim
C i w 0» 2M" '9) $i ea ,4“, (e) When = 5/ 6, calculation the mean and variance of Z . ﬁr 4’
£1" :— mﬁm Am, 9372 : 3* a; r .—2 L
a '1 . ' v 1' U U f” : s”= ﬂan 9* = 1'5“ $3, 007/
z .1:
(‘9'??"5— x/ 7. Z n" '> 5 ., f. r l
m = 93v +0BE+ $561+ (9 v1 = 65.83%!
. I\~_/‘_/,. 5. In a simple random walk, we have a sequence of independent and identicaiiy distributed
random variables Z1, 22, . . ., such that P(Z1 = 1) =1), P(Z1 = —1)= q = 1 —p. (1)
As usual, the random W’lk is deﬁned as Xn = 23.1 Zi for n 2 l, with X0 = 0.
Deﬁne (as usual) I A5? = Pm = nXH < ms < r, . . . ,X2 < r,X1< 'rlXo = 0), A51”) = P(Xn =' —r,Xn_1 > —r,Xn_2 > —r, . . . ,X2 > r, X1 > ~rlX0 = 0)
for 1" > 0. It is known that their generating functions satisfy the relationship Me) = me]? In fact, it is known dom walk throng (“9“Tq'ﬁz’pc; (b) [4] Find the generating function of {fn}1°f;o. 7(17:0 .123 mvﬂfﬂm AM”) LM 1% g,“ Ac:
ww ~a> W 9r saw. _.___.__, _ i 3‘
:Afzgx ” SVVMA” %£g§x\____j vi??? «as
: le :«WW ((3) [4] Suppose p = 1 3. Calculate the probability that the ﬁrst passage time of the hﬂﬂ 5 larger than 4. ...
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 Spring '08
 Chisholm
 Probability theory, Markov chain, Random walk, transition probability matrix, time Markov chain, discrete time Markov

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