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Unformatted text preview: EE1'32B.
 Professor Izhak Rubin Solution for Homework No. 2
SetNo. 4 Problem 1 Let F‘x(s) denote the moment generating function of the distribution function for X Fm) .—. ma“ ) 1
m __ _ 2
= / e‘szJ‘21—3e 202(3 g) dz
—m ' fa.
= j“ 21 a —E1;§[=' 202 a’s)=+u’]:
9; Id
3 ’2
= j; 21 e Tizllr(ss 05)] +2M’s «and:
V x175 l [r(# d 8)]2 e _263 dz. 1 ”Ix [O 'L': h I
2"
“In, 5.2
h Using the moment generating function, the mean and the variance of X is calculated as follows:
E(X) = 'li_rg zFxh) — 2 3_a2‘2
limde ‘2( p ) ll 1».0 d3
_ — 2 03:2
=  lim (2p 20 3):? ;( #1 )
1—0
::  ling —(2p— 2ar2 s)Fx(s). Since iim._.o ﬁx“) 2 0, we have ELK) =: p 2 d2
1'7}le = lim—F‘vfs‘l I {13
A
: [WE—J: Ee‘Ml‘ld:
0 A on
2 Lunch” Eeudxifo e"”ge_”dz
CI )1 on )1
_. _ {lIlrd / _ (»\+:3=
_ c z + 3 dz
[m2 a 2
_ A l 1
“ 2 As Jul5 '
Problem.2.4
. d 
Em  glad—ﬁns)
~_ 1' d) l + l
_, iﬂds? As A+s = —:imi[ 1 +—lL—] ”02 0—3)? 0+5)“
= 0.
a . d: 
E(X') = 'lﬂgﬁqu)
:—od33'2 A—s A+s
: limi[ 2 + 2 ]
3—02 (iv—s)“ (Ad.1)a
_ 3.
_. A3
= Vor(X). Problem 3 Set q =1 p. Problem3.1 Fo'rm=1.2,,mlandn=2,3,. P(X=m,Y=n) = P(Y=nX=m)P(X=m)
= P(Y—X=n—mX=m)P(X=m).
Set 2 = Y—X. Then, the random variable Z is the number of trials until the second head comes up after the ﬁrst. head occurs. Since a. geometric distribution has the memorylesa property, Z is governed by a geometric
distribution andis independent of X. Therefore, we have P(X=m,Y=n) =I P(Z=n—m}P(X=m)
{1 elq"'"‘" '(1 WP“
(lq)°q“". for n =2,3,.
Problem 3.2 P(X=m} flqiom”. PH” 2 n) = Z“ —qr}:'qr""'2 m=l = ('1 v 1H1 ‘ 4024"": Problem 3.3 Since P[Y = 1) = 0. the conditional probability that X = m given Y
for :1 =2.3.. = n is deﬁned onl: (1 41’4“"
{'1 1}(1 new
I nli for m=1,2.,n—l. Otherwise, PfX=me=nJ = 0. Note that given Y = n. the random variable X
has a. discrete uniform distribution in {1.2.   , n 1}. ' Problem 4 Recall the memoryleas property:r of an exponential distribution.  Problem 4.1 For an exponentially distributed random variable X with parameter A, P(X > t) = e'“. P X > 50. 'x > 30
Hit .1» 50) For > 30) e—LS —_ €419
O.G = e = 0.549. Est: nit: POI: >—5iJ—iux $36: PIX > 20}.
Problem 4.2 P(X>t+le>t) 11 II
re The solution above shows that P(X :> t+ :  X > r} = P(X > a), (i.e.. the random variable X has the
rnernorylese property}. P roblem 5(Spragins 4.1} I) The can! cncoding of states. malogoug to that in Figure 421:. is: WW Ring Gull Tilt Lina in: Cat. in: in: in:
0 0 o o o
1 I a o 1
0 2 o 1 o
1 3 o 1 1
0 4 l o o
1 s 1 o 1
0 6 1 1 o
l 7 1 1 1 ______ __ __ _ . "7; _____________________ No‘m 1mi I
0H 7.4
t: ' T V
“1.4 4—7.5 .6 new
Listening .
E. 1'
T A few comments may clarify the state diam "Tno possible sequences benedi
ately following the calling party picking up thephone (taking it oﬂ‘ book) are shown: either
thedial toneorthe userlistening mymmeﬁrnfollowedbythemhamepencﬁngonhow
quickly the caller puts the phone to his or her on). The same tinging signal has been
assumed to be observed by both the cslling party and the calledpattty.1 All conversations
have been assumed to be initiated by the celled party talking first (eg. saying 'Hello"):
although telephone company equipment does" not require this to hcppen, it is normal pro
cedure. Three possible results for an attempted cell are shown. a busy signal followed by
the calling party hanging up. no answer followed by the ceiling party. banging tip. 'or. ”suc
cessful completion of a call. If a call is successfully completed, the two partiee are
assumedmbepoumenoughthutheynevuanemptmnlkﬁmnlnneouslyiﬁmhermneh
always assumed to be talking dtning the conversation. to there are no periods with both silent. Solutions which do not involve these simplifying assumptions will be more com I
tytskestbepboneoﬂ’hoohndstartslisteningwotethedlutone). Thedinltooeister
minntedbythesurtofdialingmotshownindetnﬂ). Thisladstotheotherphonerinﬁng.
with tetminstion of ringing when it is answered. Thebn'ef conversation described then
ensues, with the eallingpartyhanging upsoonaﬁeneeognizing thmheeelled party hes
hunam Calling Party 0,—— Vain mm, W,
Vain I Rx. 3.4} (1.410.: 1.5 u 5.: 7 .s ' ,4 _
( )(H) () (.33 Ma) (7104) cured Party ma, .4 . 1. .4 .4
(3 1 (2.4) (H) ( 515”“! 0.5) a 1 a ) ProbKern ﬁgSPragin: 4.2) Although this proElem can be wind readily by use of equations in Section
4 1'1. «nth appmpriate interpretation 'a'fte'rms, it is intended L10 be solved di‘ractly. It sheuld he sokvable
Immgdtately after treatin; the material in Section 4.3. See the félltmt'tng ﬁrsurn which illustrate: wavefurma seen by transmitter and retain: af the data. message. {the ﬁgure is not drawn to sca‘laJ Aka the ﬁgure. Indicates. the receiver completes reception of the
data manage {plus overhead) at a. time equal to t, plus the time. (9M + naJJR. req'n' 24+;§§3+ 3“ + 0.001;: 2 + 11020 + 0.019;: 0'.14117(secancls}.. This yields an e‘ﬂ'ecté—ive data. rat£4: of R. z lUODlDJilIT = 7034 (bitafseaond}. ...
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 Spring '09
 IzhakRubin

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