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Massachusetts InstituteofTechnologyDepartmentofElectricalEngineering&ComputerScience

6.041/6.431: ProbabilisticSystemsAnalysis(Fall2010)

ProblemSet5: Solutions1. (a)Because of the requirednormalization propertyofany joint PDF,

2 2 2 23 13 21 = ax dy dx = ax(2 x) dx = a 22 12 + = a

3 3 3x=1 y=x x=1soa= 3/2.

(b)For1 y 2,y a 3

fY (y) = axdx = (y2 1) = (y2 1),

2 41andfY (y) = 0otherwise.

(c) Firstnoticethatfor1 x 3/2,

fX,Y (x, 3/2) (3/2)x 8xfX|Y (x| 3/2)= = 2 = .fY (3/2) 3 3 12 5

4 2Therefore,

3/2 1 8x

E[1/X| Y= 3/2] = dx = 4/5.x 51

2. (a) BydefinitionfX,Y (x, y) =fX(x)fY|X(y| x). fX(x) =axasshowninthegraph. Wehavethat

401 = axdx = 800a.

0SofX(x) =x/800.FromtheproblemstatementfY|X(y| x) =2

1x fory [0, 2x].Therefore,

1/1600, if0 x 4and0< y X.Thisoccurswithprobability

40 2x 1 1

P(Y> X) = fX,Y (x, y) dy dx = dydx = .1600 2y>x 0 x

Wecould havealsoarrivedatthisanswer byrealizingthat foreach possiblevalueofX,thereisa 1/2probabilitythatY> X.

(c)The joint density functionsatisfiesfX,Z(x, z) =fX(x) fZ|X(z|x). SinceZ isconditionally

uniformlydistributedgivenX,fZ|X(z| x) = 21x forx z x. Therefore,fX,Z(x, z) =

1/1600for0 x 40andx z x.Themarginaldensityfz(z) iscalculatedas

40 1 40|z| , if|z|

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3. (a) In order forXandYto be independent,anyobservationofXshouldnot giveany informationonY.IfXisobservedtobeequalto0,thenYmustbe0.

Inotherwords,fY|{X=0}(y| 0) fY (y).Therefore,XandY arenotindependent.=1.5

1

0, otherwise. 0.5

0

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x

2

1.8

1.6

1.4

x/2, if0 x 1,(b) fX(x) = 3x/2 + 3, if1< x 2,

fX

|Y(x |

0.

5)

fY|

X(y

|

0.

5)

fX

(x)

1.2

1fY|X(y| 0.5)=2, if0 y 1/2,0, otherwise.

1/2, if1/2 x 1,fX|Y (x| 0.5)= 3/2, if1< x 3/2,

0, otherwise.

0.8

0.6

0.4

0.2

00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

y

1.5

0.5

00 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x

1

(c)TheeventAleavesuswitharighttrianglewithaconstantheight.TheconditionalPDFisthen1/area= 8.The conditionalexpectation yields:

E[R| A] = E[XY | A]0.5 0.5

= 8xydxdy0 y

= 1/16.

(d)TheCDFofW isFW(w) =P(W w) =P(YX w) =P(Y X+w).P(Y X+w)canbecomputedbyintegratingtheareabelowthelineY=X+wforallpossiblevaluesofw.The linesY=X+wareshown belowforw= 0,w=1/2,w=1andw=3/2.TheprobabilitiesofinterestcanbecalculatedbytakingadvantageoftheuniformPDFoverthetwotriangles. Remember tomultiplytheareas bytheappropriate joint densityfX,Y (x, y)!Takenotethatthereare4regionsofinterest: w 0.

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w= 0 y= 2 x1.21

w (1, 0)y=x0.8

y

w=10.6

w (2,1)0.4

0.2

0

1 +w 2 +w0.2

1+w2+w2

ww2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x

TheCDFofW is0, ifw 0

0, ifw 0.

Asasanitycheck,FW() = 0 andFW(+) =1. Also,FW(w)iscontinuousatw=2andatw=1.

4. (a) Ifthetransmittersendsthe0symbol,thereceivedsignalisanormalrandomvariablewithameanof2andavarianceof4.Inotherwords,fY|X(y| 2)=N(2, 4).

Also,fY|X(y| 2)=N(2, 4)Theseconditionalpdfsareshowninthegraphbelow.

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than 1/2.Another wayto prove that the two are not independent is to calculate the productoftheirexpectations,andshowthatthisisnotequaltoE[XY].

(b)ApplyingthedefinitionofamarginalPDF,

for0 x 1,

fX(x) = fX,Y (x, y) dyyx+1

= 1dyx

= 1;

for0 y 1,

fY (y) = fX,Y (x, y) dxxy

= 1dx0

= y;

andfor1 y 2,

fY (y) = fX,Y (x, y) dxx

1= 1dx

y1= 2 y.

1

0 0.2 0.4 0.6 0.8 1

fY

(y)

1

0.80.8

fX

(x) 0.6

0.6

0.4

0.2

0

0.4

0.2

00 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x y

(c) Bylinearityofexpectation,theexpectedvalueofasumisthesumoftheexpectedvalues.Byinspection,E[X] = 1/2andE[Y] = 1.

Thus,E[X+Y] =E[X] +E[Y] = 3/2.

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(d)ThevarianceofX+Y is

E[(X+Y)2]E[X+Y]2 =E[X2] + 2E[XY] +E[Y2] (E[X+Y])2. (1)In part (c),E[X+Y] wascomputed,soonly theother threeexpressionsneed to becalculated.First,theexpectedvalueofX2:

1 x+1 1

E[X2] = x2 1dydx = x2 dx = 1/3.0 x 0

Also,theexpectedvalueofY2 is

1x+1 1E[Y2] = y2 dydx = (3x2+ 3x+ 1)/3dx = 7/6.

0 x 0Finally,theexpectedvalueofXY is

1 x+1

E[XY] = x y dy dx0 x

1= (2x2+x)/2dxdy = 7/12.

0Substitutingthese into (1),we getvar(X+Y) = 1/3 + 7/6 + 7/6 9/4= 5/12.

*Alternative(shortcut)solution toparts (c)and(d)*GivenanyvalueofX (in([0,1]),weobservethatY X takesvaluesbetween0and1,

andisuniformlydistributed. SincetheconditionaldistributionofYX isthesameforeveryvalueofX in [0,1],weseethatYX independentofX. Thus: (a)X isuniform,and(b)Y =X+U, whereU isalsouniformand independentofX. It follows thatE[X+Y] =E[2X+U] = 3/2.Furthermore,var(X+Y) = 4var(X) + var(U) = 5/12.

6. (a)LetAbetheeventthatthefirstcointossresultedinheads. TocalculatetheprobabilityP(A),weusethecontinuousversionof thetotalprobabilitytheorem:

1 1

P(A) = P(A| P=p)fP(p) dp= p(1+ sin(2p)) dp,0 0

whichaftersomecalculationyields

1P(A) = .

2

(b)UsingBayesrule,

P(A| P=p)fP(p)fP|A(p) = P(A) 2p(1 + sin(2p))

, if0p 1,= 10, otherwise.

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(c) LetBbetheeventthatthesecondtossresultedinheads.Wehave

1P(B| A) = P(B| P=p,A)fP|A(p) dp

01

= P(B| P=p)fP|A(p) dp02

1

= p2(1 + sin(2p)) dp. 1 0

Aftersomecalculation,thisyields

2 2 3 2 3P(B| A) = = 0.5110.

1 6 3 3

G1.Leta= (cos, sin ) andb= (bx, by).WewillshowthatnopointofRliesoutsideCifandonlyif

|b| | sin |, and |a| | cos| (2)

TheothertwoverticesofRare (cos, by) and (bx, sin ).If|bx| | cos | and|by| | sin |,theneachvertex (x, y) ofRsatisfiesx2+ y2 cos2+sin2= 1andno pointsofRcanlieoutsideofC.ConverselyifnopointsofRlieoutsideC,thenapplyingthistothetwoverticesotherthanaandb,wefind

cos2 +b2 1, and a2+sin2 1.

whichisequivalentto2.Theseconditions implythat (bx, by) liesinsideoronC,soforanygiven,theprobabilitythattherandompointb= (bx, by) satisfies (2)is

2| cos| 2| sin | 2= | sin(2)|

andtheoverallprobabilityis

1

2 2 4 /2 4| sin(2)|d= sin(2)d=

2 0 2 0 2.

Requiredfor6.431;optionalfor6.041 Page7of7

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6.041 / 6.431Probabilistic Systems Analysis and Applied Probability

Fall 2010

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MIT6_041F10_assn05_sol