All 11 Chapters Covered
h h h
SOLUTIONS
,Solutions Manual h
SUMMARY:h Inh thish chapterh weh presenth completeh solutionh toh thehexer
ciseshsethinhthehtext.
Chapterh 1
1. Problemh1.hAshdefinedhinhthehproblem,hA— h Bhishcomposedhofhthehelement
shinhAhthatharehnothinhB.h Thus,hthehitemshtohbehnotedharehtrue.h Makin
ghusehofhthehpropertieshofhthehprobabilityhfunction,hwehfindhthat:
Ph(Ah∪hB)h=hPh(A)h+hPh(Bh—hA)
andh that:
Ph(B)h=hPh(Bh—hA)h+hPh(Ah∩hB).
Combiningh theh twoh results,h weh findh that:
Ph(Ah∪hB)h=hPh(A)h+hPh(B)h—hPh(Ah∩hB).
2. Problemh 2.
(a) Ith ish clearh thathfXh(α) ≥
0.h Thus,h weh needh onlyh checkh thath th
ehintegralhofhthehPDFhishequalhtoh1.h Wehfindhthat:
∫h∞h
∫ ∞
(α)hdαh=h0.5h e−|α|hdα
fX
−∞ −∞
h ∫h 0 ∫h ∞
=h 0.5 α
e hdαh+ e−αh dα
−∞ 0
=h0.5(1h+h1)
=h 1.
ThushfXh(α)hishindeedhahPDF.
(b) BecausehfXh(α)hisheven,hitshexpectedhvaluehmusthbehzero.h Addition
-hally,hbecausehα2fXh(α)hishanhevenhfunctionhofhα,hwehfindhthat:
∫ ∞ ∫ ∞h h
α2hf X (α)hdα = 2h αh2f X (α)hdα
−∞ 0
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1
,2 RandomhSignalshandhNoise:h AhMathematicalhIntroduction
∫ ∞h
= α2e−αhdα
0
∫h ∞
byhparts
= (—αh2eh −α|h0∞
h +h αe −α dα
2 ∫
0h
∞h h
byhparts −αh ∞ −α
= 2(—αe |0h )h+h2 e dα
0
= 2.
Thus,h E(X2)h =h 2.h Ash E(X)h =h 0,h weh findh thath σ2h =h 2h andh σXh =
√ X
2.
3. Problemh 3.
Theh expectedh valueh ofh theh randomh∫variable
h
∞h h is:
E(X) = √ αe−(α− dα
1 µ)2h /(2σ2h
)
2πσ ∫h −∞
u=(α−µ)/σh h 1h h −uh2 /2
∞
= √ (σuh+hµ) dα.
2π e
−∞
2
Clearlyh theh pieceh ofh theh integralh associatedh withh ue−uh /2h ish zero.h The
remainingh integralh ish justh µh timesh theh integralh ofh theh PDFh ofh theh standar
dhnormalhRV—andhmusthbehequalhtohµhashadvertised.
NowhlethushconsiderhthehvariancehofhthehRV— —
∫h ∞h
lethushconsiderhE((Xh µ)2).hWehfindhthat:
E((Xh —hµ)2) = √ (αh—hµ)2e−(α− dα
1 µ)h2 /(2σ2h
)
2πσ −∞∫h
u=(α−µ)/σ 2hh 1h h 2h −u2h /2
∞
= σh √ uh e dα.
2π −∞
Ash thish ish justh σ2h timesh theh varianceh ofh ah standardh normalh RV,h weh fin
dhthaththehvariancehherehishσ2.
4. Problemh 4.
(a) Clearlyh(β —α)2 ≥
0.h Expandinghthishandhrearranginghithahbithweh findhtha
t:
β2h≥h2αβh—hα2.
(b) Becausehβ2 ≥ 2αβ —
α2h andhe−ah ishahdecreasinghfunctionhofha,hthehinequ
alityhmusthhold.
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, Solutionsh Manual 3
(c)
∫h ∞h 2 ∫ h ∞h 2
−βh / dβh ≤h
e 2 e−(2αβ αh )/2h
dβ
−
α α
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h h h
SOLUTIONS
,Solutions Manual h
SUMMARY:h Inh thish chapterh weh presenth completeh solutionh toh thehexer
ciseshsethinhthehtext.
Chapterh 1
1. Problemh1.hAshdefinedhinhthehproblem,hA— h Bhishcomposedhofhthehelement
shinhAhthatharehnothinhB.h Thus,hthehitemshtohbehnotedharehtrue.h Makin
ghusehofhthehpropertieshofhthehprobabilityhfunction,hwehfindhthat:
Ph(Ah∪hB)h=hPh(A)h+hPh(Bh—hA)
andh that:
Ph(B)h=hPh(Bh—hA)h+hPh(Ah∩hB).
Combiningh theh twoh results,h weh findh that:
Ph(Ah∪hB)h=hPh(A)h+hPh(B)h—hPh(Ah∩hB).
2. Problemh 2.
(a) Ith ish clearh thathfXh(α) ≥
0.h Thus,h weh needh onlyh checkh thath th
ehintegralhofhthehPDFhishequalhtoh1.h Wehfindhthat:
∫h∞h
∫ ∞
(α)hdαh=h0.5h e−|α|hdα
fX
−∞ −∞
h ∫h 0 ∫h ∞
=h 0.5 α
e hdαh+ e−αh dα
−∞ 0
=h0.5(1h+h1)
=h 1.
ThushfXh(α)hishindeedhahPDF.
(b) BecausehfXh(α)hisheven,hitshexpectedhvaluehmusthbehzero.h Addition
-hally,hbecausehα2fXh(α)hishanhevenhfunctionhofhα,hwehfindhthat:
∫ ∞ ∫ ∞h h
α2hf X (α)hdα = 2h αh2f X (α)hdα
−∞ 0
@@
SeSiesim
smiciiicsiosloaltaiotinon
1
,2 RandomhSignalshandhNoise:h AhMathematicalhIntroduction
∫ ∞h
= α2e−αhdα
0
∫h ∞
byhparts
= (—αh2eh −α|h0∞
h +h αe −α dα
2 ∫
0h
∞h h
byhparts −αh ∞ −α
= 2(—αe |0h )h+h2 e dα
0
= 2.
Thus,h E(X2)h =h 2.h Ash E(X)h =h 0,h weh findh thath σ2h =h 2h andh σXh =
√ X
2.
3. Problemh 3.
Theh expectedh valueh ofh theh randomh∫variable
h
∞h h is:
E(X) = √ αe−(α− dα
1 µ)2h /(2σ2h
)
2πσ ∫h −∞
u=(α−µ)/σh h 1h h −uh2 /2
∞
= √ (σuh+hµ) dα.
2π e
−∞
2
Clearlyh theh pieceh ofh theh integralh associatedh withh ue−uh /2h ish zero.h The
remainingh integralh ish justh µh timesh theh integralh ofh theh PDFh ofh theh standar
dhnormalhRV—andhmusthbehequalhtohµhashadvertised.
NowhlethushconsiderhthehvariancehofhthehRV— —
∫h ∞h
lethushconsiderhE((Xh µ)2).hWehfindhthat:
E((Xh —hµ)2) = √ (αh—hµ)2e−(α− dα
1 µ)h2 /(2σ2h
)
2πσ −∞∫h
u=(α−µ)/σ 2hh 1h h 2h −u2h /2
∞
= σh √ uh e dα.
2π −∞
Ash thish ish justh σ2h timesh theh varianceh ofh ah standardh normalh RV,h weh fin
dhthaththehvariancehherehishσ2.
4. Problemh 4.
(a) Clearlyh(β —α)2 ≥
0.h Expandinghthishandhrearranginghithahbithweh findhtha
t:
β2h≥h2αβh—hα2.
(b) Becausehβ2 ≥ 2αβ —
α2h andhe−ah ishahdecreasinghfunctionhofha,hthehinequ
alityhmusthhold.
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, Solutionsh Manual 3
(c)
∫h ∞h 2 ∫ h ∞h 2
−βh / dβh ≤h
e 2 e−(2αβ αh )/2h
dβ
−
α α
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