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Unformatted text preview: Section 6.4 Section 6.4 1/ a.) 0.5 b] positively dependent c) 0.2 d) 0.356 2. a] If P(AIB} = P{AB‘) then PM) = P{AB)P(B) + me‘ww‘)
= PtAIB}[P(B) + P(3‘11= PIAIB}. which implies that A and B are independent.
b) If HAIR) < P(AlB‘} then MA} = P(A13}P(B) + P(AB‘)P(B‘}
< P{AIB}[P(B}+ Ptﬂ‘ﬂ = PiAIB} :1 By interchanging the roles of B and B‘ in part b). we get the result immediately. d) Independence implies HAIR] = PM}. Plugging this into the given formula. we get
P{A{B) = MA) = P(AH)P(B) + P(AB°}P{B‘) 4:9 P{A§B){l — 19(3)] = P(AB‘}P(B‘}
=9 HAIR) = P(AB‘) so long as HS) 4. 1. e} Positive dependence implies
PUIIB) ‘2 P{A} = P{AB}P(B}+ P(AB‘)P(B‘} «=0 P{AiB}[l — P(B}] > P(AIB‘}P(B‘)
4:1 P(AB) ) P(AB‘) so long as P(B) < l. I'} This is just the same as the previous part with the inequality reversed. yﬂet Fl denote the event that the ﬁrst component fails. and W1 denote the event that the ﬁrst com
oneut works. etc. Then F.“ = W1. and P(F21F1] > PUB) implies that P
P(W2F1) = 1 — P(F2[Fi} < 1 — mm = P(W¢)
Aim.
thalwn = ——P;,‘f‘;ff;=*
_ I — P1131)” P(F2} + P(F1F2}
‘ 1— Pm} '
) 1 — Pm} — Pm} + P(F1}P{F2)
1 —P(F1)
= l — P(F2}
= Pam since Pm F3)1P[F:] = P(F1IF3) > Pm} 4. Joint distribution table for (X. Y} Section 6.4
—_————————————_—————— Margl'nal U4 U2 ”4 l
distn of X Now you can easily check that X and Y are uncorrelated but not independent: EIXY) = E[X] = 50’) = Cl — Cou(X.Y} = 0. but
P{X = 0, Y = 0) 7E P(X = 0)P(Y = D].
Sﬂaint distribution table for [X‘ Y) values I for X Matyinol distn
of Y values y 0 for Y Marginal us 1/3 113 1
diatnofx So compute ELY) = E(XY) = 0 =9 Cou{X. Y) = 0. But P{X = —~l. I” = 0} = [1 9E P(X = —1}P(Y = 0}. etc, so X and Y are not independent. In fact.
Y = X2 implies that X and Y are not independent. CautX.Y) = EﬁX. —X2](X3 +X2}] — El'X: — X2]E(X1+Xa)
= 50:? e X3} sinceE{X1} = E(X2)
= u sinceE(X12) = 5mg),
So X.Y are uncorrelated. But XJ" are not independent: For example. you can easily check that
P{X1 X2 = (LXI +Xz =3] = 0. While P{X1 — X2 =0}: li6,P{X1+X3 = 3J=1f18. Intuitively. if we know that X. — X3 = 0. then X: + X; must be even. since the two dice landed the
same way. So given that the diﬁ'erence is 0. we have some information about the sum. 7. a} Joint distribution table for (X3 + X3, X2 — X3) Marginal 1,33 [/2 lfﬁ 1
data of X2 + X; b] EM: — Xa)’ = M)" «US! + {01’ . {1x21 +0? 4113} = U6
C] E{X2X3} = PL“ =1.X.3 =1): P[X2 =1}P(X3 = I) by independence
= E(Xq)E(X3}. so X2 and X: are uncorrelated. 173 III
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n Section 6.4 8. Begin by noting that X;
= E{E(XXN1X:)]= E[XIE[XN I X1“ E{X1XN]
__ k—Xl __k3[‘Yli uut .4
‘ﬂx‘ N—rl" N—l " N—ll
* k3 _ _k_+ k“ _k(k—l}
“ N[N~1) N? N‘(N 1} ‘ N2
_ E{X1XN)’E(X!}E{Xu} ﬂ 5(XIXNI'[E{XI}}2
“’Mxhx”) ‘ SD{X1}SD{X5.') ' va'q'x‘) ' '
__ (Mk INN“) — WW ﬂ _ 1
— HI/NHUV INN) ‘ N—1 9. Use the results of Example 7 to obtain a} Hr; +1”?  i I ' W
 b) 12 n—l
Note: Brief solution in text is incorrect. ll. IVEHrdition on H100
k E(HroaH:oo} = E[E(HmuHm l Hm“ = 51310" “H3“ l H1001] l
= Etﬂmwm + 200 E1] = Ewing) + 1008mm} = (100 $12“ %+ 50') +100[50) = 7525 _ MW
corr(Hroo.Haou} — sD{Hwo}SD[H30“}
_ 7525 — 50(150) = _1_ “ (sum/i) J3 u. ”’1,” 12. X. Y, and Z are indicator random variables. EiX) = P{select. a Democrat} = or. 50’} = P{select a Republican} = 3 i h} Var(X) = ail  a), Vnr(Y) = 30—13)
 c] Cau(X.Y) = E(XY}—E(X]E{Y}=ﬂ—aﬂ
d) E{Da — Rn) = nE(X] — nE{Y] __— 11(1): — B)
e) D" — R... = ELIJX.  K}.
Var(Dn — Rn} = nVartX — Y}
= naﬂ —a} + 11.5
event that result A occurs on trial 3‘ . similarl NA =i1aﬁ I31 N9 = $13,:
3:1
NANB=ZZ ":1 J31 I1 '3
14.155 : 221mm I
a=l 1:]. == nVarlX] +nVartY] — '2nCautX.Y) [1 .3} + 2na5 = "(a + 9 — (a ‘43)?)
y for 3‘. The“ 13. For each i=1,....nlet A. denote the 174 II‘IIIIIIIIIIIIIIIII Section 6.5 Section 6.5 1. Let X and Y denote the PSAT and SAT scores in standard units. ,/
a) P(Y > o  X = —2} = P [$51331 > ° 1:01: X = —2] = P(Z > is) = 0.0663. b} PW >0X 4 a}: W =2P(X<0,Y >0).
Transform to independent random variables. as in Example 2. arctan 4 ,r' 3 P{X<0,Y>D):P{X<U.0.6X+O.BZ>0)= 2x = 0.1476. {since the corresponding region in the (X. Z) plane is a sector centered at (0,0) subtending an
angle of arctan 4H) . So the desired probability is 0.2952. c) We need HEAT score > PSAT score + 50} = P(90Y + [300 > IDDX + 1200 + 50]
= P[90{o.s.¥ + 0.32)  100x :> 5o]
= P(—46X +722 > 5t}]. Now —46X+7‘ZZ has normal distribution with mean zero and SD 85.“ {it‘s a linear combination
of independent normals); hence the desired probability is INSDIBSAQ} = 0.72. 2. Let X and Y be the heights in standard units of a randomly chosen daughtermother pair from this
/ population. Then Y = %X + 392, Iwhere X and Z are independent standard normals. Pu<X,X<Y =P{o<X.X<%X+§Z}/P(X>oi
L #3 By using the rotational symmetry of the bivariate normal distribution. the probability in the nu merator is {525}! = 5“ Hence, % of the daughters with above average height are shorter than their mothers. ll P{D<X. X<Z}/P(X>6} 3 Let X and Y denote the weight and height in standard units of a person chosen at random. The
event [person is above 90th percentile in height} is the some as (Y > 1.232]. and the event {person
is in 90th percentile in weight) is the same as (X = 1.282). Hence the desired probability is PIY >
1.282IX = 1.282). Now given X = r, Y has normal (pr. 1 — p3} distribution. so \ P(Y>zX=z]=P( Y‘“ “t” X=:) w’lmn3 > t/l—p2 :P(Z) l—pz)
1+? [l a: = 1.232 and p = 0.75 then the desired probability is P{Z } 0.484) = .314 . 4/ Var(X + 23’} = Vnr(X} + 4Var{Y} + 'ZCou{X, Y]. If X and Y are independent, then the variance
' of this sum is 5. If the correlation of X and Y is %, then the variance of the sum is 6. Therefore.
P(X +21” 5 3} is either «55) or «3(7’5. 179 ...
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 Fall '03
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