{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw14 - Section 6.4 Section 6.4 1 a 0.5 b positively...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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{A|B‘) then PM) = P{A|B)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(A|B‘)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(A|H)P(B) + P(A|B°}P{B‘) 4:9 P{A§B){l — 19(3)] = P(A|B‘}P(B‘} =9 HAIR) = P(A|B‘) so long as HS) 4. 1. e} Positive dependence implies PUIIB) ‘2 P{A} = P{A|B}P(B}+ P(A|B‘)P(B‘} «=0 P{AiB}[l — P(B}] > P(AIB‘}P(B‘) 4:1- P(A|B) ) P(A|B‘) 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(W2|F1) = 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 III III III III III II n II n I: n n n n n n n n Section 6.4 8. Begin by noting that X; = E{E(X|XN1X:)]= E[XIE[XN I X1“ E{X1XN] __ k—Xl __k3[‘Yli uu-t .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. IVEH-rdition 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 >0|X 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 daughter-mother 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>z|X=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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern