Assignment 2 - Solution.pdf - THE UNIVERSITY OF HONG KONG...

Info icon This preview shows pages 1–3. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STATlSfll Probabili and Statistics: Foundations of Actuarial Science Assignment 2 Solution 2. Denote PS as the event that the shipment is of poor quality, D1 as the event that the first drawn item is defective, and D2 as the event that the second drawn item is defective. 300 3 a PI PS =—=— () ( ) 300+1000 13 Pr(D1 |P3)Pr[Ps) (b) ”(PS “31): PrLID1 |PS)Pr[PS)+ PriDl |Elprifil = o.4><(3f13) 23205455 0.4x(3/13}+o.1x(1o/13) 11 Pr(D1, D2 |Ps)Pr(PS} (c) PI{PS ID1=D2)_ PI[D1=D2 |p3)P1-(PS)+ Pr[D1,D2 lfiipdjT} zwzfl_o_3276 [0.4)2 x(3f13)+(0.1)2 x(10/13) 29 (Note that the above calculation is based on the assumption that given a particular shipment, the events 131,132 are independent, i.e. Pr[D1,D2 |PS) = Pr[D1 |Ps)Pr[D2 |PS), and Pr[D1,D2 |E)= who, |fi)Pr(D2 | fi]. This assumption is reasonable if the number of items in each shipment is large.) 0.5-: 8. Pr(X + Y < 0.5): ESL 24xydydx = [:5 [121992 ]:'j_xdx = [:5 12x(0.5 — x)2 dx = [:5 (3x — 12::2 +12%3 )dx 3 2 '5 1 = i—4x3+3x4 =—=0.0625 2 a 16 14. (a) ZZf(x,y]=l:>ZZAx2y=l 32X1+2+3)=1 x=2 y=1 1:2 J'=1 :> A22 + 3 A = i 78 3 12 (b) g(x) = Zf(t,y)— £0 + =1 (e) Pr(X > Y): Pr(3Y 2 2):) = Pr[X g 3? 2 1 55 =—(22 x2+22x3+32><2+32x3)=—=0.8333 78 (d) The are independent as 2 I f(291)+ f(311] + f(3=2) = y 1 78 _] = f(2,2)+ f(2,3)+ f[3,2)+ f(3=3) "1'8 — — — f 11 =2,3; g(x)(y)=13x6=78x2y ora x 3 3 (e) p1=E[X)= ng(x]=Zx—=2313+3 =£=2 6923 1:2 x=213 13 3 3 4 4 4 EX2 2 2 . = x__2 +3 :2 [ ) g: gm E13 13 13 2 97 35 36 —1’3X2 — — —0.2130 01 [ ) ””1 13 [13] 169 3 3 v2 12+22+32 14 =EY= h = —= =—_2.3333 #2 ( > E” (y) z 6 6 6 3 3 3 13+23+33 E1” = 5:21:02): J"—= = ( ) g 666 6 7 2 5 or; =E(Y2)—#§ = 6—[fl =3 _ 0.5556 (1) E(2X+3Y)=2E(X)+3E(Y]=2x§+3xg=%=12.3846 i(22><1+32><1+32><2]= fl = 0.397'4 78 y 21,2,3 2 2 1?. (a) fx(x)= I:1(2x+y)dy= :[2xy+y?] =x+%, 0<x<1 _f(x,y)_1 / 1 _2x——y (b) fMLylx)— f1(x) —4[2x+y) [x+2]—4x__2, 0<y<2 (c) Pr[Y<l|X=i]=I;fm-[y|%]dy =E1+62ydy=éb1+yzh :— 20.(a) fx(x]—_[x 1 [go—1%: 1[20_x][x 11:20—17: 10<x(2() x3225 x 25 x 2 50 f[x y]_ 1 20—x / 20—x 2 x b — _ ()mele) fx(x) 25 x 50 x“ 2<y<x Hence frw(y|12)=%= 6<y<12. 121 12—8 (0) Pr(Y28|X=12)= —:=jfm[y|12)ajz I —dy=T= Lulu ...
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