Assignment 2 - Solution.pdf

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

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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 ...
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