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Midterm 1 Solution

Midterm 1 Solution - ACTSC 432/832 Midterm#1 Suggested...

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Unformatted text preview: ACTSC 432/832 Midterm #1 Suggested Answer Spring 2011 1(a) 7r(1) = 7d?) = 71(3) Conditional distributions of X2 for each urn: 0.652 0.4225, 9: = 2 fxflemi) = 2(0.65)(0.35) : 0.455, 1- = 3 0.352 01225, a.- = 4 0.52 0.25. :5- : 2 fx,le(x|2) = 2(0.5)(0.5) = 0.5, :1: = 3 0.52 0.25. 11' = 4 0.3‘2 0.09, 1‘ = 2 fx2,e(.1-|3) = 2(0.3)(0.7) = 0.42, 1' = 3 0.72 0.49. w = 4 Conditional distributions of X 1for each urn: 0.653 0.274625, 95' = 3 3(0.652)(0.35) 0.443625, :5 = 4 le'eml) Z 3(0.65)(0.352) = 0.238875, 11' = 5 0.353 0.042875, :1: = 6 0.53 0.125, 2: 2 3 , 3(0.52)(0.5) 0.375, :1: = 4 fx,le(rv|2) : 3(0.5)(0.52) : {0.375. .7: = 5 0.53 0.125, :1: = 6 0.33 0.027, .5- : 3 . 3 0.32) 0.7) 0.189, 2: z 4 11.195113): 3i0.3)((() 7‘2) : {0.441, :r. = 5 0.73 0.343. 1’ = 6 0.65 + 2(0. 5) 1.35. 0 :1 (b) 11(6) 2 0.5 —— 2(0.5) = {1.5, (2‘ = 2 0.3 —— 2(0.7) 1.7, 6 = 3 0.65 —— 22(035) ~— 1.35‘2 0.2275, 6 = 1 5(0) = 0.5 + 22(0.5)—1.52 = 0.25, 0 = 2 0.3 + 22(07) —1.72 0.21, 0 = 3 (c) welxl(1|5) : (WWW???) : 0.226449 7191X1(2|5) = mm = 0.355492 _ 0.441(1/3) _ nelxl(3|5) _ m _ 0.418059 ((0 fx,.x,(2}5) : 0.4225(0226449)+0.25(0.355492)+0.09(0.418059) = 0.222173 fX2|X1(3|5) : 0.455(0226449) + 0.5(0.355492) + 0.42(0.418059) = 0.456365 fX2|X1(4|5) = 0.1225(0.226449)+0.25(O.355492)+0.49(0.418059) : 0.321402 (e) E(X2|X1 = 5) = 2(0222173) + 3(0456365) + 4(0.321462) = 3.099289 (f) 11: E[11(@)] = (1.35 +1.5 +1.7)(1/3) = 4.55/3 2 1.516667 0 = 1510(9)] = (0.2275 + 0.25 + 0.211(1/3) = 0.6875/3 = 0.229167 a : Va7-[/1(@)] = (1.352 +1.52 + 1.72)(1/3) « 112 2 0185/32 = 0.020556 A _ ~ __ 0.6875/3 _ k _ g _ 0185/3 _ 11.148649 Z = 3—43} = 0.212034 Credibility premium = 2[Z.'"E + (1 — Zm] : 2[0.212034(5/3) + O.787966(4.55/3)} = 3.096944 J 2\/— 2 . (1 J Let t 2 1/117, then 1514 2 Hand (11') = 2tdt, thus E([email protected]— k 0)_ 0 210°" [-2m_1€_9t2l1dl : 6 [:0 [line—913(1) _ 1(2m+1) fox awozme 9' dt : [‘(2m+1) I‘ 2m+1 9-” M(9)—P(3)/192 = 2/“92 4(0) = E<X2|e = 0) — {140112 = Pom/04 — 4/04 = 20/04 _ fixen—ra 49/11 ‘ (b) E((—) 1): 5° LW‘L—dfi “’3 rm_ 7H)foo/3—o+10; 1 1 8—9/0 @- H“"l‘(a—1‘) [‘(Cx)2 () Na— r) _ Fm) #:2E(9 2): [35(a-——)12)(a z _ —4 20 ’U— 20E(® ): 83((1— l)( (x—2)(a-3)((1—4) (c) a = 130149112} — 142 = 13149-41— 112 fi 4 4 _ 134((1—1)(r11—-2)(0—3)((1—4) __ 133(u—1)7(a—2)2 _ 4 1 . _ “ 53(01— 1)7(a— 211m —'3“)(a— 4)[(“ l)(“ _ 2) " (9' “ .3)((1 _ 4)} —/j (a 1) (a 41277711413 0_ “We? 3a+2)—(a2—7a+12)} _ 40— 10 2(1— 5 * (a— 1101— 2;“: 5(0— 1)(a—2)U k~E—55(Q— l)(a— 2) '_ u _ 2(2u—5) _ -('2)('2 —5) M 2 (2 -5) ((1) Z- _ n+k n(2)(2Qn—5)+5CZu—1)(r1—2) — 211(20— 531+ Sia— 1)(a— '2) _ 271(20— 5) 5(n—1)(a 2) PC :Zi+(1 Z)/.L 211(2a— 1)+.’1Zr2— liin— 2 2:1 2(1-1)+5i(1— 1 0—2 {7((1— 12 (1—2) _ 211(20— ,1):E+10fi“2 _ ’2IL(2(1—5)+5((!—1)((!—2) (I (e) f(x1)— °° 0_C #49 _ 0 2m ' Na) 2 412° '00 . —(l+ )6 — warm) 10 00" H m d6 2 (a) E(X.,;”|(-‘) : 0) = j;;’°.1'3"f(17j|0)da;j= “60.,”‘9‘ ———\f—F7(laj : 9 0° arm—$4701”? »’ ([éL'J' _ 13'” F(a+1) 30 (73+\/1'_J')o 01+ 008 —(zr+m) ”do zmfla) (%+\/av—j)“+1 0 PM“) I = 0/3“" 2m(j§+m)a+l (f) As a function of 0, Wins) oc {H f(1'j|9)}7r(6) 71 cc {H 08—9m}6“_le_% i=1 : 60+n—le-(71;+Z;1=1 mm and so 7r(6|a:) is again a gamma pdf, but with a reaplaced by (1* = (J: + 77, andflby621/(%+Zy_1\/E—T). (g) It follows from (Eb) and (f) that E(X71+1(IX):fQOCE(X7L+1lO—_ 0) )7T(9IX)d9 __ 2 _ “7+2: 2 —f0°°.( )H9le9=m—mhfifii7) (h) It follows from ((-3) and (f) that f( $7L+1|f13): lief (-Tn+1I0)’/T (M10 Ofi: 0,. :ZJ—ccnué—I +m)n.+1 _ (-L“+")(B+Z}”=1 fl?)u+u Zmofizggl «37+ WWW 3. Let total claim experience: X17 . . . , X"; number of claims: N1, . . . , N”; and size of a claim: Y American credibility factors Based on number of claims: Z: 777777(,/— '—’\ 1), therefore 0.752 = M /\0 Au '4 ‘ - . u _ 'IL\ 2_ u\ Based on total claims. Z _ mum IW’I ), therefore () 5 _m -n-'2 'lheref‘me, +2 “'7 —,“L— 1— — 1.25 Buhlmunn c7edib7lity factors Based on number of claims: k = g = V%%, _ ._ I: A 4_ _ 2 Z— n+k— 0.6 therefOIe ——L(-/l\—= 6’1"“ 577, Based on total claims: k: : 5 = W21 __ :72 +1L2E[/\] E__[____A] __ My \ (17(A)_ _(_§Z + 1)Von'.( \) = (1.25+1)§n—%—ny and therefore 2the Buhlmann crediblity factor based on total claims is _ 7L Z _ W _ g 41.4 ...
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