ACTSC-432-1049-Midterm_exam

ACTSC-432-1049-Midterm_exam - ACTSCI 432/832 MIDTERM...

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Unformatted text preview: ACTSCI 432/832 MIDTERM EXAMINATION FALL 2004 Name: *3— I.D.#= A Aids: ' Calculator Time: 1 hour Examinar: Gordon Willmot [18] 1. For a particular line of insurance business, the insured population is composed of 25% good risks (with risk parameter 9 = 61), 50% standard risks (with risk parameter 9 = 62), and 25% poor risks (with risk parameter 6 = (93). The conditional claim distributions for these classes are given for j = 1, 2,... and 71 = 1, 2,3 by _‘ = 1—— H i As usual, the random variables {X1, X2, are independent given the risk parameter 9. (a) Give the structural probability distribution of the risk parameter 9. F(‘gn):~ Z‘rfl/Ivg; {9(62):f"z ;V‘3— 2) (fig); v.95 (b) Compute the hypothetical means for the three risk types. P331593 thslwy) PU:QWL) “((9) 1 OJ (9, my (3,) 7g (—3 94:47 (7;, 2 4:6 (9, a; Li) {- f; {,2ng (/73) (9} m“ (w (7 t2 w: (~32; (c) What net premium should be charged an insured with no past claims data avail— able? olf-F“? 03—” 7/ I (up I; E? f [.2 (d) Compute the marginal probability Pr(X1 = 6). 3 [>P(Y;:§) :: P(x,:gl(7‘_)fléé¢t‘) ('«l i Z 5 —. Z: -—— ' I -- J’ ——- g o v VS )2 x «19‘ + h? t c) 'l i Z _ .2 § -’ ,/’ /-2 X/ \ ’ fl/X/g ~ 7 i ,k/ (e) Compute the posterior distribution of (9le = 6. (DLXFéICQ. ) Wm) .— I f) L (9' I Y, =5 ) :. W~- w 2-w«~~-“""’“"‘“'W=Mwwwmw ; WWII 2:- ‘fr( y/Zé) ’73: J 2' (9 M 2101'" Wt‘:§h:é): WM“ 56”“ ((0‘) «3' M: 1 EL ‘ {W 3': Mi ) "7:3" 2, )(x :45?" “‘2?” 5' ,2 «4i.,és»2g [9( (93; 2mg» / - 3 effigntfigim m Wifjww (f) Determine the Bayesian premium E (X2|X1 = 6). .- .r 5 E€XLEP¢i=éjrn P‘fé‘gl-{yr 5 f (L, \Lug we 5'! ‘(uz 3+TQ‘5‘LTZ 3 _ _—-—- [18] 2. Consider a random sample X = (X1,X2, ...,Xn) from the (conditional on 9) distribu- tion With cumulative distribution function ’ F(:1:|6) = Pr (Xj S 1:19 =6) =1— awn: Z 0, and 6-) has probability density function (pdf) = amen-1e49, 9 > 07 With,8>0anda>2. (a) Demonstrate that E (le9) = 2/ (92. (b) Determine E (X j). %%g ELX):‘IWE(XI§)EL5}55@ I “Alf - ¢°2 fl(nafl e I , — ’5: Wm mm m a f {egg} 6’3 ‘31,“; Cay—2%! «fit/6 :33... WM} 6 p(“) D 3 a? 1/” 53/5 j : r p i v- .2 ) : WWW- We?) QyJMwm WWW-mf“ x } a: (fi-aML/‘i‘vvfiféf (0) Determine the marginal pdf of X j. ~l / “$79; {I} -520?" a )‘(*W)=mm =(H?}e ~ : «page » My My ‘ w 5727 w w a)”; V”? «6% ( ,/ g jl m : {m aw M a; w r (3 ME 9 m J Wm J9 W? any???” Wwwwwifiuamh‘m«uxiw 4” f2“ 95”} J»; 9L V ,i‘ > (d) For the random sample X = (X 1, X2, ..., Xn), show that b(6) is a conjugate prior, and identify all parameters. [‘(aéfl) .. F b“: vLOIJIf; @¢“£€“/fi£x (,9 e at Mow _ amemfi ) .3 G 6 M15 am; an fiAM dl‘5%"wéwfwm »—~'lcca)¢m;u§idfigga / M Iifi‘ if (e) Determine the Bayesian premium E (Xn+1§X ; m “ 9! ar 33*“ “f? be .. we (’2’ (if; 60-] 5 — j 2 (‘9 WWW ......................... "Wm " fipz'” v) V t 4H3 er fl 4 [’fl’%: ‘3‘ “M M”? g: a? (i p (a; ~ I Lmimimiwwm—n a fig haw") ,M o fi5”*”23 =2 1%) 3 [9] 3. Past data on a particular policyholder is given by X = (X 1,X2, ..., Xn) where the X]- are independent and identically distributed compound Poisson random variables. You are given the following information: (1) Credibility is based on the limited fiuctation approach with r = .10 and p = .95. (2) The normal approximation to the compound Poisson is assumed,*&>(1.64'5) = .95, @(196) = .975, and <I>(a:) is the cumulative distribution function of the stan-' dard normal distribution. (3) The observed past total number of claims is 300. (4) The claim size cumulative distribution function is F(y)=1-(g),y>0, where 0 > 0. Estimate the credibility factor Z. xii/\T‘: (7:394) (W at; l a - , I" I, fi if; b“ 3 J a, a-j§(~pé€g}ic!g«fw 67(3) 552-55.?(9 a? W v i - 3 —2 w 3 _r_ "1' w “2* rewvffiaag ag~5wg fig;ng a .2 , l 2:. (9 L H t (fl? 4?:Etyi}'u: EWW-‘Zfiwffg fr? WA :3? m e :11 g; gfi £5, I; f: a h L "’ i" *' 2‘ a! a x 38%: M if g , (,8 f N? 2: [f 2, ...
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ACTSC-432-1049-Midterm_exam - ACTSCI 432/832 MIDTERM...

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