act451h-d08final exam

act451h-d08final exam - University of Toronto Faculty of...

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Unformatted text preview: University of Toronto Faculty of Arts and Science '2 December Examination, 2008 . ACT 451H1F Duration - 2 hours W Aids Allowed: Non-programable calculators only. STUDENT NAME: STUDENT NUMBER: 1 . 2 1. A portfolio of insurance policies consists of two types of policies. Losses on Type 1 policies have an exponential distribution with 0 = 100. Losses on Type 2 policies have a Gamma distribution with a = 2 and 0 = 100. The policies are evenly divided between two types. A single loss arises from the portfolio. a) 5 pts Identify the hazard rate function of the loss. GONT‘D: b) 5 pts Which interval does the value at risk (VaR) of the loss at security level 90% lie in: [150, 250), [250, 350), [350, 00)? Justify your answer! CONTDQ 4 2. 5 pts The conditional random variable N [A = /\ has a Poisson distribution with mean A. The random variable A has a Gamma distribution with mean ,u, > 0 and standard deviation 0'. Identify the unconditional distribution of N and its parameters (in terms of p and a). 5 3. The distribution of losses in 1994 is Pareto with a = 3 and 0 = 1,000,000. An insurance policy calls for a deductible of 100,000 on each loss. The insurer has purchased reinsurance that pays the excess over 1,000,000 on any claim. a) 5 pts Determine the insurer’s expected cost per loss for 1994.. 6 b) 5 pts Assuming an annual inflation of 10%, determine the REINSURER’S expected cost per loss for 1995. The deductible level and the retention level remain the same. 7 4. The aggregagte claims amount 8’ has a compound negative binomial distribution with negative binomial parameters ,3 = 0,5 and r = 3. The individual claim amount distribution is discrete and has the probability function p(a:) = 0.0516,:3 = 1, 2,4, 5,8. a) 5 pts Determine the expected aggregate claims amount. b) 5 pts A gross premium of P is charged so that the probability that the aggregate claims amount would be less than the premium is 0.95. Use the normal approximation to determine P. Willi. c) 5 pts Determine the probability Pr{S S 5} using the recursive formula for the (a,b,0) class. CONTDL d) 5 pts Determine the-stop—loss premium E(S — 5).... e) 5 pts An ordinary deductible of 4 is applied to each of the individual claims. Let S L be the corresponding cost per loss random variable to S. Determine the expected aggregate cost E(SL). CGNW 10 5. An insurer issues a portfolio of 100 independent auto insiurance policies. One half of these policies have an ordinary deductible of 100 and the other half have no deductible. Assume that (i) the number of losses per year per policy has a Poisson With mean 0.02; and (ii) the loss amount X, given that it occurs, has the probability function p(50) = p(150) = p(200) = 1/3. a) 5 pts Determine the expected aggregate payments. 11 b) 5 pts Determine the probability that the aggregate payments are less than 200. ‘mm‘m; 12 6. A portfolio consists of 3 independent insurance policies with probabilities of a claim and claim amount distributions (given there is a claim) given in the following table. (B =1) .- “m- . n“ a) 5 pts Determine the mean of the aggregate claims from the portfolio. I b) 5 pts Determine the probability that the aggregate claims amount from the portfolio is equal to 3. 13 14 c) 5 pts Assuming the compound Poisson approximation with martching means, use the recursive formula for the (a,b,0) class to determine the probability that the aggregate claims amount is greater than 2. 15 d) 5 pts Assuming a normal approximation for the claim amount for each policy, determine the probability that the aggregate claims amount is greater than 2. A.8.3.1 momma—e ... I A.2.8.1 Pareto (Pu-etc Type II. Leland-«2,9 g-Il' 1(a) = T F(=) -= 1 4w" can 9 Mo) = (1-9: -1 EIX‘] a rm», 1). I: > -1 ! m ' (a? a“)°+'* F“) a 1 ' 31x1 = ml, m: is an integer u u 9"?“ +1)I'(a - k) Ems] , “1 4-1/0) I ! EIX ] a: PM . -1< k<a XA ) = a‘r k+1)r(k+1: l0)+ " "l'. k>—1 ? .. 9‘” B“ 3k] - fklg‘k+1I=/9)+:.¢_=z: I‘an . . EIX‘] I za—l)-.-(a—I; ' “khmhu‘qI. mod. -: o ‘ ElX/m] - J—[I— 9 9—1]. a A.3.2.1 GlnIunHJ } _ a — l (a + a) # 1 I 'E[XA:] II- -91n(;{¥a), 0-1 m) = HIP-Hm”) 0‘r(k+1r(a-I:) cm) I EKXMN = #31“ 1.a -Ir.=/(=+ MO) I: (1-01", Nil/0 Elxklsahgzsk). k>-a I 1110‘“ 3 0 E[X‘] = B“(a+k—1)---é, ifkhlnhtagu . ‘ 11.2.3.2 We Pasta—7,9- , - 7931-! a 1' EI(XA2)"] = Wmmdmz‘u - muss/9)]. k > _a “3) (’ +9)“ F") ' a‘r +kr1—I: - a(a+1)---(a+k-1)0‘P(a+k===/9)+z‘ll-I‘Gmlfifl. humus-r Em " “‘(T“‘—‘P(l)( )‘ "'<"<‘ mod: n 002—1), a>1.duo .. ' ElX‘] - r_:;IIIlzf+k. likinnagntivaim $33.3 Wanna—'9, 1' Kl") 1"“(1— yr‘m sl- [1 mar/sari = W]; mada :3 'OL;-1, 1>1. glue!) “5) 1’ =19)7:“"”.. '. ms) .- 1 - e-tslfr gym a 9"]‘(1+k/r), k>-7 BK}: A :9") a mu + #3111 + h/‘r: (=19)? “'24-’02 h > -r may. smuéfifiafiacu éQxJStH" _ "I'm" " ‘ r -1 7 . = .._. , o . . '0( 1. ) . 7>1 I _I_ l - g5. :>a I F(¢)_1_(o/z)a' AAJJ magma—ma (:4 can b- nus-“"1 I i I E{x~1I .3 39%. R a Em: A:)"] - silk 1 ~ I g ’ - hE‘I-l 17(1) _ 9“) mod. I- 0 fl”) " mg“ 53/2) “flag” 3 ‘7 1 Note. “though than Ippun u: be two par-mm, only a h I an EIx‘] - aq:(l:p+k_’a’/2) I“! ‘ not in Idnnu. mp: A m a “pm-:- Pea/2)! + 2"[1 — PM] I I ma. _ “Po, _ ,2') ‘ V 55.1.2 bote-a,b,9 \ I “11 a . - Ca“ b¢ VtL I I a Wufl—u) E. D<=< A'oLt- ‘co “'0. 5 I p(¢.biu) (’X" ~.ac< 7<<°° O‘I‘(a+621‘(a+k) _ Q“: JIF- C‘YC“ 2, A EV”! “ I‘(a)F(a+b+k)’ "> -“ a. la. . 7- * fl‘a(a+1)uu(c+k-1) E (X )= f4 . Var ‘ X) : 0 EV“ " —"‘(a+b)(a.+b+—‘1).--(a'+‘b+' ‘F—l)‘ 2- 1 . 9"a(a+1)m(c'+k-1) M It) = e x? C H t 4' ‘5.- G t ) EKXA')" " (n+b_)(a+' 5+ 1') »'-'(a+a+h"'—'1')‘ +z"[1 - Ha. ha «)1 ¢ix)=¥"‘) wT+L l‘so, MA 61‘ a. NORMAL DISTRIBUTION TABLE ; Entries rapnsant 1h: am uddemo WW normal Mullen a 19 z. Frtzq) m valua of: b «1. m dadml ls ulvon M1: left column. Th; aacond decimal plans ls given In tha tpp row. 3.2.1.1 Poisson—A p E FAA" k kl P0,) = eMs—I) pa = e". (1:0. bfl Elm - A. VIEW!“A 3.2.1.2 Geometric—fl 1' 5 PD 3 m can-1+5, p.:.———(1+p)»‘ Em] s-p. wm-gpuw)‘ P(x)=[1§-fl(z-1)". Thil is a spacial can ofthe negative blnomhl withr - 1. ‘ bso 3.2.1.8 Binomial—mm, (0 < q < 1. m an {nigger} a (m+ 1):] m - (1—4)". a=--‘-’-' b 1—9 1- a PI: = ('2)9"(1 — q)"”'". I: = 0.1.... .m BIN] " m?» V341“ = mm " ‘1) H3) 5 [1 + ‘I(‘ " 1)!“- B.2.1.4 Nagative binomial—fl; ~1- fl (r"1)fi Po " (1+5) . 4=m..b-=-1-+—5- r(r+1)---(r+l=—1)fl" _' P" ' " 7‘11? +‘—"‘—'g)r+u .2 EIN] - rn. Vulm-rn<1+£3"” P<s)=u—a(=—1)rn 8.8.1.8 Logarith 7" = ifi as—L b: __£_ 1 (1 +fi)1n(1'+—y,3 ' 1 +3: 1 +3. 1‘ a p" p“ k(1 + p)“ 1n(1 +‘Tsy I Em - fl/ 1n(1+fl). Vumsflflfilfiggifl' B I“ %— or __2(p;1_)' . 11‘1-5(z~1)] Pm . 1__Jm__ This is a limiting case of the zuo-txnncated negative binomial as r —o O. ...
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act451h-d08final exam - University of Toronto Faculty of...

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