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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: Nonprogramable 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 thestop—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 (Puetc Type II. Leland«2,9 gIl' 1(a) = T F(=) = 1 4w" can 9
Mo) = (19: 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 41/0) 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), 01
m) = HIPHm”) 0‘r(k+1r(aI:)
cm) I EKXMN = #31“ 1.a Ir.=/(=+
MO) I: (101", 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+k1)0‘P(a+k===/9)+z‘llI‘Gmlﬁﬂ. humusr 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  etslfr 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‘Il 17(1) _ 9“) mod. I 0
ﬂ”) " mg“ 53/2) “ﬂag” 3 ‘7 1 Note. “though than Ippun u: be two parmm, 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 botea,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+k1)
E (X )= f4 . Var ‘ X) : 0 EV“ " —"‘(a+b)(a.+b+—‘1).(a'+‘b+' ‘F—l)‘
2 1 . 9"a(a+1)m(c'+k1)
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 can1+5, p.:.———(1+p)»‘ Em] sp. wmgpuw)‘ P(x)=[1§ﬂ(z1)".
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. Vulmrn<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‘15(z~1)] Pm . 1__Jm__ This is a limiting case of the zuotxnncated negative binomial as r —o O. ...
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