Let X represent the number of customers arriving during the morning hoursand let Y represent the number of customers arriving during the afternoon hoursat a diner.This is given:

a) X and Y are Poisson distributed.b)The first moment of X is less than the first moment of Y by 8.c) The second moment of X is 60% of the second moment of Y.

Calculate the variance of Y.

i) For a Poisson random variable N, E(N) = Var(N) = λ and E(N2) = λ+λ2.ii) We are given λx = λy -8 and λx+λ2x = 0.6(λx+λ2y).iii) Combining equations, we get a quadratic in λy:λy - 8 +(λy - 8)2 = 0.6×(λy+λ2y).iv) symplifying, we get: 0.4λ2y- 15.6λy + 56 = 0.v) Two roots: λy = 35 and 4. The second root is rejected since this would makeλx negative.vi) .: the variance = 35.

Automobiles policies are separated into two groups: low-risk and high-risk. Ac-tuary Ismael examines low-risk policies, continuing until a policy with a claimis found and then stopping. Actuary Ruben follows the same procedure withhigh-risk policies. Each low-risk policy has a 10% probability of having a claim.Each high-risk policy has a 20% probability of having a claim. The claim sta-tuses of polices are mutually independent.

Calculate the probability that Actuary Ismael examines fewer policies than Ac-tuary Ruben.

The probability that Ismael examines exactly n policies is 0.1×0.9n−1. Theprobability that Ruben examines more than n policies is 0.8n. The requiredprobability is thus

∑∞n=1 0.1×0.9n−1×0.8n = 1

9

∑∞n=10.72n = 0.72

9×1−0.72 = 0.2857.Another solution:Ismael and Ruben examine policies simultaneously until at least one of themfinds a claim. At each examination there are four possible outcomes:a) Both find a claim. The probability is 0.02b) Ismael finds a claim and Ruben does not. the pro is 0.08.c) Ruben finds a claim and Ismael does not. The prob is 0.18.d) Neither finds a claim. The prob. is 0.72.Conditioning on the examination at which the process ends, the probability thatit ends with Ismael being the first to find a claim (and thus needing to examinefewer policies) is 0.08

0.02+0.08+0.18 = 828 = 0.2857.

From 27 pieces of luggage, an airline luggage handler damages a random sampleof four.The probability that exactly one of the damage pieces of luggage is insured is

1

twice the probability that none of the damage pieces are insured.

Calculate the prob that exactly two of the four damaged pieces are insured.The ratio of the probability that one of the damaged pieces is insured to the

probability ((

(r

1

)(27− r

3

))/

(27

4

))÷((

(r

0

)(27− r

4

))/

(27

4

)) = 4r

27−r , r is the

total number of pieces insured. Setting this ratio 2= 4r/27-r ⇒ r= 8.

Therefore, ((

(r

2

)(27− r

2

))/

(27

4

)) = ((

(8

2

)(19

2

))/

(27

4

)) = ( 8!

2!6!×19!2!17! )÷

27!4!23!

=0.27.

An automobile insurance company issues a one-year policy with a deductibleof 500. The prob is 0.8 that the insured automobile has no accident, and 0.0that the automobile has more than one accident. If there is an accident, theloss before application of the deductible is exponentially distributed with mean3000. Calculate the 95th percentile of the insurance company payout on thispolicy.The 95th percentile is in the range when an accident occurs. It is the 75th per-centile of the payout, given that an accident occurs, because ( 0.95-0.80)÷(1-0.80) = 0.75. Letting x be the 75th percentile of the given exponential distri-

bution, F(x) = 1 - e−x3000 = 0.75, so x = 4159. Subtracting the deductible of

500 gives 3659 as the (unconditional) 95th percentile of the insurance companypayout

A motorist makes three driving errors, each independently resulting in an ac-cident with prob 0.25. Each accident results in a loss that is exponentiallydistributed with mean 0.80. Losses are mutually independent and independentof the number of accidents.The motorist’s insurer reimburses 70% of each loss due to an accident.Calculate the variance of the total unreimburesed loss the motorist experiencesdue to accidents resulting from these driving errors.

Let N denote the number of accidents, which is binomial with parameters 14

and 3 and thus has mean 3( 14 ) = 3

4 and variance 3( 14 )( 3

4 ) = 916 . Let Xi denote

the unreimbursed loss due to the ith accident, which is 0.3 times an exponen-tially distributed random variable with mean 0.8 and therefore variance (0.8)2=0.64. i.e; Xi has mean 0.8(0.3) = 0.24 and variance 0.64(0.3)2 = 0.0576. Let Xdenote the total unreimbursed loss due to the N accidents.Using conditional variance formula. Var(X) = Var bE(X|N)c +Eb Var (X|N)c=Var[E(X1+. . .+XN )] +E[Var(X1+. . .+XN )]= Var[NE(X1)]+E[N Var(X1

= Var(0.24N) + E(0.0576N)= (0.24)2Var(N) + 0.0576E(N)=0.0576( 9

16 ) + 0.0576( 34 ) = 0.0756.

2