Download ATW316 Semester Additional Test 3 Memo

ATW316 Semester Test 3 [57 marks] [2 hours]
26 April 2012
Question 1 [16]
Consider aggregate claims over a period of 1 year, 𝑆, on a portfolio of general insurance
policies:
𝑆 + 𝑋1 + 𝑋2 + ⋯ + 𝑋𝑁
The number of claims each year, 𝑁, has a Poisson distribution with mean 12. 𝑋1 , 𝑋2 , … are
assumed to be random variables, independent of each other and independent of 𝑁, with the
following distribution:
𝑓(𝑥) = 0.01𝑒 −0.01𝑥
0 < 𝑥 < 𝑅200
𝑃(𝑋 = 𝑅200) = 𝑒 −2
The insurance company calculates premiums using a premium loading of 40% and is considering
entering into one of the following re-insurance arrangements:
(A)
No reinsurance.
(B)
Individual excess of loss insurance with retention 100 with a reinsurance company that
calculates premiums using a premium loading of 55%.
(C)
Proportional reinsurance with retention 75% with a reinsurance company that
calculates premiums using a premium loading of 45%.
i)
Find the insurance company’s expected profit under (A). [3]
𝐸(𝑃𝑟𝑜𝑓𝑖𝑡) = 𝐸(𝑃𝑟𝑒𝑚𝑖𝑢𝑚 𝑖𝑛𝑐𝑜𝑚𝑒 − 𝐶𝑙𝑎𝑖𝑚𝑠 𝑜𝑢𝑡𝑔𝑜)
= 𝐸[1.4𝐸(𝑆) − 𝐸(𝑆)]
= 1.4𝐸(𝑆) − 𝐸(𝑆)
𝐸(𝑆) = 𝐸(𝑁)𝐸(𝑋)
= 12𝐸(𝑋)
200
= 12 [∫ 𝑥(0.01)𝑒 −0.01𝑥 𝑑𝑥 + 200𝑒 −2 ]
0
200
= 12 [−𝑥𝑒 −0.01𝑥 |200
+ ∫ 𝑒 −0.01𝑥 𝑑𝑥 + 200𝑒 −2 ]
0
0
= 12 [−200𝑒 −0.01(200) −
1 −0.01(200)
1
𝑒
+
+ 200𝑒 −2 ]
0.01
0.01
= 1037.59766
𝐸(𝑃𝑟𝑜𝑓𝑖𝑡) = 0.4(1037.59766) = 415.039064
ii)
Find the premium charged by reinsurance company under (B), assuming that the
reinsurer knows nothing of claims below the retention level. [4]
𝑃𝑟𝑒𝑚𝑖𝑢𝑚 𝑜𝑓 𝑟𝑒𝑖𝑛𝑠𝑢𝑟𝑎𝑛𝑐𝑒 𝑐𝑜𝑚𝑝𝑎𝑛𝑦 = (1 + 𝜁)𝐸(𝑆𝑅 )
1.55𝐸(𝑁)𝐸(𝑍)
=
𝑃(𝑍 > 100)
200
1.55(12)
=
[ ∫ (𝑥 − 100)𝑓(𝑥)𝑑𝑥 + 100𝑃(𝑋 = 200)]
𝑃(𝑍 > 100)
100
200
=
1.55(12)
[ ∫ (𝑥 − 100)0.01𝑒 −0.01𝑥 𝑑𝑥 + 100𝑒 −2 ]
𝑃(𝑍 > 100)
100
200
1.55(12)
200
200
=
[−𝑥𝑒 −0.01𝑥 |100
+ ∫ 𝑒 −0.01𝑥 𝑑𝑥 − 100(−𝑒 −0.01𝑥 )100
+ 100𝑒 −2 ]
𝑃(𝑍 > 100)
100
=
1.55(12)
1 −0.01(200)
1 −0.01(100)
[−200𝑒 −0.01(200) + 100𝑒 −0.01(100) + (−
𝑒
+
𝑒
)
𝑃(𝑍 > 100)
0.01
0.01
− 100(−𝑒 −0.01(200) + 𝑒 −0.01(100) ) + 100𝑒 −2 ]
=
1.55(12)
1 −2
1 −1
[−200𝑒 −2 + 100𝑒 −1 −
𝑒 +
𝑒 + 100(𝑒 −2 ) − 100𝑒 −1
−2
0.01
0.01
+𝑒 ]
200
[(−𝑒 −0.01𝑥 )100
+ 100𝑒 −2 ]
=
1.55(12)
(23.25441579)
[0.367879441]
= 1175.744239
iii)
Find the probability that the insurer makes a profit of less than 500 under (C) using a
Normal approximation. [9]
𝑃[𝑃𝑟𝑜𝑓𝑖𝑡 < 500] = 𝑃[1.4𝐸(𝑆) − 1.45𝐸(𝑆𝑅 ) − 𝑆𝐼 < 500]
= 𝑃[𝑆𝐼 > 1.4𝐸(𝑆) − 1.45(0.25)𝐸(𝑆) − 500]
𝐸(𝑆) = 12𝐸(𝑋) = 1037.59766
𝑃[𝑆𝐼 > 576.5075723] = 𝑃 [𝑍 >
576.5075723 − 𝐸(𝑆𝐼 )
√𝑉𝑎𝑟(𝑆𝐼 )
]
𝐸(𝑆𝐼 ) = 0.75(1037.59766) = 778.198245
𝑉𝑎𝑟(𝑆𝐼 ) = 𝑉𝑎𝑟(0.75𝑆)
= 0.752 𝑉𝑎𝑟(𝑆)
= 0.752 (12)𝐸(𝑋 2 )
200
= 0.752 (12) [∫ 𝑥 2 (0.01)𝑒 −0.01𝑥 𝑑𝑥 + 2002 𝑒 −2 ]
0
200
(0.01)3 3−1 −0.01𝑥
𝛤(3)
= 0.752 (12) [
∫
𝑥 𝑒
𝑑𝑥 + 2002 𝑒 −2 ]
0.012
𝛤(3)
0
𝛤(3)
{𝑃[𝑋 < 200]} + 2002 𝑒 −2 ] 𝑤ℎ𝑒𝑟𝑒 𝑋~𝐺𝑎𝑚𝑚𝑎(3,0.01)
0.012
𝛤(3)
2
= 0.752 (12) [
{𝑃[𝜒2(3)
< 4]} + 2002 𝑒 −2 ]
0.012
𝛤(3)
{0.3233} + 2002 𝑒 −2 ]
= 0.752 (12) [
0.012
= 80,186.02647
= 0.752 (12) [
𝑃 [𝑍 >
576.5075723 − 778.198245
√80,186.02647
] = 𝑃[𝑍 < 0.712256576] = 0.76115
Question 2 [10]
An insurance firm models hailstorm claims according to the following assumptions:
The number of hailstorms in a year follows a Poisson (4) distribution.
The number of claims arising from the 𝑖 𝑡ℎ hailstorm is modelled as a Poisson(𝜃𝑖 ) distribution.
The parameters 𝜃𝑖 are independently and identically distributed random variables with mean 4
and standard deviation 2.
The individual claim amounts from each storm follow a Pareto(5, 𝜆𝑖 ) distribution.
The mean claim amounts 𝜇𝑖 = 0.25𝜆𝑖 are assumed to be independently and identically
distributed with mean 𝑅3000 and standard deviation 𝑅600.
𝜇𝑖 and 𝜃𝑖 are independently distributed.
Use this information to answer the following questions:
i) Calculate the mean and the variance of the annual aggregate claims outgo from all storms.
[8]
𝐸(𝑆) = 𝐸[𝐸(𝑆|𝛩𝑖 , 𝛬𝑖 )]
= 𝐸[𝐸(𝐾)𝐸(𝑆𝑖 |𝛩𝑖 , 𝛬𝑖 )]
= 𝐸[𝐸(𝐾)𝐸(𝑁𝑖 |𝛩𝑖 )𝐸(𝑋𝑖𝑗 |𝛬𝑖 )]
= 𝐸[4𝛩𝑖 𝛬𝑖 ]
= 4(4)(3,000)
= 48,000
𝑉𝑎𝑟(𝑆) = 𝐸(𝑉𝑎𝑟(𝑆|𝛩𝑖 , 𝛬𝑖 )) + 𝑉𝑎𝑟(𝐸(𝑆|𝛩𝑖 , 𝛬𝑖 ))
= 𝐸[𝐸(𝐾)𝐸(𝑆𝑖2 |𝛩𝑖 , 𝛬𝑖 )] + 𝑉𝑎𝑟[𝐸(𝐾)𝐸(𝑆𝑖 |𝛩𝑖 , 𝛬𝑖 )]
= 𝐸[4{𝑉𝑎𝑟(𝑆𝑖 |𝛩𝑖 , 𝛬𝑖 ) + 𝐸(𝑆𝑖 |𝛩𝑖 , 𝛬𝑖 )2 }] + 𝑉𝑎𝑟[4𝛩𝑖 𝛬𝑖 ]
2
= 𝐸[4{𝐸(𝑁𝑖 |𝛩𝑖 )𝐸(𝑋𝑖𝑗
|𝛬𝑖 ) + (𝛩𝑖 𝛬𝑖 )2 }] + 42 𝑉𝑎𝑟[𝛩𝑖 𝛬𝑖 ]
2
= 4𝐸 [{𝛩𝑖 [𝑉𝑎𝑟(𝑋𝑖𝑗 |𝛬𝑖 ) + 𝐸(𝑋𝑖𝑗 |𝛬𝑖 ) ] + 𝛩𝑖2 𝛬2𝑖 }] + 42 {𝐸(𝛩𝑖2 𝛬2𝑖 ) − 𝐸(𝛩𝑖 𝛬𝑖 )2 }
5
2
= 4𝐸 [{𝛩𝑖 [ 𝛬2𝑖 + 𝛬2𝑖 ] + 𝛩𝑖2 𝛬2𝑖 }] + 42 {𝐸(𝛩𝑖2 )𝐸(𝛬2𝑖 ) − (4(3,000)) }
3
8
2
= 4𝐸 [{𝛩𝑖 [ 𝛬2𝑖 ] + 𝛩𝑖2 𝛬2𝑖 }] + 42 {{22 + 42 }{6002 + 30002 } − (4(3,000)) }
3
4(8)
=
[𝐸(𝛩𝑖 )𝐸[𝛬2𝑖 ]] + 4[𝐸(𝛩𝑖2 )𝐸(𝛬2𝑖 )] + 691,200,000
3
4(8)
[(4){6002 + 30002 }] + 4[{22 + 42 }{6002 + 30002 }] + 691,200,000
3
= 399,360,000 + 748,800,000 + 691,200,000
= 1,839,360,000
=
ii) By using the normal approximation, approximate the probability that the annual aggregate
claims outgo from all storms will exceed 𝑅120 000. [2]
𝑃[𝑆 > 120,000] = 𝑃 [𝑍 >
120,000 − 48,000
√1,839,360,000
= 1 − 0.95352
= 0.04648
]
Question 3 [5]
An insurance company has insured a fleet of cars for the last four years. For year 𝑗 (𝑗 = 1, … ,4),
let 𝑌𝑗 and 𝑃𝑗 be the total amount claimed and the number of cars in the fleet, respectively. Let
𝑋𝑗 = 𝑌𝑗 /𝑃𝑗 be the average amount claimed per car in year 𝑗. Assume that the distribution of 𝑋𝑗
depends on a risk parameter 𝜃 and that the conditions of Empirical Bayes Credibility Theory
Model 2 are satisfied. The company has insured 10 similar fleets over the last four years. Using
the data from these years, 𝑚, 𝐸(𝑠 2 (𝜃)) and 𝑉(𝑚(𝜃)) are estimated to be 62.8, 106.32 and 5.8
respectively. Calculate next year’s credibility premium for a fleet of cars with claims over the
last four years given below, if the fleet will have 16 cars next year.
Year
1
2
3
4
Total amount claimed
1,000
1,200
1,500
1,400
Number of cars
15
16
18
15
Also explain how and why the credibility factor would be affected if the estimate of 𝑉(𝑚(𝜃))
increases and comment on the effect on the credibility premium.
𝑚 = 62.8
𝐸(𝑠 2 (𝜃)) = 106.32
𝑉(𝑚(𝜃)) = 5.8
𝑃̅ = 64
𝑋̅ = 65.625
64
= 0.777349639
106.32
64 +
5.8
𝑍(65.625) + (1 − 𝑍)(62.8) = 64.99601273
𝑍=
If 𝑉(𝑚(𝜃)) increases then 𝑍 will increase. If 𝑍 increases the credibility premium will also
increase, since the weight placed on data from the risk itself is higher and the mean of the data
from the risk itself is higher than the overall mean.
Question 4 [26]
Using the formula for the adjustment coefficient 𝑅
𝜆𝑀𝑋 (𝑅) = 𝜆 + 𝑐𝑅
4.1
Derive a simple upper and lower bound for 𝑅. In both cases state clearly what the
conditions are for these bounds to exist.
Claims on a portfolio of insurance policies arrive as a Poisson process with parameter 100.
Individual claim amounts follow a normal distribution with mean 30 and variance 52 . The
insurer calculates premiums using a premium loading of 20% and has initial surplus of 100.
4.2
Define carefully the ruin probabilities 𝜓(100), 𝜓(100,1) and 𝜓1 (100,1). [3]
𝜓(100) = 𝑃[𝑈(𝑡) < 0 𝑤ℎ𝑒𝑟𝑒 0 < 𝑡 < ∞]
𝜓(100,1) = 𝑃[𝑈(𝑡) < 0 𝑤ℎ𝑒𝑟𝑒 0 < 𝑡 ≤ 1]
𝜓1 (100,1) = 𝑃[𝑈(1) < 0]
4.3
Show that for this portfolio the value of the adjustment coefficient 𝑅 is 0.011 correct to
3 decimal places. [5]
1 + (1 + 𝜃)𝑚1 𝑅 = 𝑀𝑋 (𝑅)
1
2 )𝑅 2
1 + 1.2(30)𝑅 = 𝑒 30𝑅+2(5
𝐿. 𝐻. 𝑆 = 1 + 1.2(30)(0.011) = 1.396
1
2 )(0.011)2
𝑅. 𝐻. 𝑆 = 𝑒 30(0.011)+2(5
= 1.39307356
𝐿. 𝐻. 𝑆 ≅ 𝑅. 𝐻. 𝑆
4.5
Calculate an upper bound for 𝜓(100) and an estimate of 𝜓1 (100,1). [5]
𝜓(100) ≤ 𝑒 −𝑅(100) = 0.332871083
𝜓1 (100,1) = 𝑃[𝑈(1) < 0]
= 𝑃[100 + 1.2(30)(100) − 𝑆(1) < 0]
= 𝑃[𝑆(1) > 3700]
3700 − 𝐸[𝑆(1)]
= 𝑃 [𝑍 >
]
√𝑉𝑎𝑟[𝑆(1)]
𝐸[𝑆(1)] = 𝐸[𝑁(1)]𝐸(𝑋) = 100(30) = 3000
𝑉𝑎𝑟[𝑆(1)] = 𝐸[𝑁(1)]𝐸(𝑋 2 ) = 100(52 + 302 ) = 92,500
𝑃[𝑍 > 2.301585822] = 1 − 0.98928 = 0.01072
Suppose that the insurer now takes out a proportional reinsurance contract with retention level
𝛼. The premium charged by the reinsurer has a loading of 30%.
4.6
State the conditions that must hold in respect of the retention level 𝛼 to ensure that the
net premium income of the insurer is positive and exceeds the expected aggregate claims per
unit time [2], and hence: {Counts 4}
𝑐𝑁𝑒𝑡 > 0
(1 + 𝜃)𝐸(𝑆) − (1 + 𝜉)𝐸(𝑆𝑅 ) > 0
1.2(30)𝜆 > 1.3(1 − 𝛼)(30)𝜆
𝛼 > 0.076923076
And
𝑐𝑁𝑒𝑡 > 𝐸(𝑆𝐼 )
1.2(100)(30) − 1.3(100)(30)(1 − 𝛼) > (30)(100)𝛼
𝛼>
1
3
4.7
Maximise the adjustment coefficient R under this proportional reinsurance contract and
state the upper bound for the insurer’s ultimate probability of ruin. [5] {Only counts 3}
1
From 2.4 we can see that 𝛼 must lie in the interval, (3 , 1]. At some value for 𝛼 in this interval, it
is reasonable to suppose that 𝑅 will be maximised. However substituting the appropriate values
into the equation 𝜆 + 𝑐𝑁𝑒𝑡 𝑅 = 𝜆𝑀𝑌 (𝑅) gives:
100 + (1.2(30)100 − 1.3(30)(100)(1 − 𝛼))𝑅 = 100𝑀𝑌 (𝑅)
Now
1
2 )(𝛼𝑅)2
𝑀𝑌 (𝑅) = 𝐸(𝑒 𝑅𝑌 ) = 𝐸(𝑒 𝑅(𝛼𝑋) ) = 𝑀𝑋 (𝛼𝑅) = 𝑒 30𝛼𝑅+2(5
So
1
2
2
100 + (−200 + 3900𝛼)𝑅 = 𝑒 30𝛼𝑅+2(5 )(𝛼𝑅)
This must now be written in the form 𝑅 = 𝑓(𝛼) and then the derivative must be taken with
respect to 𝛼 to find the value of 𝛼 for which 𝑅 is maximised, i.e.
𝑑𝑅
𝑑𝛼
= 𝑓 ′ (𝛼) = 0. If this value
1
for 𝛼 lies within the required interval, (3 , 1], then we can say that there exists a retention level
𝛼 for which the adjustment coefficient is maximised, in other words, there exists a reinsurance
agreement under the premium loading structure specified which would improve the “safety
coefficient” of the company and hence reduce the probability of ruin.
However, there is no explicit expression of the form 𝑅 = 𝑓(𝛼) in this case and numerical
methods need to be employed.