ATW 316 & ATW 314 Class Test 3 Memo [15 marks] [27 minutes]
The following information is available from an investigation where an actuarial assistant attempted to fit different distributions to claim data.
Loss Amount
[0 2347]
[2347 4694]
[4694 7041]
[7041 9388]
[9388 11735]
[11735 14082]
[14082 16429]
[16429 18776]
[18776 21123]
[21123 23470]
Number of Losses
(Observed)
8510
1075
280
92
28
9
2
3
0
1
Exponential (
(Expected)
A 𝜆 ) Pareto ( 𝛼, 𝛽)
(Expected)
8659.5643
1003.6369
219.2564
67.2664
25.5513
11.2471
5.5156
2.9381
1.6709
1.0020
B
Weibull ( 𝑐, 𝛾 )
(Expected) 𝜒 2
𝐸𝑥𝑝
= 7542.2283
𝜒
2
𝑃𝑎𝑟 𝜒
2
𝑊𝑏𝑙
= 𝐶
= 6.4355
Where ∑ 𝑥 𝑖
= 11913172 upper quartile is 𝑞
2
and ∑ 𝑥
= 1562.3698
𝑖
2
.
= 39854334971 . The lower quartile is 𝑞
1
= 187.9746
and the i) ii)
Fit an exponential distribution to the data by estimating 𝜆 using maximum likelihood estimation and find the value of A. 𝑛
𝐿(𝜆) = ∏ 𝜆𝑒 −𝜆𝑥 𝑖 = 𝜆 𝑛 𝑒 −𝜆 ∑ 𝑛 𝑖=1 𝑥 𝑖 𝑖=1 𝑛
𝐼(𝜆) = 𝑛𝑙𝑛(𝜆) − 𝜆 ∑ 𝑥 𝑖 𝑖=1 𝑛 𝑑𝐼(𝜆)
= 𝑛
− ∑ 𝑥 𝑖
= 0 𝑑𝜆 𝜆̂ 𝑖=1 𝜆̂ = 𝑛
∑ 𝑛 𝑖=1 𝑥 𝑖
=
10,000
11,913,172
= 0.0008394070026
Fit a Weibull distribution to the data by estimating 𝑐 and 𝛾 using the method of percentiles.
𝑃(𝑋 < 187.9746) = 0.25 = 1 − 𝑒
−𝑐187.9746
𝛾
𝑃(𝑋 < 1562.3698) = 0.75 = 1 − 𝑒
−𝑐1562.3698
𝛾 𝑒
[ 𝑐 = − 𝑙𝑛0.75
187.9746
𝛾 𝑙𝑛0.75
187.9746
𝛾
]1562.3698
𝛾
= 0.25
0.75𝑒
[
1562.3698
187.9746
] 𝛾
= 𝑙𝑛0.25

iii) iv) ln ( ln(0.25) ln(0.75)
) 𝛾 = ln (
1562.3698
187.9746 ) ln(0.75) 𝑐 = −
187.9746
0.742583499
= 0.742583499
= 0.005891206949
Find the values of A and B.
7041
𝐴 = 10,000 ∫ 𝜆𝑒
−𝜆𝑥 𝑑𝑥 = 10,000(𝑒
−(0.0008394070026)4694
− 𝑒
−(0.0008394070026)7041
)
4694
= 167.3331397
𝐵 = 10,000(𝑒
−𝑐(9388) 𝛾
− 𝑒
−𝑐(11735) 𝛾
) = 32.0987394
By using the data in the table, find the value of C and comment on the fit of the three proposed models in terms of all the data and especially the tails. Which model would you propose and why? [3,3,5,4] 𝜒 2
𝑃𝑎𝑟
= ∑
10 𝑖=1
(𝑂 − 𝐸) 2
𝐸
= 38.17718033
The degrees of freedom for the exponential model are 10-1-1=8 and the critical value at 5% significance is 15.51. The calculated chi-square statistic is given as 𝜒
2
𝐸𝑥𝑝
= 7542.2283
. Thus we reject the null hypothesis that the data follow an exponential distribution. The exponential tail is decreasing much too fast.
The degrees of freedom for the Pareto model are 10-1-2=7 and the critical value at 5% significance is 14.07. The calculated chi-square statistic is 38.17718033
. Thus we reject the null hypothesis that the data follow a Pareto distribution. It looks as if the tail of the Pareto is not decaying fast enough.
The degrees of freedom for the Weibull model are 10-1-2=7 and the critical value at 5% significance is 14.07. The calculated chi-square statistic is 𝜒
2
𝑊𝑏𝑙
= 6.4355
and we do not reject the null-hypothesis that the data can be approximated by a Weibull distribution. The extra flexibility that is added to the Weibull distribution due to the 𝛾 parameter allows the
Weibull distribution to fit the tail of the data quite well.
NB! Notice that we did not reject the null-hypothesis for the Weibull distribution. However not rejecting the null-hypothesis does not give us incontrovertible proof that the data are observations from a Weibull distribution. The null-hypothesis reads: “The data can be
approximated by a Weibull distribution.” For all we know there is another model that fits the data even better. But among those we tested, the Weibull provides the “best fit” to the data.