Losses occur with the following distribution.
(
0 if x < 0
FX(x) = 1 − e−x/θ otherwise
Suppose we have an ordinary deductible of $d. Please find the following:
(1) The per-payment Distribution Function
(2) The per-loss Survival Function
(3) The Expected Cost per loss
Problem 2. Use the same loss distribution as the prvious problem, with θ = 40 but that our policy hasAdd no deductible, a limit of $200 and a coinsurance of 75%. We
Problem 3. Determine the first four probabilities for an ETNB distribution with r = 2 and β = 1. Do this both for the truncated version and for the modified version, with
z 0 +0.01 +0.02 +0.03 +0.04 +0.05 +0.06
0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764
Problem 4. Consider a Poisson random variable with parameters λ = 2. Determine the first four probabilities for this random variable. Then determine the corresponding probabilities for the zero-truncated and zero-modified (with 2) versions.
1
2
Standard Normal Distribution Table
Φ(−z) = 1 − Φ(z)
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315
1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770
1.2
1.3 0.8849Assign 0.8869me 0.8888nt oje0.8925ct xam0.89
1.4
0.9032 0.9049 0.9066 0.9099 0.9131
1.5 Pr E
0.9192 0.9207 0.9222 0.8907 0.9251 0.8944 0.9279
1.6
0.9332
1.7 0.9082 0.9115
0.9452 htt ps:/ wco co
1.8 0.9345 0.9357 0.9236 0.9382 0.9265 0.9406
0.9554
1.9
0.9641 0.9463 0.9564 0.9474 0.9573 0.9495 0.9515 0.960
2 0.9649 dd /po0.9370 0.9591 der.0.93940.9686
0.9713
0.9772
2.2
0.9821
2.1 0.97190.9778 0.9826A 96569726 0.9783 0.9830 0.94840.9582 hat 0.9671 0.9599 0.9678po0.9505 wcod 0.97
2.3
0.9861
2.4 0.9864 0.9896 0.9868 0.9898 0.9664 0.9738 0.9744 0.9803 0.984
0.9893
2.5 0.9920 0.9940 0.9922 0.9941WeC 0.97320.9793 0.9798 0.9881 0.990
0.9918
2.6 0.9955 0.9966 0.9956 0.9967 0.9838 0.9931 0.994
0.9938 0.9842
2.7 0.9975 0.9982 0.9976 0.9982 0.9788 0.9875 0.9961 0.997
0.9953 0.9878
2.8 0.9834 0.9904 0.9979 0.998
0.9965 0.9987 0.9987 0.9906
2.9 0.9871 0.9927 0.9989
0.9974 0.9929
3 0.9981 0.9901 0.9945 0.9946
0.9925
0.9987 0.9959 0.9960
0.9943 0.9969
0.9957 0.9977 0.9970
0.9968 0.9984 0.9978
0.9977 0.9988 0.9984
0.9983 0.9989
0.9988
3
Solutions
This is an Exponential(θ) distribution
Problem 1.
(1) Per-Payment Distribution Function:
(2) Per-Loss Survival Function:
SY L(y) = SX(y + d)
= e−(y+d)/θ
(3) Expected Cost per Loss for ordinary deductible:
Add E(X) −E(X ∧ d) = θ − θ(1 − e−d/θ)
= θe−d/θ
Problem 2.
(1) Per-Payment density Function:
(2) Per-Loss hazard rate Function:
(
0 0 < y < d
hY L(y) =
hX(y) y > d
−y/θ y/θ
(assume
4
(3) Expected Cost per payment for franchise deductible:
Problem 3.
(1) The Loss Elimination Ratio:
(2) The expected cost per loss is defined as:
(1 + r){E[X] −E[X ∧ d/(1 + r)]} = (1.02){E[X] −E[X ∧ d/1.02]}
= 1.02(θ − θ(1 − e−(d/1.02θ)))
= 1.02θe−(d/1.02θ)
Problem 4.
Based on our loss distribution and policy details, we have the following per-loss random variable: Add
We need the 1st and 2nd moments of our new variable Y L. Note that because d = 0, E[X ∧ d] = E[X ∧ 0] = 0 because we always have 0 ≤ X. It follows that:
and (again, because d = 0)
E[Y L2] = α2(1 + r)2E[(X ∧ (u/1.02))2]
= α2(1 + r)2θ2Γ(3)Γ(3;u/(1.02θ)) + (u/1.02θ)2e−(u/1.02θ) = 2032.625 Combining these, we have:
Var(Y L) = 2032.625 − 30.3732 = 1110.132
