Problem 1. 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 Copyright By PowCoder代写 加微信 powcoder
(3) The Expected Cost per loss
Problem 2. Use the same loss distribution as the prvious problem, with θ = 40 but that our policy has no deductible, a limit of $200 and a coinsurance of 75%. We assume an annual inflation rate of 2% What is the variance of each loss? You may assume that Γ(3; 200/1.02) = 1.
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 pM0 = 0.1
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 pM0 = 0.2) versions.
Standard Normal Distribution Table
√ e 2 dx, Φ(−z)=1−Φ(z)
Φ(z)=P(Z ≤z)=
z 0 +0.01 +0.02 +0.03 +0.04 +0.05 +0.06 +0.07 +0.08 +0.09
0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890 2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974 2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981 2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
3 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990
Problem 1.
This is an Exponential(θ) distribution
(1) Per-Payment Distribution Function:
FYp(y)= FX(y+d)−FX(d)
1 − FX(d) = e−d/θ − e−(y+d)/θ
= e−d/θ − e−y/θe−d/θ
e−d/θ = 1 − e−y/θ
(2) Per-Loss Survival Function:
SY L (y) = SX (y + d)
= e−(y+d)/θ
(3) Expected Cost per Loss for ordinary deductible:
E(X) − E(X ∧ d) = θ − θ(1 − e−d/θ) = θe−d/θ
Problem 2.
e−y/θ ed/θ e(d−y)/θ
=θ=θ (2) Per-Loss hazard rate Function:
0 0
e−y/θ ey/θ (assume y > d) = θ = 1
(1) Per-Payment density Function:
fYP(y)= fX(y), y>d
(3) Expected Cost per payment for franchise deductible: E(X)−E(X∧d)+d=(θ−θ(1−e−d/θ))ed/θ +d
Problem 3.
(1) The Loss Elimination Ratio:
= θ(e−d/θed/θ) + d = θ + d
E(X ∧d) = θ(1−e−d/θ) = 1−e−d/θ E(X) θ
(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 ran- dom variable:
α(1+r)X 0≤X< u YL = 1.02
αu X≥ u 1.02
We need the 1st and 2nd moments of our new variable Y L. Note that because d = 0, E[X∧d]=E[X∧0]=0becausewealwayshave0≤X. Itfollowsthat:
Lu E[Y ]=α(1+r) E[X∧1.02]
= α(1 + r)θ(1 − e−(u/1.02θ)) = 30.373
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
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