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Here are a set of exercises that guide the viewer through some of the theoretical foundations of Loss Data Analytics. Each tutorial is based on one or more questions from the professional actuarial examinations – typically the Society of Actuaries Exam C.
Tutorial Structure. Each guided tutorial has a strategy set that describes the context. When you hit the “Start quiz” button, you begin the tutorial that is comprised of a series of mini-questions designed to lead you to the target question. At each stage, hints are provided as well as feedback on the correct solution of each mini-question.
Your Assignment. In reviewing these exercises, ideally the viewer will:
- Work the problem posed referring only to basic theory
- Even if you get the answer correct, review the strategy for this type of problem by clicking (revealing) the Strategy for … header
- If you feel comfortable with the strategy and got the problem correct, then you may choose to move on. However, you might also decide to follow the step-by-step process for solving the problem by clicking on the “Start Quiz” button. It is not really a quiz — it is a guided tutorial.
Collective Risk Model: General considerations
Consider the collective risk model, which has the representation;
$$S=X_1+…+X_N$$
Where (S) the aggregate loss of random number (N) of individual claims (X_1,…,X_N). Here, we assume that conditional on (N=n) , (X_1,…,X_n) are (i.i.d) random variables. Also, the frequency (N) and severity (Y) are independent
Using the law of iterated expectations, the mean of (S) is given by;
$$begin{array}{ll}
E(S)
&=E[E(S|N)]=E[NE(X)]=E(N)E(X) \
&=mu E(N)quad quad end{array}$$
Where (mu=E(X_i)).
Using the law of total variation, the variance of (S) is given by;
$$begin{array}{ll}
Var(S)
&=E[Var(S|N)]+Var[E(S|N)] \
&=E[N Var(X)]+Var[N E(X)] \
&=E(N)Var(X)+Var(N)E(X)^2 \
&=sigma^2E(N)+mu^2Var(N) quad quad end{array}$$
Where (Var(X_i)=sigma^2).
Special case: Poisson distributed frequency. (Nsim Poisson(lambda))
$$begin{array}{ll}
E(N)&=Var(N)=lambda \
Var(S)&=lambda(sigma^2+mu^2)=lambda E(X^2)
end{array}$$
Strategy for Aggregate Claim Distribution Problems
The pdf(pmf) of (S) is given by;
$$f_S(s)=sum_{n=0}^{infty}Pr(N=n)cdot f_X^{*n}(s) quad quad (1)$$
This means that to calculate (f_S(s)), we must sum up over all possibilities of (n) claims summing up to (s). The n-fold convolution of (f_X) is (f_X^{*n}) and (f_X^{*0}(0)=1). To obtain the cdf of (S), its given by;
$$F_S(s)=sum_{i=0}^{s}f_S(i)quad quad (2)$$
To apply these rules,
- From the question, identify the distribution of the claim frequency (N) and severity (X).
- If the question requires you to calculate the pdf(pmf) of (S) ,(f_S(s)), its is given by equation (1) above.
- If the question requires you to calculate the cdf, (F_S(s)) it is given by equation (2).
Aggregate Claim Distribution SOA #113
The number of claims , (N), made on an insurance portfolio follows the following distribution:
$$
{scriptsize
begin{matrix}
begin{array}{ccc}
hline
n & Pr(N=n) \
hline
0 & 0.7 \
2 & 0.2 \
3 & 0.1 \
hline \
end{array}
end{matrix}
}
$$ If a claim occurs, the benefit is 0 or 10 with probability 0.8 and 0.2, respectively. The number of claims and the benefit for each claim are independent.
(a) Calculate the expected value of (S), (E(S)), and variance of (S) ,(Var(S))
(b) Calculate the probability that aggregate benefits will exceed expected benefits by more than 2 standard deviations.
[WpProQuiz 82]
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Aggregate Claim Distribution SOA #95
The number of claims in a period has a geometric distribution with mean 4.The amount of each claim (X) follows (Pr(X=x)=0.25), (x=1,2,3,4). The number of claims and the claim amounts are independent. (S) is the aggregate claim amount in the period.
Calculate (F_S(3))
[WpProQuiz 81]
Strategy for Aggregate Loss Distribution Approximation Problems
When the number of observed losses (n),is large, the observed aggregate losses are approximated normally distributed. To use normal approximation for aggregate loss distributions, (F_S(s)), we use the formula below:
$$F_S(s)=Pr(S leq s)=Phileft(frac{s-mu_s}{sigma_s}right) quad quad (1)$$
Where;
$$mu_s=E(S) quad text{and} quad sigma_s=sqrt{Var(S)}$$
When (n) is small then the distribution of the aggregate losses is positively skewed and hence its better to approximate aggregate losses with a distribution that is positively skewed (e.g LogNormal distribution ). To use LogNormal approximations:
$$F_S(s)=Pr(S leq s)=Pr(ln S leq ln s)=Phileft(frac{ln s-mu_s}{sigma_s}right) quad quad (2)$$
Where (u_s) and (sigma_s) are found using method of moments :
$$
left{begin{array}{ll}
exp{mu_s+sigma_s^2/2}=E(S)\
exp{2mu_s+2sigma_s^2}=E(S^2)=E(S)^2+Var(S) end{array}right. $$
To apply these rules,
- From the question, identify the distribution of the claim frequency (N) and severity (X).
- Calculate (E(S)) and (Var(S)) using the appropriate formulae
- Using equation (1) compute the aggregate loss distribution function (F_S(s)).
- When (n) is small,use equation (2) for a LogNormal approximation for the aggregate loss distribution function (F_S(s)) .
Normal Approximation SOA #118
For an individual over 65:
(i) The number of pharmacy claims is a Poisson random variable with mean 25.
(ii) The amount of each pharmacy claim is uniformly distributed between 5 and 95.
(iii) The amounts of the claims and the number of claims are mutually independent.
Determine the probability that aggregate claims for this individual will exceed 2000 using the normal approximation.
[WpProQuiz 83]
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Strategy for Stop Loss Premium Problems
The stop loss premium is the expected aggregate losses above of aggregate deductible. Here, aggregate loss means the sum of individual claims payments after individual claim modifications such as policy limits and deductibles but before aggregate modifications. An aggregate deductible is a deductible applied to aggregate losses rather than individual losses.
Assuming the frequency (N) and severity (X) are independent, the mean of aggregate losses,(S) , is given by;
$$E(S)=E(N)E(X) quad quad (1)$$
With an aggregate deductible (d), the stop loss premium is the mean aggregate loss (E(S)) minus the expected value of the aggregate loss limited to (d) , (E(Swedge d)). The stop loss premium is given by :
$$E[(S-d)_+]=E(S)-E(Swedge d) quad quad (2)$$
For discrete aggregate loss distributions:
$$E(Swedge d)=sum_{slt d} s f_S(s) + d (1-P(Slt d)) $$
For continuous aggregate loss distributions:
$$E(Swedge d)=int_0^d s f_S(s) ds + int_d^{infty} d f_S(s) ds \
= int^d_0 (1-F(s)) ds . $$
To apply these rules,
- From the question, identify the distribution of the claim frequency (N) and severity (X).
- Calculate the mean aggregate loss (E(S)) and the expected value of the aggregate loss limited to (d), (E(Swedge d)), using the appropriate formulae
- Using equation (2) compute the stop loss premium (E[(S-d)_+]) .
Stop loss premium SOA #99
For a certain company, losses follow a Poisson frequency distribution with mean 2 per year, and the amount of a loss is 1, 2, or 3, each with probability 1/3. Loss amounts are independent of the number of losses, and of each other.
An insurance policy covers all losses in a year, subject to an annual aggregate deductible of 2.
Calculate the expected claim payments for this insurance policy.
[WpProQuiz 84]
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Strategy Coverage Modifications Problems
Consider an insurance contract with deductible d. This causes the number of claims for amounts greater than zero to change. Let (N^L) be the number of losses and given (X_i), the ground-up losses for (i=1,…,n^L), denote the number of claims by (N^P) . We assume , given (N^L=n^L) , (X_i) are (i.i.d) with common distribution (X). Then;
$$N^P=I(X_1 gt d)+…+I(X_{N^L}gt d)$$
It can be shown that for distributions of the ((a,b,0)) class (N_L) and (N_P) follow the same distribution but with different parameters. Let (v=Pr(X gt d)), then we have:
$$
{scriptsize
begin{matrix}
begin{array}{lll}
hline
text{Distribution} & N^L & N^P \
hline
text{Poisson} & lambda & vlambda\
text{Binomial} & n, p & n, vp\
text{Negative binomial} & r,beta & r, vbeta \
hline \
end{array}
end{matrix}
}
$$
Also, recall that (X^L) , the random variable for the payment per loss with deductible (d) is given by :
$$X^L=
left{begin{array}{ll}
0 & X_i lt d \
X_i-d & X_i geq d end{array} right. $$
And (X^P),the random variable for the per payment on an insurance with deductible (d) is the payment per loss variable conditioned on (X>d) or
$$X^P=
left{begin{array}{ll}
undefined & X_i lt d \
X_i-d & X_i geq d end{array} right. $$
In modeling aggregate losses with per loss deductibles, one way is to model the number of losses (N^L) in which no modification is needed for the frequncy model and use the payment per loss variable ,(X^L). Another way is to use the modified frequency model (N^P) with the per payment random variable (X^P). Then, we can express the aggregate loss below:
$$begin{array}{ll}
S
&=X_1^L+…+X_{N^L}^L\
&=X_1^P+…+X_{N^P}^P
end{array}$$
If (X) follows uniform, exponential or Pareto then using the per payment approach makes calculation easier.
To apply these rules,
- From the question, identify coverage modification, the distribution of the claim frequency (N) and severity (X).
- Model the aggregate claims using the per-loss apporach or using the per-payment approach
Coverage Modifications SOA #212
For an insurance:
(i) The number of losses per year has a Poisson distribution with (lambda= 10 )
(ii) Loss amounts are uniformly distributed on (0, 10).
(iii) Loss amounts and the number of losses are mutually independent.
(iv) There is an ordinary deductible of 4 per loss.
(a) Calculate the expected value of aggregate payments in a year.
(b) Calculate the variance of aggregate payments in a year.
[WpProQuiz 85]