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bonfire09
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Homework Statement
There are ##N = 10000## clients of an insurance company. One-half of them will file claims with probability ##p_1 = .05##, another half of them will file claims with probability ##p_2 = .03##. Each claim is worth ##$1000##. Find the Value-at-Risk at the level α = 0.99, that is, the amount of money the company should accumulate to be able to pay its customers with probability greater than or equal to α
Homework Equations
The Attempt at a Solution
What I did was let ##\sum_{i=1}^{5000} X_i## where ##X_i \sim## Bernoulli(##p_1##) be the proportion of 5000 customers with probability of filing a claim with ##p_1=.05## probability and ##\sum_{i=1}^{5000} Y_i## be the proportion of the other 5000 customers with probability ##p_2=.03## of filing a claim. Again ##Y_i\sim##Bernoulli(##p_2##).Then we know ##\sum_{i=1}^{5000} X_i## and ##\sum_{i=1}^{5000} Y_i## both follow a binomial distribution. Since ##p_1## and ##p_2## are large enough with each sample size being large enough we can use a normal approximation. From here
##\sum_{i=1}^{5000} X_i\sim\text{N(5000*.05,5000*.05*.95)=N(250,237.5)}##
##\sum_{i=1}^{5000} Y_i\sim \text{ N(5000*.03,5000*.03*.97)=N(150,145.5)}##
Then we let ##Z=\sum_{i=1}^{5000} X_i+\sum_{i=1}^{5000} Y_i\sim\text{N(400,383)}##
From here since each claim is ##1000## dollars then ##1000Z## is total amount in claims and 10000p is the amount the company should collect where p is the amount of money we collect from each individual. which is a constant value to make this problem work. We want to find
##P(1000Z \leq 10000p)=.99 \implies P(Z\leq 10p)=.99 \implies P(\dfrac{z-400}{\sqrt{383}}\leq\dfrac{10p-400}{\sqrt{383}})=.99\implies \dfrac{10p-400}{\sqrt{383}}=2.33\implies p=44.559##. So we should collect about $44.56 from each person to keep the value at risk at 99%. But I am not sure if this is correct or not. Any help would be greatly appreciated. Thanks