what premium should the company charge each policy holder to assure that the premium income will cover the...
$begingroup$
A car insurance company has $2,500$ policy holders. The expected claim
paid to a policy holder during a year is $1,000$ with a standard
deviation of $900$. What premium should the company charge each policy
holder to assure that with probability $0.999$ the premium income will
cover the cost of the claims?
I'm having trouble with interpreting this problem. I want to correctly denote the random variables to use CLT or Chebyshev's inequality.
So I suppose $X$ to be the cost of the claims, then we have $E[X]=1000$, $Var(X) = 900^2$.
Let $y$ be the cost of the premium, so we need to find $ P(Y-Xgeq 0)$ right?
probability variance standard-deviation
edited Dec 7 '18 at 6:07
idea
2,15841125
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
$endgroup$
add a comment |
$begingroup$
A car insurance company has $2,500$ policy holders. The expected claim
paid to a policy holder during a year is $1,000$ with a standard
deviation of $900$. What premium should the company charge each policy
holder to assure that with probability $0.999$ the premium income will
cover the cost of the claims?
I'm having trouble with interpreting this problem. I want to correctly denote the random variables to use CLT or Chebyshev's inequality.
So I suppose $X$ to be the cost of the claims, then we have $E[X]=1000$, $Var(X) = 900^2$.
Let $y$ be the cost of the premium, so we need to find $ P(Y-Xgeq 0)$ right?
probability variance standard-deviation
edited Dec 7 '18 at 6:07
idea
2,15841125
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
$endgroup$
add a comment |
$begingroup$
A car insurance company has $2,500$ policy holders. The expected claim
paid to a policy holder during a year is $1,000$ with a standard
deviation of $900$. What premium should the company charge each policy
holder to assure that with probability $0.999$ the premium income will
cover the cost of the claims?
I'm having trouble with interpreting this problem. I want to correctly denote the random variables to use CLT or Chebyshev's inequality.
So I suppose $X$ to be the cost of the claims, then we have $E[X]=1000$, $Var(X) = 900^2$.
Let $y$ be the cost of the premium, so we need to find $ P(Y-Xgeq 0)$ right?
probability variance standard-deviation
edited Dec 7 '18 at 6:07
idea
2,15841125
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
$endgroup$
A car insurance company has $2,500$ policy holders. The expected claim
paid to a policy holder during a year is $1,000$ with a standard
deviation of $900$. What premium should the company charge each policy
holder to assure that with probability $0.999$ the premium income will
cover the cost of the claims?
I'm having trouble with interpreting this problem. I want to correctly denote the random variables to use CLT or Chebyshev's inequality.
So I suppose $X$ to be the cost of the claims, then we have $E[X]=1000$, $Var(X) = 900^2$.
Let $y$ be the cost of the premium, so we need to find $ P(Y-Xgeq 0)$ right?
probability variance standard-deviation
probability variance standard-deviation
edited Dec 7 '18 at 6:07
idea
2,15841125
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
edited Dec 7 '18 at 6:07
idea
2,15841125
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
edited Dec 7 '18 at 6:07
idea
2,15841125
edited Dec 7 '18 at 6:07
idea
2,15841125
edited Dec 7 '18 at 6:07
idea
2,15841125
2,15841125
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
asked Dec 7 '18 at 5:58
dxdydzdxdydz
2989
2989
add a comment |
add a comment |
1 Answer
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$begingroup$
No; you need to look at the aggregate claims random variable. The idea is that the insurer needs to collect sufficient premiums to cover the aggregate claims with at least $0.999$ probability.
Assuming each claim is independent, then the aggregate claims $S = sum_{i=1}^{2500} X_i$ where $X_i$ is the annual claim size for policyholder $i$, for the book of business is approximately normally distributed with mean $$operatorname{E}[S] = 2500 operatorname{E}[X_i] = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] = 2250000.$$ Consequently, $$Z = frac{S - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}$$ is approximately standard normal, and we want to find some number $P$ such that $$Pr[S le P] = 0.999;$$ equivalently, $$Prleft[ Z le frac{P - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}right] = 0.999.$$ Since the $99.9^{rm th}$ percentile of the standard normal is approximately $3.09023$, it follows that $$frac{P - 2500000}{1500} = 3.09023,$$ or $$P = 2504635.35.$$ This represents the total premiums the insurer needs to collect; dividing by $2500$ gives the per-policyholder premium, assuming that all policyholders are rated equally. What is the excess above the expected loss per policyholder?
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
$endgroup$
add a comment |
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$begingroup$
No; you need to look at the aggregate claims random variable. The idea is that the insurer needs to collect sufficient premiums to cover the aggregate claims with at least $0.999$ probability.
Assuming each claim is independent, then the aggregate claims $S = sum_{i=1}^{2500} X_i$ where $X_i$ is the annual claim size for policyholder $i$, for the book of business is approximately normally distributed with mean $$operatorname{E}[S] = 2500 operatorname{E}[X_i] = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] = 2250000.$$ Consequently, $$Z = frac{S - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}$$ is approximately standard normal, and we want to find some number $P$ such that $$Pr[S le P] = 0.999;$$ equivalently, $$Prleft[ Z le frac{P - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}right] = 0.999.$$ Since the $99.9^{rm th}$ percentile of the standard normal is approximately $3.09023$, it follows that $$frac{P - 2500000}{1500} = 3.09023,$$ or $$P = 2504635.35.$$ This represents the total premiums the insurer needs to collect; dividing by $2500$ gives the per-policyholder premium, assuming that all policyholders are rated equally. What is the excess above the expected loss per policyholder?
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
$endgroup$
add a comment |
$begingroup$
No; you need to look at the aggregate claims random variable. The idea is that the insurer needs to collect sufficient premiums to cover the aggregate claims with at least $0.999$ probability.
Assuming each claim is independent, then the aggregate claims $S = sum_{i=1}^{2500} X_i$ where $X_i$ is the annual claim size for policyholder $i$, for the book of business is approximately normally distributed with mean $$operatorname{E}[S] = 2500 operatorname{E}[X_i] = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] = 2250000.$$ Consequently, $$Z = frac{S - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}$$ is approximately standard normal, and we want to find some number $P$ such that $$Pr[S le P] = 0.999;$$ equivalently, $$Prleft[ Z le frac{P - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}right] = 0.999.$$ Since the $99.9^{rm th}$ percentile of the standard normal is approximately $3.09023$, it follows that $$frac{P - 2500000}{1500} = 3.09023,$$ or $$P = 2504635.35.$$ This represents the total premiums the insurer needs to collect; dividing by $2500$ gives the per-policyholder premium, assuming that all policyholders are rated equally. What is the excess above the expected loss per policyholder?
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
$endgroup$
add a comment |
$begingroup$
No; you need to look at the aggregate claims random variable. The idea is that the insurer needs to collect sufficient premiums to cover the aggregate claims with at least $0.999$ probability.
Assuming each claim is independent, then the aggregate claims $S = sum_{i=1}^{2500} X_i$ where $X_i$ is the annual claim size for policyholder $i$, for the book of business is approximately normally distributed with mean $$operatorname{E}[S] = 2500 operatorname{E}[X_i] = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] = 2250000.$$ Consequently, $$Z = frac{S - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}$$ is approximately standard normal, and we want to find some number $P$ such that $$Pr[S le P] = 0.999;$$ equivalently, $$Prleft[ Z le frac{P - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}right] = 0.999.$$ Since the $99.9^{rm th}$ percentile of the standard normal is approximately $3.09023$, it follows that $$frac{P - 2500000}{1500} = 3.09023,$$ or $$P = 2504635.35.$$ This represents the total premiums the insurer needs to collect; dividing by $2500$ gives the per-policyholder premium, assuming that all policyholders are rated equally. What is the excess above the expected loss per policyholder?
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
$endgroup$
No; you need to look at the aggregate claims random variable. The idea is that the insurer needs to collect sufficient premiums to cover the aggregate claims with at least $0.999$ probability.
Assuming each claim is independent, then the aggregate claims $S = sum_{i=1}^{2500} X_i$ where $X_i$ is the annual claim size for policyholder $i$, for the book of business is approximately normally distributed with mean $$operatorname{E}[S] = 2500 operatorname{E}[X_i] = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] = 2250000.$$ Consequently, $$Z = frac{S - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}$$ is approximately standard normal, and we want to find some number $P$ such that $$Pr[S le P] = 0.999;$$ equivalently, $$Prleft[ Z le frac{P - operatorname{E}[S]}{sqrt{operatorname{Var}[S]}}right] = 0.999.$$ Since the $99.9^{rm th}$ percentile of the standard normal is approximately $3.09023$, it follows that $$frac{P - 2500000}{1500} = 3.09023,$$ or $$P = 2504635.35.$$ This represents the total premiums the insurer needs to collect; dividing by $2500$ gives the per-policyholder premium, assuming that all policyholders are rated equally. What is the excess above the expected loss per policyholder?
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
answered Dec 7 '18 at 6:15
heropupheropup
63.3k762101
63.3k762101
add a comment |
add a comment |
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