Chebyshev's inequality to compute the premium income
This is a problem 9.7 from Anderson "Introduction to probability"
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? Compute the answer with both Chebyshev's inequality
and CLT.
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 $i$, then $$operatorname{E}[S] = 2500 * operatorname{E}[X_i] = 2500 * 1000 = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] =2500 * 900^2.$$
We want to fins s (the premium income), such that $P(Sleq s)=0.999.$ This is equivalent to $P(Sgeq s)=0.001.$ By Chebyshev's inequality
$$P(Sgeq s) = P(S-operatorname{E}[S]geq s- operatorname{E}[S])leq P(|S-operatorname{E}[S]|geq s-operatorname{E}[S])leq frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2}$$
My question is how do we get the desired equality from this inequality? Thanks
probability
asked Dec 7 at 21:24
dxdydz
1949
add a comment |
This is a problem 9.7 from Anderson "Introduction to probability"
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? Compute the answer with both Chebyshev's inequality
and CLT.
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 $i$, then $$operatorname{E}[S] = 2500 * operatorname{E}[X_i] = 2500 * 1000 = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] =2500 * 900^2.$$
We want to fins s (the premium income), such that $P(Sleq s)=0.999.$ This is equivalent to $P(Sgeq s)=0.001.$ By Chebyshev's inequality
$$P(Sgeq s) = P(S-operatorname{E}[S]geq s- operatorname{E}[S])leq P(|S-operatorname{E}[S]|geq s-operatorname{E}[S])leq frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2}$$
My question is how do we get the desired equality from this inequality? Thanks
probability
asked Dec 7 at 21:24
dxdydz
1949
add a comment |
This is a problem 9.7 from Anderson "Introduction to probability"
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? Compute the answer with both Chebyshev's inequality
and CLT.
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 $i$, then $$operatorname{E}[S] = 2500 * operatorname{E}[X_i] = 2500 * 1000 = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] =2500 * 900^2.$$
We want to fins s (the premium income), such that $P(Sleq s)=0.999.$ This is equivalent to $P(Sgeq s)=0.001.$ By Chebyshev's inequality
$$P(Sgeq s) = P(S-operatorname{E}[S]geq s- operatorname{E}[S])leq P(|S-operatorname{E}[S]|geq s-operatorname{E}[S])leq frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2}$$
My question is how do we get the desired equality from this inequality? Thanks
probability
asked Dec 7 at 21:24
dxdydz
1949
This is a problem 9.7 from Anderson "Introduction to probability"
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? Compute the answer with both Chebyshev's inequality
and CLT.
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 $i$, then $$operatorname{E}[S] = 2500 * operatorname{E}[X_i] = 2500 * 1000 = 2500000,$$ and variance $$operatorname{Var}[S] = 2500 operatorname{Var}[X_i] =2500 * 900^2.$$
We want to fins s (the premium income), such that $P(Sleq s)=0.999.$ This is equivalent to $P(Sgeq s)=0.001.$ By Chebyshev's inequality
$$P(Sgeq s) = P(S-operatorname{E}[S]geq s- operatorname{E}[S])leq P(|S-operatorname{E}[S]|geq s-operatorname{E}[S])leq frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2}$$
My question is how do we get the desired equality from this inequality? Thanks
probability
probability
asked Dec 7 at 21:24
dxdydz
1949
asked Dec 7 at 21:24
dxdydz
1949
asked Dec 7 at 21:24
dxdydz
1949
asked Dec 7 at 21:24
dxdydz
1949
asked Dec 7 at 21:24
dxdydz
1949
1949
add a comment |
add a comment |
1 Answer
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I guess that the answer consists in solving the inequality:
$$frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2} leq 0.001,$$
which solution is:
$$s geq sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$$
or
$$s leq sqrt{frac{operatorname{Var}[S]}{0.001}} - operatorname{E}[S].$$
So the solution is $ sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$.
answered Dec 7 at 22:32
dallonsi
1187
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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I guess that the answer consists in solving the inequality:
$$frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2} leq 0.001,$$
which solution is:
$$s geq sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$$
or
$$s leq sqrt{frac{operatorname{Var}[S]}{0.001}} - operatorname{E}[S].$$
So the solution is $ sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$.
answered Dec 7 at 22:32
dallonsi
1187
add a comment |
I guess that the answer consists in solving the inequality:
$$frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2} leq 0.001,$$
which solution is:
$$s geq sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$$
or
$$s leq sqrt{frac{operatorname{Var}[S]}{0.001}} - operatorname{E}[S].$$
So the solution is $ sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$.
answered Dec 7 at 22:32
dallonsi
1187
add a comment |
I guess that the answer consists in solving the inequality:
$$frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2} leq 0.001,$$
which solution is:
$$s geq sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$$
or
$$s leq sqrt{frac{operatorname{Var}[S]}{0.001}} - operatorname{E}[S].$$
So the solution is $ sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$.
answered Dec 7 at 22:32
dallonsi
1187
I guess that the answer consists in solving the inequality:
$$frac{operatorname{Var}[S]}{(s-operatorname{E}[S])^2} leq 0.001,$$
which solution is:
$$s geq sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$$
or
$$s leq sqrt{frac{operatorname{Var}[S]}{0.001}} - operatorname{E}[S].$$
So the solution is $ sqrt{frac{operatorname{Var}[S]}{0.001}} + operatorname{E}[S]$.
answered Dec 7 at 22:32
dallonsi
1187
answered Dec 7 at 22:32
dallonsi
1187
answered Dec 7 at 22:32
dallonsi
1187
answered Dec 7 at 22:32
dallonsi
1187
1187
add a comment |
add a comment |
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