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Percent loss covered on average with a deductible
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Suppose you have auto insurance with a deductible of $200, and with no restriction on maximal payment. The probability of a loss is 0.10, and suppose that the distribution of the loss is exponential with a mean of $1000.
What percentage of the loss does the insurance cover on average? The answer in the back of the book is 57.4%, but I am not sure how to reproduce it. This is how I thought of doing the problem:
Say that $1000 = 1 unit of money (for ease of calculation), so that the deductible $d = 0.20$
The payment function may be described as $$r(X) = begincases 0, quad quad x leq 0.20 \ x-0.2, quad x > 0.20 endcases $$
where $x$ is the realization of the stochastic variable $$X = begincases xi, quad p = 0.10 \ 0, quad p = 0.90 endcases $$
Here, $xi$ is the exponentially distributed loss. Since we are taking $1000 to be a single unit of money, this just has a mean of 1.
Since we are interested in the percentage of the loss covered, we are assuming that a loss has occurred and that that loss is larger than the deductible. So, the expected value of the loss under this condition is
$$E[xi | xi > 0.20] = 0.2 + E[xi] = 0.2 + 1 = 1.2$$
which is just a result of the shifted exponential distribution. In this case, the insurance will cover 1 unit of money, or $1000, and so we have 83% coverage on average when there is a loss. (1 divided by 1.2).
What is the mistake?
probability conditional-expectation actuarial-science risk-assessment
asked Jul 20 at 6:45
Marcel
1123
add a comment |Â
up vote
0
down vote
favorite
Suppose you have auto insurance with a deductible of $200, and with no restriction on maximal payment. The probability of a loss is 0.10, and suppose that the distribution of the loss is exponential with a mean of $1000.
What percentage of the loss does the insurance cover on average? The answer in the back of the book is 57.4%, but I am not sure how to reproduce it. This is how I thought of doing the problem:
Say that $1000 = 1 unit of money (for ease of calculation), so that the deductible $d = 0.20$
The payment function may be described as $$r(X) = begincases 0, quad quad x leq 0.20 \ x-0.2, quad x > 0.20 endcases $$
where $x$ is the realization of the stochastic variable $$X = begincases xi, quad p = 0.10 \ 0, quad p = 0.90 endcases $$
Here, $xi$ is the exponentially distributed loss. Since we are taking $1000 to be a single unit of money, this just has a mean of 1.
Since we are interested in the percentage of the loss covered, we are assuming that a loss has occurred and that that loss is larger than the deductible. So, the expected value of the loss under this condition is
$$E[xi | xi > 0.20] = 0.2 + E[xi] = 0.2 + 1 = 1.2$$
which is just a result of the shifted exponential distribution. In this case, the insurance will cover 1 unit of money, or $1000, and so we have 83% coverage on average when there is a loss. (1 divided by 1.2).
What is the mistake?
probability conditional-expectation actuarial-science risk-assessment
asked Jul 20 at 6:45
Marcel
1123
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose you have auto insurance with a deductible of $200, and with no restriction on maximal payment. The probability of a loss is 0.10, and suppose that the distribution of the loss is exponential with a mean of $1000.
What percentage of the loss does the insurance cover on average? The answer in the back of the book is 57.4%, but I am not sure how to reproduce it. This is how I thought of doing the problem:
Say that $1000 = 1 unit of money (for ease of calculation), so that the deductible $d = 0.20$
The payment function may be described as $$r(X) = begincases 0, quad quad x leq 0.20 \ x-0.2, quad x > 0.20 endcases $$
where $x$ is the realization of the stochastic variable $$X = begincases xi, quad p = 0.10 \ 0, quad p = 0.90 endcases $$
Here, $xi$ is the exponentially distributed loss. Since we are taking $1000 to be a single unit of money, this just has a mean of 1.
Since we are interested in the percentage of the loss covered, we are assuming that a loss has occurred and that that loss is larger than the deductible. So, the expected value of the loss under this condition is
$$E[xi | xi > 0.20] = 0.2 + E[xi] = 0.2 + 1 = 1.2$$
which is just a result of the shifted exponential distribution. In this case, the insurance will cover 1 unit of money, or $1000, and so we have 83% coverage on average when there is a loss. (1 divided by 1.2).
What is the mistake?
probability conditional-expectation actuarial-science risk-assessment
asked Jul 20 at 6:45
Marcel
1123
Suppose you have auto insurance with a deductible of $200, and with no restriction on maximal payment. The probability of a loss is 0.10, and suppose that the distribution of the loss is exponential with a mean of $1000.
What percentage of the loss does the insurance cover on average? The answer in the back of the book is 57.4%, but I am not sure how to reproduce it. This is how I thought of doing the problem:
Say that $1000 = 1 unit of money (for ease of calculation), so that the deductible $d = 0.20$
The payment function may be described as $$r(X) = begincases 0, quad quad x leq 0.20 \ x-0.2, quad x > 0.20 endcases $$
where $x$ is the realization of the stochastic variable $$X = begincases xi, quad p = 0.10 \ 0, quad p = 0.90 endcases $$
Here, $xi$ is the exponentially distributed loss. Since we are taking $1000 to be a single unit of money, this just has a mean of 1.
Since we are interested in the percentage of the loss covered, we are assuming that a loss has occurred and that that loss is larger than the deductible. So, the expected value of the loss under this condition is
$$E[xi | xi > 0.20] = 0.2 + E[xi] = 0.2 + 1 = 1.2$$
which is just a result of the shifted exponential distribution. In this case, the insurance will cover 1 unit of money, or $1000, and so we have 83% coverage on average when there is a loss. (1 divided by 1.2).
What is the mistake?
probability conditional-expectation actuarial-science risk-assessment
asked Jul 20 at 6:45
Marcel
1123
asked Jul 20 at 6:45
Marcel
1123
asked Jul 20 at 6:45
Marcel
1123
asked Jul 20 at 6:45
Marcel
1123
1123
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
The error comes from the statement in bold:
Since we are interested in the percentage of the loss covered, we are
assuming that a loss has occurred and that that loss is larger than
the deductible.
This is a mistaken interpretation of the question. Instead, the question is implying that when a loss happens, some of these losses are below the deductible, and therefore not covered. In such a case, the percentage of loss covered is $0%$.
Consequently, the calculation of the expected proportion of loss covered should be $$operatornameEleft[frac(X - 0.2)_+Xright],$$ where $(X - 0.2)_+ = max(0, X - 0.2)$ is the claim payment and $$X sim operatornameExponential(1)$$ is the ground-up loss given that a loss occurs. Note we do not need $xi$ nor do we need the information that a loss occurs with probability $0.1$, because the first part of your statement "we are assuming a loss has occurred" is in fact correct.
Then the calculation proceeds as
$$beginalign*
operatornameEleft[frac(X - 0.2)_+Xright]
&= operatornameE[0 mid X le 0.2]Pr[X le 0.2] + operatornameEleft[1 - frac15X ,bigg| ; X > 0.2 right]Pr[X > 0.2] \
&= e^-1/5 - int_x=0.2^infty frac15x e^-x , dx \
&approx 0.574201.
endalign*$$
answered Jul 20 at 7:12
heropup
59.7k65895
add a comment |Â
up vote
0
down vote
I don't see a basis in the problem statement for conditioning on $xigt0.20$. Quite to the contrary, it seems to me that if you want to know the average percentage of the loss covered by your insurance, the fact that it covers $0%$ if the loss is below the deductible seems like a very relevant piece of information to include in the calculation.
Without the conditioning, you get
$$E[r(X)]=mathrm e^-0.2cdot1=mathrm e^-0.2approx81.9%;,$$
slightly less than your value under conditioning. This, however, is not the average percentage of loss covered; it's the average loss covered, expressed as a percentage of the average loss incurred. The average percentage of loss covered is
$$
left(1-mathrm e^-0.2right)cdot0+mathrm e^-0.2int_0^inftyfrac xx+0.2,mathrm e^-xmathrm dx=0.2operatornameEi(-0.2)+mathrm e^-0.2approx57.4%;,
$$
where $operatornameEi$ denotes the exponential integral.
answered Jul 20 at 7:02
joriki
164k10180328
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The error comes from the statement in bold:
Since we are interested in the percentage of the loss covered, we are
assuming that a loss has occurred and that that loss is larger than
the deductible.
This is a mistaken interpretation of the question. Instead, the question is implying that when a loss happens, some of these losses are below the deductible, and therefore not covered. In such a case, the percentage of loss covered is $0%$.
Consequently, the calculation of the expected proportion of loss covered should be $$operatornameEleft[frac(X - 0.2)_+Xright],$$ where $(X - 0.2)_+ = max(0, X - 0.2)$ is the claim payment and $$X sim operatornameExponential(1)$$ is the ground-up loss given that a loss occurs. Note we do not need $xi$ nor do we need the information that a loss occurs with probability $0.1$, because the first part of your statement "we are assuming a loss has occurred" is in fact correct.
Then the calculation proceeds as
$$beginalign*
operatornameEleft[frac(X - 0.2)_+Xright]
&= operatornameE[0 mid X le 0.2]Pr[X le 0.2] + operatornameEleft[1 - frac15X ,bigg| ; X > 0.2 right]Pr[X > 0.2] \
&= e^-1/5 - int_x=0.2^infty frac15x e^-x , dx \
&approx 0.574201.
endalign*$$
answered Jul 20 at 7:12
heropup
59.7k65895
add a comment |Â
up vote
0
down vote
accepted
The error comes from the statement in bold:
Since we are interested in the percentage of the loss covered, we are
assuming that a loss has occurred and that that loss is larger than
the deductible.
This is a mistaken interpretation of the question. Instead, the question is implying that when a loss happens, some of these losses are below the deductible, and therefore not covered. In such a case, the percentage of loss covered is $0%$.
Consequently, the calculation of the expected proportion of loss covered should be $$operatornameEleft[frac(X - 0.2)_+Xright],$$ where $(X - 0.2)_+ = max(0, X - 0.2)$ is the claim payment and $$X sim operatornameExponential(1)$$ is the ground-up loss given that a loss occurs. Note we do not need $xi$ nor do we need the information that a loss occurs with probability $0.1$, because the first part of your statement "we are assuming a loss has occurred" is in fact correct.
Then the calculation proceeds as
$$beginalign*
operatornameEleft[frac(X - 0.2)_+Xright]
&= operatornameE[0 mid X le 0.2]Pr[X le 0.2] + operatornameEleft[1 - frac15X ,bigg| ; X > 0.2 right]Pr[X > 0.2] \
&= e^-1/5 - int_x=0.2^infty frac15x e^-x , dx \
&approx 0.574201.
endalign*$$
answered Jul 20 at 7:12
heropup
59.7k65895
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The error comes from the statement in bold:
Since we are interested in the percentage of the loss covered, we are
assuming that a loss has occurred and that that loss is larger than
the deductible.
This is a mistaken interpretation of the question. Instead, the question is implying that when a loss happens, some of these losses are below the deductible, and therefore not covered. In such a case, the percentage of loss covered is $0%$.
Consequently, the calculation of the expected proportion of loss covered should be $$operatornameEleft[frac(X - 0.2)_+Xright],$$ where $(X - 0.2)_+ = max(0, X - 0.2)$ is the claim payment and $$X sim operatornameExponential(1)$$ is the ground-up loss given that a loss occurs. Note we do not need $xi$ nor do we need the information that a loss occurs with probability $0.1$, because the first part of your statement "we are assuming a loss has occurred" is in fact correct.
Then the calculation proceeds as
$$beginalign*
operatornameEleft[frac(X - 0.2)_+Xright]
&= operatornameE[0 mid X le 0.2]Pr[X le 0.2] + operatornameEleft[1 - frac15X ,bigg| ; X > 0.2 right]Pr[X > 0.2] \
&= e^-1/5 - int_x=0.2^infty frac15x e^-x , dx \
&approx 0.574201.
endalign*$$
answered Jul 20 at 7:12
heropup
59.7k65895
The error comes from the statement in bold:
Since we are interested in the percentage of the loss covered, we are
assuming that a loss has occurred and that that loss is larger than
the deductible.
This is a mistaken interpretation of the question. Instead, the question is implying that when a loss happens, some of these losses are below the deductible, and therefore not covered. In such a case, the percentage of loss covered is $0%$.
Consequently, the calculation of the expected proportion of loss covered should be $$operatornameEleft[frac(X - 0.2)_+Xright],$$ where $(X - 0.2)_+ = max(0, X - 0.2)$ is the claim payment and $$X sim operatornameExponential(1)$$ is the ground-up loss given that a loss occurs. Note we do not need $xi$ nor do we need the information that a loss occurs with probability $0.1$, because the first part of your statement "we are assuming a loss has occurred" is in fact correct.
Then the calculation proceeds as
$$beginalign*
operatornameEleft[frac(X - 0.2)_+Xright]
&= operatornameE[0 mid X le 0.2]Pr[X le 0.2] + operatornameEleft[1 - frac15X ,bigg| ; X > 0.2 right]Pr[X > 0.2] \
&= e^-1/5 - int_x=0.2^infty frac15x e^-x , dx \
&approx 0.574201.
endalign*$$
answered Jul 20 at 7:12
heropup
59.7k65895
answered Jul 20 at 7:12
heropup
59.7k65895
answered Jul 20 at 7:12
heropup
59.7k65895
answered Jul 20 at 7:12
heropup
59.7k65895
59.7k65895
add a comment |Â
add a comment |Â
up vote
0
down vote
I don't see a basis in the problem statement for conditioning on $xigt0.20$. Quite to the contrary, it seems to me that if you want to know the average percentage of the loss covered by your insurance, the fact that it covers $0%$ if the loss is below the deductible seems like a very relevant piece of information to include in the calculation.
Without the conditioning, you get
$$E[r(X)]=mathrm e^-0.2cdot1=mathrm e^-0.2approx81.9%;,$$
slightly less than your value under conditioning. This, however, is not the average percentage of loss covered; it's the average loss covered, expressed as a percentage of the average loss incurred. The average percentage of loss covered is
$$
left(1-mathrm e^-0.2right)cdot0+mathrm e^-0.2int_0^inftyfrac xx+0.2,mathrm e^-xmathrm dx=0.2operatornameEi(-0.2)+mathrm e^-0.2approx57.4%;,
$$
where $operatornameEi$ denotes the exponential integral.
answered Jul 20 at 7:02
joriki
164k10180328
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
add a comment |Â
up vote
0
down vote
I don't see a basis in the problem statement for conditioning on $xigt0.20$. Quite to the contrary, it seems to me that if you want to know the average percentage of the loss covered by your insurance, the fact that it covers $0%$ if the loss is below the deductible seems like a very relevant piece of information to include in the calculation.
Without the conditioning, you get
$$E[r(X)]=mathrm e^-0.2cdot1=mathrm e^-0.2approx81.9%;,$$
slightly less than your value under conditioning. This, however, is not the average percentage of loss covered; it's the average loss covered, expressed as a percentage of the average loss incurred. The average percentage of loss covered is
$$
left(1-mathrm e^-0.2right)cdot0+mathrm e^-0.2int_0^inftyfrac xx+0.2,mathrm e^-xmathrm dx=0.2operatornameEi(-0.2)+mathrm e^-0.2approx57.4%;,
$$
where $operatornameEi$ denotes the exponential integral.
answered Jul 20 at 7:02
joriki
164k10180328
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I don't see a basis in the problem statement for conditioning on $xigt0.20$. Quite to the contrary, it seems to me that if you want to know the average percentage of the loss covered by your insurance, the fact that it covers $0%$ if the loss is below the deductible seems like a very relevant piece of information to include in the calculation.
Without the conditioning, you get
$$E[r(X)]=mathrm e^-0.2cdot1=mathrm e^-0.2approx81.9%;,$$
slightly less than your value under conditioning. This, however, is not the average percentage of loss covered; it's the average loss covered, expressed as a percentage of the average loss incurred. The average percentage of loss covered is
$$
left(1-mathrm e^-0.2right)cdot0+mathrm e^-0.2int_0^inftyfrac xx+0.2,mathrm e^-xmathrm dx=0.2operatornameEi(-0.2)+mathrm e^-0.2approx57.4%;,
$$
where $operatornameEi$ denotes the exponential integral.
answered Jul 20 at 7:02
joriki
164k10180328
I don't see a basis in the problem statement for conditioning on $xigt0.20$. Quite to the contrary, it seems to me that if you want to know the average percentage of the loss covered by your insurance, the fact that it covers $0%$ if the loss is below the deductible seems like a very relevant piece of information to include in the calculation.
Without the conditioning, you get
$$E[r(X)]=mathrm e^-0.2cdot1=mathrm e^-0.2approx81.9%;,$$
slightly less than your value under conditioning. This, however, is not the average percentage of loss covered; it's the average loss covered, expressed as a percentage of the average loss incurred. The average percentage of loss covered is
$$
left(1-mathrm e^-0.2right)cdot0+mathrm e^-0.2int_0^inftyfrac xx+0.2,mathrm e^-xmathrm dx=0.2operatornameEi(-0.2)+mathrm e^-0.2approx57.4%;,
$$
where $operatornameEi$ denotes the exponential integral.
answered Jul 20 at 7:02
joriki
164k10180328
answered Jul 20 at 7:02
joriki
164k10180328
answered Jul 20 at 7:02
joriki
164k10180328
answered Jul 20 at 7:02
joriki
164k10180328
164k10180328
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
add a comment |Â
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
Thanks for the answer, I was somehow only thinking a "loss" had occurred if the insurance company was required to pay for it. I suppose I was not thinking from the perspective of the person who suffered the loss! It is not obvious to me how you obtained your last series of equations. Is "average percentage of loss covered" a standard formula?
– Marcel
Jul 20 at 7:21
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
@Marcel: A better term would actually have been "expected percentage of loss". It's just the expected value of the percentage of loss. The percentage of loss is $0$ with probability $1-mathrm e^-0.2$ (loss below the deductible) and $x/(x+0.2)$ if the insurance pays $x$ for a loss of $x+0.2$, with probability density $mathrm e^-(x+0.2)$. I got the exponential integral from Wolfram|Alpha; you can get it by transforming to $u=x+0.2$ and decomposing the fraction into $1+0.2/u$ (I should have written it in terms of the loss $u$ instead of the coverage $x$ in the first place.)
– joriki
Jul 20 at 7:28
add a comment |Â
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