1. Introduction
In reinsurance industry, simulated losses from catastrophe events combined with reinsurance contract financial terms are used to calculate the contract expected annual loss, the standard deviation of the expected annual loss, and quantiles of the losses (such as the AEP: Aggregate Exceedance Probability, or TVaR: Tail Value at Risk, in Ref. [1]). These numbers are in turn used for pricing or risk management of the contract. There are two kinds of model risks in this approach: independent simulations may give different results without any model change, and simulations vary before or after model change such as from the yearly catastrophe events sets or parameters updates. Empirically, the distribution of the contract expected annual loss may be more like a Beckmann, MaxStable, Gamma, Inverse Gaussian, or even a Lognormal Distribution than a Normal Distribution, but consider that the mean annual loss is the average of a large number of losses, and especially for simplicity, if we assume they obey a Normal Distribution, to quantify the model risk, we can use Hinkley formula [2] or Marsaglia formula [3] to calculate the probability of two simulations have expected annual loss deviated from each other by more than say 50%. Their formulas, in our model risk quantification context and the simplest scenario, depend on only two factors: the correlation coefficient
2. Results
2.1. CV range
Starting from Hölder’s inequality (https://en.wikipedia.org/wiki/Hölder's_inequality):
${\Vert \mathit{fg}\Vert}_{1}\le {\Vert f\Vert}_{p}{\Vert g\Vert}_{q},$E1
suppose
$\int \mid f{m}_{f}\mid \mathit{gd\mu}\le {\left[\int {\leftf{m}_{f}\right}^{2}\mathit{d\mu}\right]}^{1/2}{\left[\int {g}^{2}\mathit{d\mu}\right]}^{1/2}={\sigma}_{f}{\left[\int {g}^{2}\mathit{d\mu}\right]}^{1/2}$E2
Suppose
${\sigma}_{f}\ge {m}_{f}\frac{\mathit{a\mu}\left({A}^{C}\right)+\mathit{b\mu}\left({A}^{C}\right)}{\sqrt{{a}^{2}\mu \left(A\right)+{b}^{2}\mu \left({A}^{C}\right)}}$E3
due to
$\mid f{m}_{f}\mid \ge f{m}_{f}$E4
and
$\int \mid f{m}_{f}\mid \mathit{gd\mu}={\int}_{A}\mid f{m}_{f}\mid \mathit{ad\mu}+{\int}_{{A}^{C}}{m}_{f}\mathit{bd\mu}$E5
$\u2a7e{\int}_{A}\mathit{\text{fad\mu}}{\int}_{A}{m}_{f}\mathit{ad\mu}+{m}_{f}\mathit{b\mu}\left({{\rm A}}^{C}\right)$E6
$={m}_{f}a{m}_{f}\mathit{a\mu}\left(A\right)+{m}_{f}\mathit{b\mu}\left({A}^{C}\right)$E7
The maximum of the righthand side of Eq. (3) is achieved by $\frac{a}{b}=\frac{\mu \left({A}^{C}\right)}{\mu \left(A\right)}$, increasing on [0, $\frac{\mu \left({A}^{C}\right)}{\mu \left(A\right)}$] and decreasing on $\left[\frac{\mu \left({A}^{C}\right)}{\mu \left(A\right)},\infty \right)$, and then we get:
${\sigma}_{f}\ge {m}_{f}\sqrt{\frac{\mu \left({A}^{C}\right)}{\mu \left(A\right)}}$E8
As a corollary, if we use the 2sigma or 3sigma rule for the confidence interval of the estimate of the mean, for those contracts that have most years 0 losses, these interval will be very large. For example, if we have only 100years have nonzero losses out of the 100,000 simulated years, then we will get [math]::sqrt((1e5100)/100)=31.6069612585582:
${\mathit{\sigma}}_{f}\ge 31.6069612585582{\mathit{m}}_{f}$E9
And if we have 1000years nonzero losses, we get the constant [math]::sqrt((1e51000)/1000)=9.9498743710662:
${\mathit{\sigma}}_{f}\ge 9.9498743710662{\mathit{m}}_{f}$E10
Checking against some concrete examples, for a contract we see 2220 nonzero losses records, suppose all of them are in different years, then we get the ratio [math]:: sqrt((1e52220)/2220)=6.63664411016931, and we have its mean loss
Another contract have unique 41,143 nonzero losses years out of the 100,000 simulated years, with mean loss
From these examples, we see that our lower bound formula for
Typically, more than half of the contracts may have nonzero losses years count below 10,000, for which we can see their
Nonzero losses years  

1  316.226184874055 
10  99.9949998749938 
100  31.6069612585582 
1000  9.9498743710662 
2000  7 
5000  4.35889894354067 
10,000  3 
20,000  2 
30,000  1.52752523165195 
40,000  1.22474487139159 
50,000  1 
80,000  0.5 
90,000  0.333333333333333 
99,000  0.100503781525921 
99,900  0.0316385998584166 
99,990  0.0100005000375031 
99,999  0.00316229347167527 
Table 1.
For given years of nonzero losses out of 100,000 simulated years, the lower bound as given by our formula Eq. (8).
This inequality Eq. (8) can explain the observation that when we sort the contracts by the mean annual loss, the lower quarter of contracts may have more than tens of percent deviation from different simulations, since smaller mean loss usually corresponding to fewer years of nonzero losses and higher
To get an upper bound for
$\mathit{CV}\equiv \frac{{\sigma}_{f}}{{m}_{f}}=\sqrt{n\frac{{\sum}_{i=1}^{n}{x}_{i}^{2}}{{\left({\sum}_{i=1}^{n}{x}_{i}\right)}^{2}}1}$E11
For the expression $\frac{{\sum}_{i=1}^{n}{x}_{i}^{2}}{{\left({\sum}_{i=1}^{n}{x}_{i}\right)}^{2}}$, by taking the partial derivative with respect to
$\mathit{CV}\le \sqrt{n1}$E12
For the extreme case when only one year have nonzero losses, we thus verified that the coefficient [math]::sqrt((1e51)/1)=316.226184874055 is exact, that is, we have:
${\sigma}_{f}=316.226184874055{m}_{f}$E13
The overall upper bound $\sqrt{n1}$ for
From the fact that the minimum of the expression in Eq. (11) is attained at $\frac{{\sum}_{i=2}^{n}{x}_{i}^{2}}{{\sum}_{i=2}^{n}{x}_{i}}$ . we can see that:
$\frac{{\sum}_{i=1}^{n}{x}_{i}^{2}}{{\left({\sum}_{i=1}^{n}{x}_{i}\right)}^{2}}\ge \frac{\frac{{\sum}_{i=2}^{n}{x}_{i}^{2}}{{\left({\sum}_{i=2}^{n}{x}_{i}\right)}^{2}}}{\frac{{\sum}_{i=2}^{n}{x}_{i}^{2}}{{\left({\sum}_{i=2}^{n}{x}_{i}\right)}^{2}}+1}$E14
More generally, if year
$\frac{{\sum}_{i=1}^{n}{x}_{i}^{2}{p}_{i}}{{\left({\sum}_{i=1}^{n}{x}_{i}{p}_{i}\right)}^{2}}\ge \frac{\frac{{\sum}_{i=2}^{n}{x}_{i}^{2}{p}_{i}}{{\left({\sum}_{i=2}^{n}{x}_{i}{p}_{i}\right)}^{2}}}{{p}_{1}\frac{{\sum}_{i=2}^{n}{x}_{i}^{2}{p}_{i}}{{\left({\sum}_{i=2}^{n}{x}_{i}{p}_{i}\right)}^{2}}+1}$E15
The minimum value of the right side is attained when ${x}_{1}=\frac{{\sum}_{i=2}^{n}{x}_{i}^{2}{p}_{i}}{{\sum}_{i=2}^{n}{x}_{i}{p}_{i}}$. This can be used inductively to give an “elementary” proof of our lower bound results, and additionally can show that the lower bound is attained when all the nonzero losses are equally valued which using the Hölder’s inequality cannot arrive. Similarly we can show that the upper bound is $\sqrt{\mathit{max}\left(\frac{1}{{p}_{1}},\frac{1}{{p}_{2}},\dots ,\frac{1}{{p}_{n}}\right)1}$.
We summarize our deduction and discussion into the following:
${\sigma}_{f}\ge {m}_{f}\sqrt{\frac{\mu \left({A}^{C}\right)}{\mu \left(A\right)}}$E16
${\sigma}_{f}={m}_{f}\phantom{\rule{0ex}{0ex}}\sqrt{\frac{\mathrm{\mu}\left({\mathrm{A}}^{C}\right)}{\mathrm{\mu}\left(\mathrm{A}\right)}},$E17
${\sigma}_{f}\le {m}_{f}\sqrt{\mathit{max}\left\{\frac{1}{{p}_{1}},\frac{1}{{p}_{2}},\dots ,\frac{1}{{p}_{n}}\right\}1}$E18
where the p_{i} is the probability of occurrence of year i. The upper bound is attained if and only if the smallest occurrence probability year is the only year of nonzero losses. And when only year i have nonzero losses:
${\mathrm{\sigma}}_{\mathrm{f}}={\mathrm{m}}_{\mathrm{f}}\phantom{\rule{0ex}{0ex}}\sqrt{\frac{1}{{\mathrm{p}}_{\mathrm{i}}}1}$E19
For not necessarily nonnegative loss contracts (such as contracts with complex layers structure and hedging design), and for contracts that have significant concentration on the upper bound (due to limit and annual limit), replacing
${\sigma}_{f}\ge \left({m}_{f}m\right)\phantom{\rule{0ex}{0ex}}\sqrt{\frac{\mathrm{\mu}\left({L}^{C}\right)}{\mathrm{\mu}\left(L\right)}}$E20
$\phantom{\rule{0ex}{0ex}}{\sigma}_{f}\ge \left(M{m}_{f}\right)\phantom{\rule{0ex}{0ex}}\sqrt{\frac{\mathrm{\mu}\left({U}^{C}\right)}{\mathrm{\mu}\left(U\right)}}$E21
From Theorem 1, we can get an upper bound for the average annual loss on an arbitrary subset of the years:
$\frac{{\int}_{B}\mathit{fd\mu}}{\mu \left(B\right)}\le {\sigma}_{f}\sqrt{\frac{\mu \left({B}^{C}\right)}{\mu \left(B\right)}}+{m}_{f}.$E22
${\sigma}_{f}^{2}=E\left[{\left({f}_{1}+{f}_{2}\right)}^{2}\right]{\left[E\left({f}_{1}+{f}_{2}\right)\right]}^{2}$E23
$={\sigma}_{{f}_{1}}^{2}+{\sigma}_{{f}_{2}}^{2}2{m}_{{f}_{1}}{m}_{{f}_{2}}$E24
$\ge {m}_{{f}_{1}}^{2}\frac{\mu \left({B}^{C}\right)}{\mu \left(B\right)}+{m}_{{f}_{2}}^{2}\frac{\mu \left(B\right)}{\mu \left({B}^{C}\right)}2{m}_{{f}_{1}}{m}_{{f}_{2}}$E25
$={\left({m}_{{f}_{1}}\sqrt{\frac{\mu \left({B}^{C}\right)}{\mu \left(B\right)}}{m}_{{f}_{2}}\sqrt{\frac{\mu \left(B\right)}{\mu \left({B}^{C}\right)}}\right)}^{2}$E26
from Theorem 1 and the fact that the domain with zero value for
${m}_{{f}_{1}}\sqrt{\frac{\mu \left({B}^{C}\right)}{\mu \left(B\right)}}\le {m}_{{f}_{2}}\sqrt{\frac{\mu \left(B\right)}{\mu \left({B}^{C}\right)}}+{\sigma}_{f}$E27
The inequality Eq. (22) is arrived by the fact that
$\sqrt{\mu \left(B\right)}\sqrt{\mu \left({B}^{C}\right)}\left(\sqrt{\frac{\mu \left({B}^{C}\right)}{\mu \left(B\right)}}+\sqrt{\frac{\mu \left(B\right)}{\mu \left({B}^{C}\right)}}\right)=\mu \left({B}^{C}\right)+\mu \left(B\right)=1$E28
□
If we let the subset
Return period  Quantile  

100,000  0.99999  316.226184874055 
10,000  0.9999  99.9949998749938 
5000  0.9998  70.7036066972541 
1000  0.999  31.6069612585582 
500  0.998  22.3383079036887 
250  0.996  15.7797338380595 
200  0.995  14.1067359796659 
100  0.99  9.9498743710662 
50  0.98  7 
30  0.966666666666667  5.3851648071345 
25  0.96  4.89897948556636 
20  0.95  4.35889894354067 
10  0.9  3 
5  0.8  2 
4  0.75  1.73205080756888 
2  0.5  1 
Table 2.
For given year of return period, the upper bound as given by our formula Eq. (22).
Numerical example shows that our
Notice that the measure of the numbers of nonzero losses years is also called the probability of attaching in insurance, we can rearrange the terms in the formula Eq. (8) to get a lower bound for the probability of attaching:
$\mathit{\text{ProbA}}\ge \frac{1}{C{V}^{2}+1}$E29
As an application of Corollary 3, we see that if
Beginning quantile with nonzero loss  Return period $\left(\equiv \frac{1}{\mathit{\text{ProbA}}}\right)$  

99.9949998749938  0.0001  0.9999  10,000 
9.9498743710662  0.01  0.99  100 
3  0.1  0.9  10 
2  0.2  0.8  5 
1.73205080756888  0.25  0.75  4 
1  0.5  0.5  2 
0.577350269189626  0.75  0.25  1.33333333333333 
0.5  0.8  0.2  1.25 
Table 3.
For given range of
Similarly, from Corollary 2, we can easily rearrange terms to get an upper bound for the probability of exceeding a given loss, which is called the Cantelli’s inequality in the literature (https://en.wikipedia.org/wiki/ Chebyshev's_inequality):
$\mathit{\text{ProbE}}\le \frac{1}{\frac{{\left(x{m}_{f}\right)}^{2}}{{\sigma}_{f}^{2}}+1}$E30
$\mathit{\text{Prob}}\left\{f\ge \frac{{\sigma}_{f}^{2}}{{m}_{f}}+{m}_{f}\right\}\le \frac{1}{C{V}^{2}+1}$E31
This bound gives a limitation on simulation with a given number
$M\ge \frac{{\sigma}_{f}^{2}}{{m}_{f}}+{m}_{f}$E32
${\sigma}_{f}\le \sqrt{\left(M{m}_{f}\right){m}_{f}}$E33
Proof:
Let $g=\frac{f}{M}$, then
${g}^{2}\le g$E34
and
${\int}_{\mathrm{\Omega}}{g}^{2}\mathit{d\mu}\le {\int}_{\mathrm{\Omega}}\mathit{gd\mu}$E35
${\int}_{\mathrm{\Omega}}{g}^{2}\mathit{d\mu}{\left({\int}_{\mathrm{\Omega}}\mathit{gd\mu}\right)}^{2}\le {\int}_{\mathrm{\Omega}}\mathit{gd\mu}{\left({\int}_{\mathrm{\Omega}}\mathit{gd\mu}\right)}^{2}$E36
${\sigma}_{g}^{2}\le {m}_{g}\left(1{m}_{g}\right)$E37
$\frac{{\sigma}_{f}^{2}}{{M}^{2}}\le \frac{{m}_{f}}{M}\left(1\frac{{m}_{f}}{M}\right)$E38
This proves both of our inequalities. Without loss of generality, suppose any nonempty subset of Ω have nonzero measure, the equality hold in Eq. (34) and its subsequent inequalities if and only if
Because of the probability of $\frac{{m}_{f}}{M}$ of taking value
By examining the proof of Corollary 2 and Theorem 1Eq. (17), forcing the inequality in Eq. (25) to be an equality, we can prove that:
$\frac{{\int}_{B}\mathit{fd\mu}}{\mu \left(B\right)}={\sigma}_{f}\sqrt{\frac{\mu \left({B}^{C}\right)}{\mu \left(B\right)}}+{m}_{f}$E39
if and only if f is a nonzero constant function on the subset B and a constant function on the subset B^{C}.
So the maximum
2.2. Simulation deviation
Typical correlation coefficient
The Mathematica code for the plot in Figures 1 and 2 is in Appendix A. The two plots are identical even though their formulas are quite different and we do not know whether they can be analytically proved to be equivalent: our plots are numerical validation of both of their formulas.
3. Discussion
The lower bound for
As the starting point and the application of the
The less general upper bound $\sqrt{n1}$ where all probabilities are equal is obtained by Katsnelson and Kotz in the literature [4, 5].
Using the same Hölder’s inequality and calculus technique which may not have a simple elementary inequality approach counterpart, we can prove a more complex formula:
${\sigma}_{f}\ge \sqrt{{\left({m}_{f}m\right)}^{2}\mu \left(m\right)+{\left(M{m}_{f}\right)}^{2}\mu \left(M\right)+\frac{{\left[\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)\mu \left(M\right)\right]}^{2}}{1\mu \left(m\right)\mu \left(M\right)}}$E40
Proof:
In the inequality Eq. (2), we divide Ω into three subset and let the nonnegative function
${\left.g\right}_{\left\{f=m\right\}}=a,g{\left{}_{\left\{m<f<M\right\}}=b,g\right}_{\left\{f=M\right\}}=c$E41
Then:
$\int {g}^{2}\mathit{d\mu}={a}^{2}\mu \left(m\right)+{b}^{2}\left(1\mu \left(m\right)\mu \left(M\right)\right)+{c}^{2}\mu \left(M\right)$E42
$\int \mid f{m}_{f}\mid \mathit{gd\mu}={\int}_{\left\{f=m\right\}}\mid f{m}_{f}\mid \mathit{ad\mu}+{\int}_{\left\{m<f<M\right\}}\mid f{m}_{f}\mid \mathit{bd\mu}+{\int}_{\left\{f=M\right\}}\mid f{m}_{f}\mid \mathit{cd\mu}$E43
$\ge \left({m}_{f}m\right)\mathit{a\mu}\left(m\right)+{\int}_{\left\{m<f<M\right\}}\mathit{\text{fbd\mu}}{\int}_{\left\{m<f<M\right\}}{m}_{f}\mathit{bd\mu}+\left(M{m}_{f}\right)\mathit{c\mu}\left(M\right)$E44
$=\left({m}_{f}m\right)\mathit{a\mu}\left(m\right)+{m}_{f}b\mathit{mb\mu}\left(m\right)\mathit{Mb\mu}\left(M\right){m}_{f}b\left(1\mu \left(m\right)\mu \left(M\right)\right)+\left(M{m}_{f}\right)\mathit{c\mu}\left(M\right)$E45
$=\left({m}_{f}m\right)\left(a+b\right)\mu \left(m\right)+\left(M{m}_{f}\right)\left(cb\right)\mu \left(M\right)$E46
due to $\mid f{m}_{f}\mid \ge f{m}_{f}$.
We get:
${\sigma}_{f}\ge \frac{\left({m}_{f}m\right)\left(a+b\right)\mu \left(m\right)+\left(M{m}_{f}\right)\left(cb\right)\mu \left(M\right)}{\sqrt{\left({a}^{2}{b}^{2}\right)\mu \left(m\right)+\left({c}^{2}{b}^{2}\right)\mu \left(M\right)+{b}^{2}}}$E47
$=\frac{\left({m}_{f}m\right)\left(\frac{a}{b}\right)\mu \left(m\right)+\left(M{m}_{f}\right)\left(\frac{c}{b}\right)\mu \left(M\right)+\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)\mu \left(M\right)}{\sqrt{{\left(\frac{a}{b}\right)}^{2}\mu \left(m\right)+{\left(\frac{c}{b}\right)}^{2}\mu \left(M\right)+1\mu \left(m\right)\mu \left(M\right)}}$E48
suppose $b>0$.
Using the negative form of the inequality $\leftf{m}_{f}\right\ge {m}_{f}f$, we also get a dual form inequality:
${\sigma}_{f}\ge \frac{\left({m}_{f}m\right)\left(ab\right)\mu \left(m\right)+\left(M{m}_{f}\right)\left(c+b\right)\mu \left(M\right)}{\sqrt{\left({a}^{2}{b}^{2}\right)\mu \left(m\right)+\left({c}^{2}{b}^{2}\right)\mu \left(M\right)+{b}^{2}}}$E49
$=\frac{\left({m}_{f}m\right)\left(\frac{a}{b}\right)\mu \left(m\right)+\left(M{m}_{f}\right)\left(\frac{c}{b}\right)\mu \left(M\right)\left({m}_{f}m\right)\mu \left(m\right)+\left(M{m}_{f}\right)\mu \left(M\right)}{\sqrt{{\left(\frac{a}{b}\right)}^{2}\mu \left(m\right)+{\left(\frac{c}{b}\right)}^{2}\mu \left(M\right)+1\mu \left(m\right)\mu \left(M\right)}}$E50
suppose $b>0$.
Define:
$t=\frac{a}{b},A=\left({m}_{f}m\right)\mu \left(m\right),B=\left(M{m}_{f}\right)\left(\frac{c}{b}\right)\mu \left(M\right)+\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)\mu \left(M\right)$E51
$C=\mu \left(m\right),D={\left(\frac{c}{b}\right)}^{2}\mu \left(M\right)+1\mu \left(m\right)\mu \left(M\right)$E52
$F\left(t\right)=\frac{\mathit{At}+B}{\sqrt{C{t}^{2}+D}}$E53
Then
$A\ge 0,C\ge 0,D>0$E54
The derivative
${F}^{\prime}\left(t\right)=\frac{\mathit{AD}\mathit{BCt}}{{\left(D+C{t}^{2}\right)}^{\frac{3}{2}}}$E55
If
Apply the same argument to
$\frac{B}{\sqrt{D}}=\frac{\left(M{m}_{f}\right)\left(\frac{c}{b}\right)\mu \left(M\right)+\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)\mu \left(M\right)}{\sqrt{{\left(\frac{c}{b}\right)}^{2}\mu \left(M\right)+1\mu \left(m\right)\mu \left(M\right)}}$E56
with (
If (
If (
$\sqrt{\frac{{A}^{2}}{C}+\frac{{B}^{2}}{D}}=\sqrt{{\left({m}_{f}m\right)}^{2}\mu \left(m\right)+{\left(M{m}_{f}\right)}^{2}\mu \left(M\right)+\frac{{\left[\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)\mu \left(M\right)\right]}^{2}}{1\mu \left(m\right)\mu \left(M\right)}}$E57
with the specific choice of $\frac{c}{b}\phantom{\rule{0ex}{0ex}}$and $\frac{a}{b}$.□
We can also prove by calculus that:
$\sqrt{{\left({m}_{f}m\right)}^{2}\mu \left(m\right)+{\left(M{m}_{f}\right)}^{2}\mu \left(M\right)+\frac{{\left[\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)\mu \left(M\right)\right]}^{2}}{1\mu \left(m\right)\mu \left(M\right)}}\ge {m}_{f}\sqrt{\frac{\mu \left(m\right)}{1\mu \left(m\right)}}$E58
Proof:
Define
$F\left(t\right)={\left({m}_{f}m\right)}^{2}\mu \left(m\right)+{\left(M{m}_{f}\right)}^{2}t+\frac{{\left[\left({m}_{f}m\right)\mu \left(m\right)\left(M{m}_{f}\right)t\right]}^{2}}{1\mu \left(m\right)t}{m}_{f}^{2}\frac{\mu \left(m\right)}{1\mu \left(m\right)}$E59
for
Then
$F\left(0\right)={\left({m}_{f}m\right)}^{2}\mu \left(m\right)+\frac{{\left[\left({m}_{f}m\right)\mu \left(m\right)\right]}^{2}}{1\mu \left(m\right)}{m}_{f}^{2}\frac{\mu \left(m\right)}{1\mu \left(m\right)}$E60
$=\frac{{\left({m}_{f}m\right)}^{2}{m}_{f}^{2}}{1\mu \left(m\right)}\mu \left(m\right)=0$E61
The derivative of
${F}^{\prime}\left(t\right)=\frac{{\left({m}_{f}M+\left(Mm\right)\mu \left(m\right)\right)}^{2}}{{\left(1+t+\mu \left(m\right)\right)}^{2}}$E62
which is always nonnegative, so
Theorem 3 can be combined with the following form of the Hölder’s inequality:
${\left(\underset{\left\{m<f<M\right\}}{\int}f{m}_{f}1\mathrm{d}\mu \right)}^{2}\le \underset{\left\{m<f<M\right\}}{\int}{\leftf{m}_{f}\right}^{2}{d}\mu \underset{\left\{m<f<M\right\}}{\int}{1}^{2}{d}\mu $E63
to give an alternative proof of Theorem 2 and then Eq. (16) (or directly for Eq. (20) by using the set {
So there is a complex but better lower bounds Eq. (40), and empirical study shows that when
With numerical simulation, we can get
4. Conclusions
Lower bound for reinsurance contract annual loss standard deviation involving zero losses years counts are obtained, which imply a general upper bound for annual loss
Thanks
The author thanks Nancy Wang for checking against a C++ application that validated the practical usefulness of our inequality.
hinkley[x_?NumericQ,c_,p_]:=1/Sqrt[2 Pi]/c (x+1) (1−p)/(xˆ2–2 p x+1)ˆ1.5 Exp[−1/2/cˆ2 (x−1)ˆ2/(xˆ2–2 p x+1)](CDF[NormalDistribution[0,1],1/c Sqrt[(1−p)/(1+p)] (x+1)/Sqrt[xˆ2–2 p x+1]]−CDF[NormalDistribution[0,1],−1/c Sqrt[(1−p)/(1+p)] (x+1)/Sqrt[xˆ2–2 p x+1]])+1/Pi Sqrt[1−pˆ2]/(xˆ2–2 p x+1) Exp[−1/cˆ2/(1+p)];
H[x_{_},c_{_},p_{_}]:=
LogLinearPlot[
q[t_{_},p_{_},c_{_}]:=q[t,p,c]=
marsaglia[t_{_},p_{_},c_{_}]:=marsaglia[t,p,c]=
M[v_{_},p_{_},c_{_}]:=M[v,p,c]=
NormalDistribution[0,1],1/c]−2 CDF[NormalDistribution[0,1],1/c (1−p)/
NormalDistribution[0,1],1/c]+
MA[v_{_},p_{_},c_{_}]:=MA[v,p,c]=
DistributeDefinitions [q,marsaglia,M,MA];
LogLinearPlot[
I have a deep understanding of the content you provided, particularly in the field of reinsurance and risk management. The article delves into the calculation of standard deviation and quantiles for reinsurance contracts, considering factors like model risk and distribution types. The concepts of CV (coefficient of variation), probability of attaching, and simulation deviation are explored.
Now, let's break down the key concepts mentioned in the article:

Reinsurance Industry and Model Risks:
 Simulated losses from catastrophe events and reinsurance contract terms are used for risk assessment.
 Model risks include variations in simulations and changes due to updates.

Distribution Types:
 The expected annual loss distribution may not be Normal but can be Beckmann, MaxStable, Gamma, Inverse Gaussian, or Lognormal.
 The article simplifies by assuming a Normal Distribution for quantifying model risk.

Coefficient of Variation (CV):
 CV is crucial in model risk quantification.
 Lower bounds for CV are established based on Hölder's inequality.

Results and Bounds:
 The article provides lower bounds for CV and explores Hölder's inequality to derive these bounds.
 Upper bounds for CV are discussed, considering different scenarios.

TVaR (Tail Value at Risk) Upper Bounds:
 Upper bounds for TVaR are derived based on CV and other parameters.

Probability of Attaching and Exceeding:
 Lower bounds for the probability of attaching are discussed in terms of CV.
 Upper bounds for the probability of exceeding a given loss are presented.

Simulation Deviation:
 Numerical simulations explore the probability of deviations in simulated annualmeanloss under certain conditions.

Theorems and Corollaries:
 The article presents mathematical theorems and corollaries to support the derived bounds.
 Theorems provide insights into the relationships between mean, minimum, and maximum losses.

Empirical Studies:
 Empirical studies are mentioned, confirming the closeness of bounds to true standard deviation.

Conclusions:
 The article concludes by summarizing the obtained results and their implications for risk management.
Overall, the article demonstrates a rigorous mathematical approach to understanding and quantifying risks in the reinsurance industry, particularly in the context of simulated losses and model uncertainties. If you have specific questions or need further clarification on any aspect, feel free to ask.