Question

192 Marine losses for an oil company. The frequency distribution shown in the table depicts the property and marine losses incurred by a large oil company over the last 2 years.This distribution can be used by the company to predict future losses and to help determine an appropriate level of insurance coverage. In analyzing the losses within an interval of the distribution, for simplification, analysts may treat the interval as a uniform probability distribution (Research Review, Summer 1998). In the insurance business, intervals like these are often called layers.

Layer Property and Marine Losses(millions of \() Frequency

1 0.00–0.01 0.668

2 0.01–0.05 0.38

3 0.05–0.10 7

4 0.10–0.25 4

5 0.25–0.50 2

6 0.50–1.00 1

7 1.00–2.50 0

Source: Based on J. M. Cozzolino and P. J. Mikolaj, “Applications of the Piecewise Constant Pareto Distribution,” Research Review,Summer 1998, pp. 39–59.

a. Use a uniform distribution to model the loss amount in layer 2. Graph the distribution. Calculate and interpret its mean and variance.

b. Repeat part a for layer 6.

c. If a loss occurs in layer 2, what is the probability that it exceeds \)10,000? That it is under \(25,000?

d. If a layer-6 loss occurs, what is the probability that it is between \)750,000 and \(1,000,000? That it exceeds \)900,000? That it is exactly $900,000?

Short answer

⦁ Mean of the loss for layer 2 is 0.03, variance of the loss for the layer 2 is 0.00013.

⦁ Mean of the loss for layer 6 is 0.75, variance of the loss for the layer 6 is 0.0208.

⦁ The probability if loss exceeds $10000 for the layer 2 is 0.375.

⦁ The probability that it is between $750,000$ and $1000,000 if the loss occurs in layer 6 is 0.5.The probability at the particular point equals 0. Hence, the probability is exactly $900,000 is

Step-by-step solution

Given information

The frequency distribution given depicts the property and marine losses incurred by a large oil company over the last 2 years.

This distribution can be used by the company to predict future losses and to help determine an appropriate level of insurance coverage.

Use uniform distribution to model the loss amount in layer 2.

⦁

The random variable x denotes the amount loss which follows the uniform distribution between 0.01 and 0.05 million dollars.

Pdf of uniform distribution can be given as

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{b}-\mathrm{a}},\mathrm{a}\le \mathrm{x}\le \mathrm{b}$

Here, b = 0.05 and c = 0.01

Plot the graph of the distribution.

The relevant graph will be as shown below.

From the graph, it is can be seen that the distribution function of the uniform distribution lies between area 0.01 and 0.05.

Calculate mean and variance.

The expected value of the distribution can be obtained as

$\mu =\frac{a+b}{2}$

That is.

$$\begin{gathered} =\frac{0.01+0.05}{2} \\ =\frac{0.06}{2} \\ =0.03 \end{gathered}$$

Hence, we can say that the loss for the layer 2 on an average is $0.03 million.

The variance can be given by${\sigma }^2=\frac{a-b}{\sqrt{12}}$

Hence,

$$\begin{gathered} \sigma =\frac{0.05-0.01}{\sqrt{12}} \\ =\frac{0.04}{3.4641} \\ =0.0115 \end{gathered}$$

Therefore,

$$\begin{gathered} {\sigma }^2={\left(0.0115\right)}^2 \\ =0.00013 \end{gathered}$$

Hence, variance of the loss for the layer 2 is 0.00013.

Use uniform distribution to model the loss amount in layer 6.

⦁

X denotes the amount loss which follows the uniform distribution between 0.50 and 1.00

Here b=1.00 and a=0.50

Therefore,

$$\begin{gathered} f\left(x\right)=\frac{1}{1.00-0.50} \\ =\frac{1}{0.50} \\ =20.50\le x\le 1.00 \\ \mathrm{Thus}f\left(x\right)=\left\{\begin{array}{ll}2& 0.50\le x\le 1.00\\ 0& otherwise\end{array}\right. \end{gathered}$$

Calculate mean and variance

The expected value of the distribution can be obtained as

$\mu =\frac{a+b}{2}$

Therefore

$$\begin{gathered} =\frac{0.50+1.00}{2} \\ =0.75 \end{gathered}$$

Hence, we can say that the loss for the layer 6 on an average is 0.75 million dollars.

The variance can be given by ${\sigma }^2=\frac{a-b}{\sqrt{12}}$

Hence,

$$\begin{gathered} \sigma =\frac{1.00-0.50}{\sqrt{12}} \\ =\frac{0.50}{3.4641} \\ =0.1443 \end{gathered}$$

Therefore,

$$\begin{gathered} {\sigma }^2={\left(0.1443\right)}^2 \\ =0.0208 \end{gathered}$$

Hence, variance of the loss for the layer 6 is 0.0208.

calculating the probability loss exceeds 10,000$ for the layer 2

⦁

Loss of $10,000 corresponds t the value x=0.01, which is obtained by the ratio of $10,000 divided by the $1 milllion and a loss of $25,000 corresponds to the value x=0.025. This can be obtained by the ratio of $25000 divided by the $1 million.

Hence,

$$\begin{gathered} P\left(x<0.025\right)=\frac{b-a}{d-c} \\ =\frac{.025-0.01}{0.05-0.01} \\ =\frac{0.015}{0.04} \\ =0.375 \end{gathered}$$

Hence, the probability if loss exceeds $10000 for the layer 2 is 0.375.

calculating the probability if it is between 750,000$ and 1000,000$ if the  toss occurs in layer 6

⦁

Toss of $75,000 corresponds to the value x=0.75, which can be obtained from the ratio of $75,000 divided by 1 million dollar

Toss of $1000,000 corresponds to the value x=1, which can be obtained from the ratio of $1000,000 divided by 1 million dollar

Consider,

$$\begin{gathered} \mathrm{P}\left(0.75<\mathrm{x}<1\right)=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{d}-\mathrm{c}} \\ =\frac{1-0.75}{1.00-.050} \\ =\frac{0.25}{0.50} \\ =0.5 \end{gathered}$$

Thus, the probability that it is between $750,000 and $1000,000 if the loss occurs in layer 6 is 0.5.

calculating the probability if loss occurs and exceeds 900,000$ in layer 6

Loss of $900,000 corresponds to the value x=0.9 which has been obtained by the ratio of $900,000 divided by the 1 million dollars.

Consider

$$\begin{gathered} \mathrm{P}\left(\mathrm{x}>0.9\right)=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{d}-\mathrm{c}} \\ =\frac{1-0.90}{1.00-.050} \\ \mathrm{P}\left(\mathrm{x}>0.9\right)=\frac{0.10}{0.50} \\ =0.2 \end{gathered}$$

Since uniform distribution is continuous distribution, the probability at the particular point equals 0. Hence, the probability is exactly $900,000 is $\mathrm{P}\left(\mathrm{x}>0.90\right)=0$.