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Insurance Risk Insurance companies know the risk of insurance is greatly reduced if the company insures not just one person, but many people. How does this work? Let \(x\) be a random variable representing the expectation of life in years for a 25 -year-old male (i.e., number of years until death). Then the mean and standard deviation of \(x\) are \(\mu=50.2\) years and \(\sigma=11.5\) years (Vital Statistics Section of the Statistical Abstract of the United States, 116th edition). Suppose Big Rock Insurance Company has sold life insurance policies to Joel and David. Both are 25 years old, unrelated, live in different states, and have about the same health record. Let \(x_{1}\) and \(x_{2}\) be random variables representing Joel's and David's life expectancies. It is reasonable to assume \(x_{1}\) and \(x_{2}\) are independent. $$\begin{aligned} &\text { Joel, } x_{1}: \mu_{1}=50.2 ; \sigma_{1}=11.5\\\ &\text { David, } x_{2}: \mu_{2}=50.2 ; \sigma_{2}=11.5 \end{aligned}$$ If life expectancy can be predicted with more accuracy, Big Rock will have less risk in its insurance business. Risk in this case is measured by \(\sigma\) (larger \(\sigma\) means more risk). (a) The average life expectancy for Joel and David is \(W=0.5 x_{1}+0.5 x_{2}\) Compute the mean, variance, and standard deviation of \(W\). (b) Compare the mean life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W).\) (c) Compare the standard deviation of the life expectancy for a single policy \(\left(x_{1}\right)\) with that for two policies \((W).\) (d) The mean life expectancy is the same for a single policy \(\left(x_{1}\right)\) as it is for two policies \((W),\) but the standard deviation is smaller for two policies. What happens to the mean life expectancy and the standard deviation when we include more policies issued to people whose life expectancies have the same mean and standard deviation (i.e., 25 -year-old males)? For instance, for three policies, \(W=(\mu+\mu+\mu) / 3=\mu\) and \(\sigma_{W}^{2}=(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}+(1 / 3)^{2} \sigma^{2}=\) \((1 / 3)^{2}\left(3 \sigma^{2}\right)=(1 / 3) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{3}} \sigma .\) Likewise, for \(n\) such policies, \(W=\mu\) and \(\sigma_{W}^{2}=(1 / n) \sigma^{2}\) and \(\sigma_{W}=\frac{1}{\sqrt{n}} \sigma .\) Looking at the general result, is it appropriate to say that when we increase the number of policies to \(n\), the risk decreases by a factor of \(\sigma_{W}=\frac{1}{\sqrt{n}} ?\)

Step by step solution

01

Calculate the Mean of W

To find the mean of \(W\), use the formula for the mean of a weighted sum of independent random variables. For \(W = 0.5x_1 + 0.5x_2\), the mean \(\mu_W\) is:\[ \mu_W = 0.5\mu_1 + 0.5\mu_2 \]Given \(\mu_1 = 50.2\) and \(\mu_2 = 50.2\),\[ \mu_W = 0.5(50.2) + 0.5(50.2) = 50.2 \]

02

Calculate the Variance of W

The variance of \(W\) can be calculated using the formula for the variance of the sum of two independent random variables:\[ \sigma_W^2 = (0.5^2)\sigma_1^2 + (0.5^2)\sigma_2^2 \]Given \(\sigma_1 = 11.5\) and \(\sigma_2 = 11.5\), compute:\[ \sigma_W^2 = 0.25(11.5)^2 + 0.25(11.5)^2 = 0.25 \times 2 \times 132.25 = 66.125 \]

03

Calculate the Standard Deviation of W

The standard deviation \(\sigma_W\) is the square root of the variance:\[ \sigma_W = \sqrt{\sigma_W^2} = \sqrt{66.125} \approx 8.13 \]

04

Compare Mean Expectations

The mean life expectancy for a single policy \(\mu_x = 50.2\) is the same as the mean for two policies \(\mu_W = 50.2\). There is no change in mean when we average the expectations.

05

Compare Standard Deviations

The standard deviation for a single policy \(\sigma = 11.5\) is larger than the standard deviation for two policies \(\sigma_W \approx 8.13\). This demonstrates a reduction in uncertainty.

06

Analyze Effect of Increasing Number of Policies

With more policies, the mean life expectancy \(\mu\) remains unchanged. The standard deviation is reduced by a factor depending on the number of policies. It shrinks by \(\frac{1}{\sqrt{n}}\), which reflects decreasing risk with increasing policies. Thus, the standard deviation becomes \(\sigma_W = \frac{1}{\sqrt{n}} \sigma\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distributions

In the world of insurance risk management, understanding the concept of probability distributions is crucial. These distributions help detail how likely different outcomes are, which in turn affects predictions about future events, such as a person's life expectancy. In the case of Joel and David, both represent separate, independent probability distributions, each describing possible life expectancies.

We often represent these distributions using random variables like \(x_1\) and \(x_2\), each with their own mean (average outcome, \(\mu\)) and standard deviation (measure of spread, \(\sigma\)). For independent random variables, their means add up in a weighted way to form the mean of a new random variable \(W\) that represents the average life expectancy of two people. Importantly, if both random variables have the same mean, as is the case with Joel and David, the mean for \(W\) remains the same.

• Joel: \(x_1\) with \(\mu_1 = 50.2\)
• David: \(x_2\) with \(\mu_2 = 50.2\)
• Weighted mean for two policies \(W: \mu_W = 50.2\)

Probability distributions provide a systematic way to assess insurance risks and understand how variations in life expectancy across several individuals might sum up, providing a realistic scenario for risk management.

Variance and Standard Deviation

The concepts of variance and standard deviation are fundamental to grasping variability and risk, especially in insurance. Variance, \(\sigma^2\), indicates the extent to which outcomes differ from the mean, while the standard deviation, \(\sigma\), is the square root of the variance, offering a practical measure of spread.

In our example with Joel and David, we calculated the variance of their combined life expectancy \(W = 0.5x_1 + 0.5x_2\) using the formula for independent random variables:

• Variance formula: \(\sigma_W^2 = (0.5^2)\sigma_1^2 + (0.5^2)\sigma_2^2\)
• Resulting variance: \(\sigma_W^2 = 66.125\)

The standard deviation of \(W\), obtained by taking the square root of the variance, is \(\sigma_W \approx 8.13\). This smaller standard deviation compared to that of a single policy signifies that combining policies generally reduces risk.

Standard deviation provides insurers with insight into risk management by showing how much variation is expected in future claims. The fundamental understanding that a lower standard deviation leads to less risk is used by insurance companies to set premiums and make informed decisions.

Law of Large Numbers

The law of large numbers is a principle that states as the number of policies (or samples) increases, the average of these samples will converge to the expected mean of the entire population. In an insurance context, this rule implies that the more individuals covered, the more accurately an insurer can predict outcomes, such as the average life expectancy.

When insurance companies increase the number of policies, two important effects occur:

• The average life expectancy \(\mu\) remains unchanged, regardless of the number of individuals.
• The standard deviation reduces, proportional to \(\frac{1}{\sqrt{n}}\), where \(n\) is the number of policies.

This reduction in standard deviation means that as the insurer covers more people, the risk associated with variance in individual life expectancies decreases. Hence, while the average prediction remains the same, the outcomes become more reliable and consistent, aligning closer to the predicted mean.

Insurance companies leverage this law by insuring large groups, thereby minimizing unforeseen risks and stabilizing their overall risk portfolio. This strategy is pivotal for ensuring financial stability and providing fair premiums for policyholders.