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 $\sigma_{W}=\frac{1}{\sqrt{n}} ?$