I'm trying to understand the question from the 'Probability for Risk Management 2e' book and the solution I found online vs. the solution on the answer key does not match.
For the insurance policy that pays $1000$ for an injury and $10000$ for
death, what is the standard deviation for the claim amount on $5$ units
of insurance? (Note: Some employees receive $0 of claim payment. This
value of the random variable must be included in your calculation).
https://fac.ksu.edu.sa/sites/default/files/sol_hw2_math380.pdf (Question #5)
$\mathrm{sd}(\text{claim}) = 0.5017$.
The above looks like it's doing the standard deviation calculation based on the binomial but that formula hasn't been covered up to this point of the book.
The answer in the book is 3,421.84.
I know that $\mathrm{E}[x] = 114$.
so,
$$
V(x) = ((1000-114)^2 \cdot 7.3\%) + ((10000-114)^2 \cdot 0.41\%) + ((0-114)^2 \cdot 92.19\%)
\approx 469991.004
$$, taking square root $\approx 685.558899001$
Is the 685.55 the std on 1-unit of insurance? How do I get value for 5 units?
Best Answer
I had to read the whole question because the part you quoted provides insufficient information. I will also say that this is a rather poor quality question and solution for reasons we will see shortly.
Let's look at the variance of the claim amount on $5$ units of insurance, with the additional (implied, but required) assumption that:
Then the claim amount random variable $L$ for a single unit of insurance has PMF $$\begin{align*} \Pr[L = 0] &= 0.92\color{red}{2}9 \\ \Pr[L = 1000] &= 0.073 \\ \Pr[L = 10000] &= 0.0041 \\ \end{align*}$$ The expected value of $L$ is $$\operatorname{E}[L] = (0)(0.9229) + (1000)(0.073) + (10000)(0.0041) = 114$$ which is the pure premium. The expected value of $L^2$ is $$\operatorname{E}[L^2] = (0)^2(0.9229) + (1000)^2(0.073) + (10000)^2(0.0041) = 483000.$$ Therefore, the variance on a single unit of insurance is $$\operatorname{Var}[L] = \operatorname{E}[L^2] - \operatorname{E}[L]^2 = 470004.$$ Now, in addition to the implied but unstated assumptions above, we encounter a problem regarding the manner in which the $5$ units of insurance are issued. If they are issued to a single employee, then the variance is $5^2 = 25$ times the variance of $L$, since $$\operatorname{Var}[5L] = 5^2 \operatorname{Var}[L],$$ hence the standard deviation of $5L$ is $5$ times the standard deviation of $L$. This results in an answer closest to that provided by the text: $$\operatorname{SD}[5L] = 5 \operatorname{SD}[L] = 5\sqrt{470004} \approx 3427.84.$$ However, this is not explicitly stated. One unit can be assigned to each of $5$ employees, in which case the variance is the sum of $5$ independent and identically distributed claim amount variables: $$\operatorname{Var}[L_1 + L_2 + L_3 + L_4 + L_5] = 5\operatorname{Var}[L],$$ and the standard deviation is only $\sqrt{5 \operatorname{Var}[L]} \approx 1532.98$.