(Example 10-11 continued) Consider the economic evaluation of collision and comprehensive (fire, theft, etc.) insurance for an auto. One example was described in Figure 10-6. The probabilities and outcomes are summarized in the calculation of the expected values, which was done using Equation 10–5.

Question

Consider the economic evaluation of collision and comprehensive (fire, theft, etc.) insurance for an auto. One example was described in Figure 10-6. The probabilities and outcomes are summarized in the calculation of the expected values, which was done using Equation 10–5.

[latex] ext { Expected value }= ext { Outcome } A imes P( A )+ ext { Outcome }_{ B } imes P( B )+\cdots[/latex] (10-5)

[latex]\begin{aligned}EV _{ ext {accident w/ins. }} &=(0.9)(0)+(0.07)(300)+(0.03)(500)=\$ 36 \EV _{ ext {accident w/o ins. }} &=(0.9)(0)+(0.07)(300)+(0.03)(13,000)=\$ 411\end{aligned}[/latex]

Calculate the standard deviations for insuring and not insuring.

Accepted Answer

The first step is to calculate the EV([latex] ext { outcome } \left.^{2} ight)[/latex] for each.

[latex]\begin{aligned}EV _{ ext {accident w/ins. }}^{2} &=(0.9)\left(0^{2} ight)+(0.07)\left(300^{2} ight)+(0.03)\left(500^{2} ight)=13,800 \EV _{ ext {accident w/o ins. }}^{2} &=(0.9)\left(0^{2} ight)+(0.07)\left(300^{2} ight)+(0.03)\left(13,000^{2} ight)=5,076,300\end{aligned}[/latex]

Then the standard deviations can be calculated.

[latex]\begin{aligned}\sigma_{ w / ext { ins. }} &=\sqrt{ EV _{ w / ext { ins. }}^{2}-\left( EV _{ w / ext { ins. }} ight)^{2}} \&=\sqrt{\left(13,800-36^{2} ight)}=\sqrt{12,504}=\$ 112\end{aligned}[/latex]

[latex]\begin{aligned}\sigma_{ w / o ext { ins. }} &=\sqrt{ EV _{ w / o ext { ins. }}^{2}-\left( EV _{ w / o ext { ins. })^{2}} ight.} \&=\sqrt{\left(5,076,300-411^{2} ight)}=\sqrt{4,907,379}=\$ 2215\end{aligned}[/latex]

As described in Example 10-11, the expected value cost of insuring is $836 (=$36 + $800) and the expected value cost of self-insuring is $411. Thus the expected cost of not insuring is about half the cost of insuring. But the standard deviation of self-insuring is 20 times larger. It is clearly riskier.

Which choice is preferred depends on how much risk one is comfortable with. As stated before, this is an example of expected values alone not determining the decision. Buying insurance has an expected cost that is $425 per year higher, but that insurance limits the maximum loss to $500 rather than $13,000. The $425 may be worth spending to avoid that risk.

Question 10.12: (Example 10-11 continued) Consider the economic evaluation o...

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