I have been reading a paper by Bove & Elia (2011), where they quote a definition of the value of statistical life from Bellavance, Dionne & Lebeau (2009). I have tried making my peace with the necessity of this concept, but I have been rather confused by the used definition. It says that the value of a statistical life is a function of the willingness to pay to reduce the risk of death and the risk of death itself. However, the risk of death reduces the value of a statistical life. Why would it do that? Is it a weight as in an NPV calculation, where the risk of default reduces the expected cash flow?
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If you're willing to pay \$10 to reduce your risk of death by 1 in 1M, then your VSL is \$10M.†
Example. Say there are two bike helmets. One costs \$20 while the other costs \$30. The two helmets are exactly identical, except that the more expensive one has been proven to reduce your risk of death by an additional 1 in 1M.
If you're willing to buy the more expensive helmet, then your VSL is (at least) \$10M. (If you're exactly indifferent about buying the more expensive helmet, then your VSL is exactly \$10M.)
I'm not sure what you mean by "the risk of death reduces the value of a statistical life."
†At least under the simplest approach, which ignores other important considerations like risk aversion and the discrepancy between willingness-to-pay and willingness-to-accept.
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$\begingroup$ What I mean with "the risk of death reduces the value of a statistical life." comes from the formula, which I failed to post. It is that the VSL=WTP/R, so value of statistical life is willingness to pay over risk of death. If the risk of death is high, the denominator is big and the WTP/R is small. VSL thus is small as well, and that seems counterintuitive to me. $\endgroup$– JotHaCommented Apr 18, 2018 at 1:22
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2$\begingroup$ @user9504090: Ah I see. Your mistake here is to think of WTP as a constant. It is not. Instead, it is a variable that depends on R. Here your error is to implicitly assume that if I'm willing to pay at most WTP = \$10 to reduce my risk of death by R = 1 in 1M, then I'm also willing to pay only at most WTP = \$10 to reduce my risk of death by R = 1 in 100. But this is false. $\endgroup$– user18Commented Apr 18, 2018 at 1:25
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$\begingroup$ Ah ok! That’s great and I see where my mistake was. To clear out all of my misunderstanding: Would there be a positive or a negative relationship between R and VSL? To me it could go either way. Really depends on how WTP reacts to R, i.e. the elasticity of WTP with respect to R. Is there any research on this or anyone who has provided a logic for thinking about it? $\endgroup$– JotHaCommented Apr 18, 2018 at 1:45
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1$\begingroup$ @user9504090: If WTP(R) is defined as the maximum one is willing to pay to reduce one's risk of death by probability R, then WTP(R) is certainly increasing in R. So, the relationship between WTP and R is positive. There is a vast literature on VSL which you can explore. $\endgroup$– user18Commented Apr 18, 2018 at 1:51