The King on Nostr: nprofile1q…0qmju I'm pretty sure with a more realistic model of risk-aversion, the ...
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I'm pretty sure with a more realistic model of risk-aversion, the conclusion is actually reversed! That is, generally speaking, the sound advice is that the more ambient risk there is the *worse* additional risk is compared to safe options.
Here is the most popular model among economists. Someone who wants money, but is risk-averse, is maximizing the expected value of some concave function of money. A common example is the logarithm (although empirically it appears that you need something even *more* risk-averse than the logarithm). The important thing is that the more money you have, the less sensitive the utility function is to changes in money. (See https://en.wikipedia.org/wiki/Risk_aversion#Measures_of_risk_aversion_under_expected_utility_theory)
Say that your wealth is currently $100 but is going to change by either +x or -x (50:50 odds). You are considering a decision that could either give you $1 or -$3. How does this depend on x?
If you are risk-neutral, then 3:1 odds that things go well are sufficient for this to be a good decision. If you have logarithmic utility (natural log of dollars), look at this graph: https://www.wolframalpha.com/input?i=%28%280.5+ln%28100+%2B+x%29+%2B+0.5+ln%28100+-+x%29%29+-+%280.5+ln%2897+%2B+x%29+%2B+0.5+ln%2897+-+x%29%29%29%2F+%28%280.5+ln%28101+%2B+x%29+%2B+0.5+ln%28101+-+x%29%29+-+%280.5+ln%28100+%2B+x%29+%2B+0.5+ln%28100+-+x%29%29%29+from+x%3D-97+to+97
Even at x=0 there is some risk aversion, you want at least 3.05:1 odds for the expected utility to be positive. As x increases, you actually become less bold. At about x=$80 you need 3.2:1 odds. As it approaches $97 the odds required go to infinity, since at x=$97 bad luck would result in ruin.
Basically, using standard deviation to calculate a "value at risk" is not compatible with the expected utility model. Risks being independent is nice (correlated risks would make the above graph much worse), but not sufficient for your argument to carry.
I'm pretty sure with a more realistic model of risk-aversion, the conclusion is actually reversed! That is, generally speaking, the sound advice is that the more ambient risk there is the *worse* additional risk is compared to safe options.
Here is the most popular model among economists. Someone who wants money, but is risk-averse, is maximizing the expected value of some concave function of money. A common example is the logarithm (although empirically it appears that you need something even *more* risk-averse than the logarithm). The important thing is that the more money you have, the less sensitive the utility function is to changes in money. (See https://en.wikipedia.org/wiki/Risk_aversion#Measures_of_risk_aversion_under_expected_utility_theory)
Say that your wealth is currently $100 but is going to change by either +x or -x (50:50 odds). You are considering a decision that could either give you $1 or -$3. How does this depend on x?
If you are risk-neutral, then 3:1 odds that things go well are sufficient for this to be a good decision. If you have logarithmic utility (natural log of dollars), look at this graph: https://www.wolframalpha.com/input?i=%28%280.5+ln%28100+%2B+x%29+%2B+0.5+ln%28100+-+x%29%29+-+%280.5+ln%2897+%2B+x%29+%2B+0.5+ln%2897+-+x%29%29%29%2F+%28%280.5+ln%28101+%2B+x%29+%2B+0.5+ln%28101+-+x%29%29+-+%280.5+ln%28100+%2B+x%29+%2B+0.5+ln%28100+-+x%29%29%29+from+x%3D-97+to+97
Even at x=0 there is some risk aversion, you want at least 3.05:1 odds for the expected utility to be positive. As x increases, you actually become less bold. At about x=$80 you need 3.2:1 odds. As it approaches $97 the odds required go to infinity, since at x=$97 bad luck would result in ruin.
Basically, using standard deviation to calculate a "value at risk" is not compatible with the expected utility model. Risks being independent is nice (correlated risks would make the above graph much worse), but not sufficient for your argument to carry.