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In Prof. Andrew Lo's book, Adaptive Markets: Financial Evolution at the Speed of Thought, in the section The Origin of Risk Aversion in Chapter 6, he uses the tribble example to illustrate how risk aversion comes about. But I am a bit confused by the math here. He supposes that each tribble is faced with two possible actions, either having three offsprings for sure or having two/four offsprings with 50/50 chance. He goes on to calculate that after two generations, the number of risk-averse tribbles from a single risk-averse grandparent will be 9 for certain, but claims that the number of risk-neutral tribbles from a single risk-neutral grandparent will, on average, be 8. Why is this the case? Isn't the expected number of offsprings after two generations also 9 for the risk-neutral tribbles? I hope someone who has read this book can enlighten me.

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I wouldn't sweat this one. He's just making a simple "on average" assumption and not examining the wings of the actual distribution. Below is the actual quote you're referring to; and why this one isn't quite right in this case.

But the point he's trying to make isn't invalid just because he chose a bad example to try to make it. In his defence, consider the following alternative similar dilemma. You are offered a perpetual guaranteed 5% annual return, versus a 50:50 of 0% or 10% each year. Your CAGR will be 1.10^0.5 = (1+ 4.88%). IE there IS an actual not just a certainty value to risk aversion here (when presented with fair bets).

Lo just chose a bad example here (following on from his previous use of tribbles earlier that chapter). I strongly suspect he knew this, but suspected few people would pick him up on details, when his narrative was sailing along neatly at a good pace of knots.


Lo, p.205

Why does this happen? Once again, it involves the Law of Averages. After just two generations, the number of risk- averse tribbles arising from a single risk- averse grandparent will be 3 × 3 = 9 for certain (recall that these tribbles always have three off spring), while the number of risk-neutral tribbles from a single risk- neutral grandparent will, on average, be 2 × 4 = 8 (these tribbles can have two or four offspring, so sometimes it’s two and sometimes it’s four, with equal likelihood), which is 11 percent fewer. This is a small diff erence, but it occurs across the entire population of grandparents. As a result, the Law of Averages tells us that the gap between the number of risk- averse and risk- neutral tribbles will grow over time, until eventually risk aversion becomes the dominant type of behavior in the overall population

Except the lower median ignores the outliers who have 16 grandkids... Below is the actual distribution, and the average is 9!

enter image description here


EDIT - To @noob's question, and I've worked out where the 8 figure comes from.

Lo's arguing that the "Law of Averages" gives an expected Risk-Neutral population of E(RN) after n generations: E(RN) = 4^(0.5n) . 2^(0.5n) log(ERN) = 0.5n * (log(2) + log(4)) n = 2, log(ERN) = 2.079, ERN = 8.000.

Where this, I think, goes wrong is that it assumes that all the tribbles in half the generations will have in one generation, and all have 4 in a different generation. Over time, this will indeed average out to 8.

However, if you allow different tribbles in the same generation to have different numbers of offspring, then the reproduction rate remains three.

Below, I've double-checked this by Monte-Carlo'ing every tribble for 10 generations, 4000 times. Lo's expectation of 2.8 -> 8 after 2, is about 20 standard errors away from the MC simulation after 10 generations...

enter image description here

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  • $\begingroup$ Thanks. Could you explain your example? Which choice has the CAGR you mentioned? $\endgroup$ Feb 1, 2021 at 5:07
  • $\begingroup$ (I am disappointed if such an illustrious professor gave an example in his book which is wrong). It is really so? $\endgroup$
    – nbbo2
    Feb 2, 2021 at 10:43
  • $\begingroup$ See additional comments in edited version above @noob2. I'm wary of trash-talking the guy, as I 100% appreciate he walks the walk.... but I cannot see how he is right here. And would love to find out how I'm wrong for my own education! ;-) $\endgroup$
    – demully
    Feb 2, 2021 at 17:47
  • $\begingroup$ Could you explain what you mean by the CAGR being 1.0488? Is this for the guaranteed return or the random return? I am a bit confused how this example illustrates risk-aversion. $\endgroup$ Feb 7, 2021 at 2:55
  • $\begingroup$ Hi, sorry. It's a simple illustrative example of how geometric averages fall short of arithmetic averages, when there's volatility/variance. So if a (theoretical) investment had a 50:50 chance of a zero and +10% return, then your CAGR would be only 4.88%, versus the 5% you would get with a 100% probability of a +5% return. Their "average" return is the same; but the "long-run" return will be lower under higher volatility. And so investors will indeed be more risk-averse towards volatile routes to the same average expected return. Which is Lo's basic point, even if his example here was wrong. $\endgroup$
    – demully
    Feb 7, 2021 at 23:44

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