Now I think that paul was right. How is it that when Paul writes on one thing what he says is true of lots of things ? Does God talk to Paul. If so, Paul sure talks back.
My old thought was if Blaise was wrong then when you lose you can not lose real big. Does this mean that the one strange rule guys were right ? No. If Blaise was wrong then when you win with one strange rule you do not care, so when you win you win real small. Thus while when you lose, you don't lose real big the one strange rule is close to right for you aim is close to log wealth. This is just what Paul said.
To show this you need to have that the risk is just the same sort of risk each time.
OK so what is that in bigger words ? Paul is Paul Samuelson, who once argued that it is not necessarily a good idea to buy the portfolio with the highest geometric mean return even if you have a very long time horizon. He chose to argue this in a paper which contained no words of more than one syllable (except for the last word which was syllable). He called geometric mean portfolio approach proposed by Latane and (independently) Haakonsen "one strange rule". Their argument was that, by the law of large numbers, one would almost surely have higher end of period wealth if, each period, one maximized geometric mean return than if one always bought a different portfolio. this is true, but, as Samuelson pointed out, "if you lose you could lose real big".
In an earlier post, I wondered if the right (possibly changing) portfolio would have to be close to geometric mean maximizing almost all of the time if the utility function were bounded above and below. This means that, although you can lose real big in Euros or dollars, you can't lose real big in utils. With one syllable words "if the utility function is bounded above and below becomes "if Blaise [Pascal] was wrong".
Recently I thought again. The formal problem is time is discrete, each period there are risky assets with iid returns. The player aims to maximize a concave function of time T wealth which is bounded above and below. Does the geometric mean argument make sense in this case ? That is, is the time varying optimal portfollio chosen in t close to geometric mean maximizing for almost all t ? In the old post, I concluded that either it was or the derivative of the utility function with respect to wealth at T was almost certainly very close to zero.
So assume that the utility function is bounded below (wlog by zero) and above by Umax
To make the question simple, assume that one must distribute wealth across assets the same way each period so an iid random variable is added to log wealth each period. Is the right portfollio geometric mean maximizing ? Well , for any other portfollio P, by the law of large numbers as T goes to infinity, the chance that the geometric mean maximizing portfolio will give higher end of period wealth than P goes to 1. Chose T so this chance is 1-epsilon/umax for some small epsilon. If the utility function is bounded above and below, what happens in the other cases can give P utility minus expected geometric mean maximizing utility of no more than epsilon.
with probability 1-epsilon/umax the geometric mean maximizing approach gives higher wealth. This means that, on average, this can matter only
epsilon/(1-epsilon/umax) utils. does this mean that the optimal portfollio P must be very close to geometric mean maximizing ?
Already in the old post, I realised that if the marginal utility of wealth was very low, then the difference in wealth might be very large but matter very little.
Now I realise that bounded utility, iid returns and constant portfolios,means that the marginal utility of Log wealth is almost certainly tiny that is (Wealth)(dU/dWealth)must be tiny. For utility to be monotonic in wealth and bounded above and below (Wealth)(dU/dWealth) must go to zero as Wealth goes to zero and infinity.
With iid returns (with finite variance) (log Wealth - log initial wealth)/root(T) will have a normal distribution so |log(Wealth)| will almost certainly go to infinity so (Wealth)(dU/dWealth) will almost certainly go to zero. This means that there is no implication for optimal P must be close to geometric mean maximizing.
The assumption that "if you lose you can lose real big" is sufficient but not necessary for Samuelson's conclusion. posted by Robert
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