# Is the geometric mean of the price really equal to 1?

If we assume that the geometric mean of the price is 1, then the probability of a price drop of 20% should be equal to the probability of an increase of 25% (0.8*1.25=1). If so, then we have a trading system with a positive expectation. If this is not the case, and, for example, a price drop of 20% corresponds to an increase of 20%, then the price will fall over time, and then there is also a positive expectation. What is the problem with such reasoning?

• Firstly, why should the mean be constant? Secondly, past performance has little to do with future performance. Commented Jul 1, 2023 at 22:37
• If we assume that the geometric mean of the price is 1 - why would we assume that? If the assumption is flawed then the whole reasoning is flawed. Commented Jul 2, 2023 at 2:03
• "If so, then we have a trading system with a positive expectation" , no, if the geometric mean is 1, then the average gains and losses cancel out (as you show), and you have a neutral expectation. Why would you think there's a positive expectation? Commented Jul 2, 2023 at 2:19
• I think I see where you're flawed - a geometric mean of 1 does not mean that each of those outcomes has an equal probability at any given time, it means that gains and losses over time cancel each other out. Commented Jul 2, 2023 at 2:20
• @DStanley You have \$1, leverage \$4 price = \$1 50%: price = 0.8, profit=-\$1 50%: price = 1.25, profit=\$2.25 Expectation = 0.5 * 2.25 = 1.125 Optimal f according to the Kelly criterion = (0.5*2.25-1)/(2.25-1) = 0.1 Geometric mean of the system ≈ 1.006 Commented Jul 2, 2023 at 7:11

## 1 Answer

You seem to be wondering how a positive logarithmic expectation for a trading system can be compatible with the efficient market hypothesis (EMH). Note that EMH says no system can beat "the market". But you are measuring returns relative to a particular investment (e.g., USD cash) that is not itself "the market". The optimally weighted system you are deriving is "the market" (or would be, under the hypothetical assumptions you are making).

It may be counterintuitive that two investments, each with logarithmic expectation ≤0 (geometric mean ≤1), can in some cases be combined to give positive logarithmic expectation -- but this is the miracle of diversification.

What you can't beat, according to EMH, is a portfolio that is already optimally diversified. Any deviation from the weights of "the market" will result in a lower logarithmic expectation.

In your hypothetical where stocks are equally likely to fall 20% or rise 25%, the optimal portfolio is half stocks and half cash (not surprising since the two choices are symmetric -- each has zero logarithmic expectation relative to the other). You indirectly obtained this in a comment where you started with 5x leverage on stocks and then found that the Kelly criterion gave 0.1x on that, i.e., overall 0.5x.

In your hypothetical where stocks are equally likely to fall 20% or rise 20%, the optimal portfolio is all cash. Again we can use the Kelly criterion, giving zero weight on stocks. It is not correct that shorting will help, because that will just reverse the arithmetic returns -- it will still be equally likely to fall X% or rise X%, and the logarithmic expectation will still be negative. This is an unrealistic scenario because, with no reason to invest in them, stocks would not exist.

A general statement of EMH is that every investment or trading system has zero arithmetic expectation relative to the optimal portfolio (the market). Thus, the Kelly criterion says the weight that should be shifted to any investment (relative to the market weighting) is zero. Equivalently, any investment other than the market itself (e.g., a component of the market) will have a negative logarithmic expectation relative to the market.

Using arithmetic expectation relative to the market as an empirical criterion for violating EMH is discussed here.

• so well-written ! Commented Jul 3, 2023 at 0:38