Fernholz and Karatzas have published various papers about so called stochastic portfolio theory. Basically they say that the return to be expected from a portfolio on the long run is rather the growth rate $$ \gamma = \mu - \frac12 \sigma^2 $$ than $\mu$, where $\mu$ is the drift coefficient of the price process $S_t$ which solves the following SDE: $$ dS_t = \mu S_t dt + \sigma S_t dB_t. $$

One can argue with Ito's lemma, with the geometric mean of a lognormal random variable and similar - but what is the intuition behind this?

As references see Stochastic Portfolio Theory and Stock Market Equilibrium by Fernholz and Shay for the first paper on this and Does a Low Volatility Portfolio Need a “Low Volatility Anomaly?” by Meidan as a more recent reference.

If I am not mistaken then the above SDE would look like this $$ dS_t = (\mu-\sigma^2/2) S_t dt + \sigma S_t \circ dB_t $$ in Stratonovich form and one sees the "correct" growth rate... which is another link. But what is the big picture of all this?

  • $\begingroup$ My understanding of SPT is that the relative market capitalization of stocks in the market is stable. Therefore if a firm is successful, its market share will increase, but to maintain this stability its drift will go down to allow others to catch up. I am not sure where you got this drift expression from, do you have a reference? $\endgroup$
    – Kiwiakos
    Nov 11, 2015 at 11:24
  • $\begingroup$ SPT says something about this. But isn't the start of this line of thoughts in the growth rate observation? I added 2 references above. $\endgroup$
    – Richi Wa
    Nov 11, 2015 at 11:45
  • $\begingroup$ Thx for the refs, I think I understand your question. Imagine a BGM with zero drift starting from 100. For large horizons the expected value might be 100, but if you look at specific paths they either go to zero or blow up, in fact as time inceases all paths will eventually go to zero. So they say that the mean is not representative and the look at the median instead. $\endgroup$
    – Kiwiakos
    Nov 11, 2015 at 12:09
  • $\begingroup$ No, there should be more to it ... I will do some more research but what they look at is somehing like the average annual return over longer periods which is the geometric mean of the GBM ... $\endgroup$
    – Richi Wa
    Nov 11, 2015 at 12:27
  • 2
    $\begingroup$ I am sure @MarkJoshi could help us out here ;) $\endgroup$
    – Richi Wa
    Nov 11, 2015 at 13:30

2 Answers 2


This will depend on the definition of "return on the long run". If we define the annualized return on the long run by $\frac{1}{T}\ln \frac{S_T}{S_0}$ for a certain time $T$ in the future, then \begin{align*} E\left( \frac{1}{T}\ln \frac{S_T}{S_0} \right) = \mu-\frac{1}{2}\sigma^2, \end{align*} as claimed. Note that $\mu$ is the instant, or instantaneous, return.

  • $\begingroup$ Yes and if I rewrite your expression then we come back to the (stochastic) geometric mean. But it is more straight forward. Thank you. $\endgroup$
    – Richi Wa
    Nov 14, 2015 at 9:25

Trying to shed some light here:

What we also see using this here, is that if returns are log-normally distributed, ie. $$ 1 + r = \exp(\mu + \sigma Z), $$ with $Z$ standard-normal, then $$ E[1+r] = \exp(\mu + \frac 12 \sigma^2) $$ holds. But the geometric mean $GM$ is given by $\exp(\mu)$ and we have $$ \log(GM) = \mu = \log(E[1+r]) - \sigma^2 /2 $$

and the average annual return $aan$ over $n$ years $$ 1 + aan = (\prod_{i=1}^n (1+r_i))^{1/n} $$ is just the same as the geometric mean. Finally after $n$ years we have $$ \prod_{i=1}^n (1+r_i) = (1 + aan)^n $$ we should care about the geometric mean most.

One more observation: One cans show that for $x$ close to zero it holds that $$ \log(1+x) \approx x - \frac{x^2}{2}. $$ Then taking the expecation we get $$ E\log(1+x) \approx \mu - \frac{\sigma^2}{2}. $$


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