Why Long-Run Portfolio Growth Is Below Expected Return

# Intuitive explanation of stochastic portfolio theory


# Intuitive explanation of stochastic portfolio theory












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?

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/21742

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.

## Answer by Richi Wa (score 4)

https://quant.stackexchange.com/a/21722

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 $$ E\log(1+x) \approx \mu - \frac{\sigma^2}{2}. $$