There are two ways to think about investment returns and randomness.

First is sort of like 'bank interest', with randomness. Suppose we invest 100 units of currency. Suppose each year there is a normally-distributed 'market return' with mean $\mu$ and s.d. $\sigma$. And let's assume $\mu=.10$ and $\sigma=.15$ for concreteness. Then, what we do is add the normally distributed return, which in this case is distributed as $N(10, 15)$ to the initial 100. This gives us a random value distributed as $N(110, 15)$. Then we do it again the next year, and it will be distributed as $N(121,\cdot)$, etc. After T years, the distribution is $N(100\cdot 1.1^T,\cdot)$. This means the expected value, mean, median, and mode are all the same: $100\cdot 1.1^T$

The other approach is to use GBM from stochastic calculus. To make it all correspond when there is not randomness (i.e., if $\sigma=0$), the appropriate value of $\mu$ for the normal distribution underlying the log-normal price distribution assumed in GBM is $ln(1.1)$, which is something like 9.53% (just due to the difference with continuous compounding). Now, if there is no uncertainty (or very, very little), everything corresponds, and the value at time T is $100\cdot e^{.0953T}$, which is the same as $100\cdot 1.1^T$.

But, if there is uncertainty, GBM tells us that the end result is a log-normal price distribution, which means that, if we call $\mu$ the value .0953, (1) the expected value is $100\cdot e^{(\mu-\frac{\sigma^2}{2})T+\sigma W_t}$ This also means that, compared to the 'bank interest' model, the return profile doesn't have the same mode or median as the GBM solution.

My question is this: How does one explain these discrepancies in terms of 'real money'?

Suppose we looked at 10 years of annual returns, and they really were exactly 10% per year on average, and that was a perfect predictor of future average returns. That is an observed number, and gives us $\mu$ (as described above) for either model.

If we imagine 10,000 investors all investing in an independent but identically distributed investment as described above, I can see two ways to reason about it.

1) That the market, on average, delivers 10%, and that the expected value across investors must be the same under either model. But, the 'bank account' approach is more correct, since adding normals must give a result which is normal, and the mean, mode (maximum likelihood), and median are all the same since the result is symmetrical.

2) The other way to view it is that one of these models is 'wrong' in some way. But which? And why?

My intuition is that the GBM is more 'correct'; given the convexity and the like, these people with different but identically distributed return streams when observed would have a tail to the right, and the mode would be below the mean (or expected) value of the investment. Some people explain the form of the solution (having the $(\mu-\frac{\sigma^2}{2})t$ form) as reflecting 'volatility drag', but the GBM solution actually has a tail to the right which, it seems, grows with increasing volatility - making the average outcome more and more above the most likely outcome. But why is that right?

  • $\begingroup$ By the way, see my comment below. In fact,$ E[S_t]=S_0e^{\mu t}$ because the $\sigma W_t$ term in the exponent exactly offsets the $-\frac{\sigma^2}{2}$ in the first term. $\endgroup$ – eSurfsnake May 25 '18 at 20:16

The expected value under GBM is:

$$\mathbb{E}(S_t) = S_0 e^{\mu t}$$

No $+\frac{\sigma^2}{2}$ in the exponent.

For reference, read this helpful math.SE answer.

Thus there is no 'extra value' to query.


Firstly, as pointed out by @Phil H, your average is wrong.

Now, maybe this will address your question. Recall that the solution to the GBM SDE with drift coefficient $\mu$ and diffusion coefficient $\sigma$ is:

$$ S_t=S_0\exp\left\{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t\right\} $$

Note that the drift has been corrected by a negative term. Where does it come from? Suppose a stock depreciates by a percentage $r \in [0,1]$ one day, only to recover by the same percentage next day. Then the total return is:

$$ (1-r)(1+r) = 1-r^2<1 $$

Therefore volatility of returns has an inherent cost. This is sometimes called "volatility drag" (see for example this question) and indeed, as you can see in my example above, it does result in "real money".

  • $\begingroup$ OK, but $E[S_t]$, I think, becomes $e^\mu$, since - and maybe I am very wrong here - we could re-write as $S_0e^{(\mu -\frac{\sigma^2}{2})t}e^{\sigma W_t}$, and (I think) $E[e^{\sigma W_t}]=\frac{\sigma^2}{2}t$, so the end result is $E[S_t]=e^{\mu t}$? $\endgroup$ – eSurfsnake May 25 '18 at 19:36
  • $\begingroup$ The second answer (above) seems to give this result. $\endgroup$ – eSurfsnake May 25 '18 at 19:48
  • $\begingroup$ @eSurfsnake That is correct, but what you have written in your question is not. $\endgroup$ – Daneel Olivaw May 26 '18 at 16:10
  • $\begingroup$ I updated the question. What is still wrong? $\endgroup$ – eSurfsnake May 27 '18 at 4:33

It is "real money". Just think of a contract that pays $-\ln(S_T/S_0)$, which incidentally is used to replicate variance swaps. Such a contract is worth (undiscounted) $\frac{1}{2}\int_0^TE[\sigma_t^2] dt $.


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