Why Volatility Lowers Long-Run Compounded Growth
Summary
The document distinguishes an asset’s expected arithmetic return from its compounded growth rate. Under the return model described in the question, the expected return over a short interval is given by the drift, while the continuously compounded rate includes a volatility adjustment that reduces growth. The answer emphasizes that these are different measures, so an expected arithmetic return does not imply that wealth typically compounds at that same rate.
To make the effect intuitive, the response compares rectangles with equal average side lengths but different areas: variation between the sides reduces the area relative to a square with the same average side length. This illustrates why variability can lower compounded growth. The cited stock example and distribution motivate the question, but the answer does not develop a derivation, discuss estimation, or address how the effect changes across investment horizons. Its explanation is conceptual and applies within the stated model assumptions; it is not a general forecast of realized returns for a particular asset.
Key ideas
- Expected arithmetic return and continuously compounded growth are distinct quantities.
- In the stated model, volatility lowers the growth rate relative to the drift parameter.
- A sequence of variable returns can compound to less growth than its arithmetic average suggests.
- The rectangle analogy illustrates how variability reduces an aggregate outcome relative to equal average inputs.
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# Answer by Charles Fox (score 1, accepted)
# Which are the practical implications that the continuously compounded rate of return can be smaller than the expected rate of return?
I'm reading Hull's Options, Futures and other Derivatives and it intrigues me that the distribution of the continuously compounded rate of return x is: $x \sim \phi(\mu - \frac{\sigma^2}{2}, \frac{\sigma^2}{T})$
This happens when: $\frac{\Delta S}{S} \sim \phi(\mu \Delta t, \sigma^2\Delta t)$
My question is: what are the practical implications of $\mu - \frac{\sigma^2}{2} < \mu$? Does it mean that although most of the times your return will be less than $\mu$, your expected return is $\mu$?
What does Ernest P. Chan means when he says:
> Suppose a certain stock exhibits a true (geometric) random walk, by which I mean there is a 50-50 chance that the stock is going up 1% or down 1% every minute. If you buy this stock, are you most likely, in the long run, to make money, lose money, or be flat? ... Most traders will blurt out the answer “Flat!”, and that is wrong. The correct answer is you will lose money, at the rate of 0.005% every minute!
Does it mean that although most of the times your return will be less than μ, your expected return is μ?
## Answer by Charles Fox (score 1, accepted)
https://quant.stackexchange.com/a/45123
The key word in your question is compounded. The expected arithmetic return for each $\Delta t$ is $\mu$, but the growth rate is $\mu - \frac{\sigma^2}{2}$. As others mentioned, volatility reduces the growth rate.
This is similar to the area of a rectangle. If one rectangle has sides 3 and 1, its average side length is 2, and its area is 3. If another is a square with length 2, its average side length is also 2, but the area is 4. Variability reduces area relative to the average side length.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.