Are asset return means difficult to predict because they have no lower bound?

In finance, it is widely known that the volatility of asset returns ($$\sigma$$) are easier to predict than the expected value of asset returns ($$\mu$$) , otherwise known as the average return or mean.

Is this partly due to the fact that asset volatility is restricted to be a positive value ($$\sigma \in (0,+\infty)$$), whereas asset returns and mean can take on negative percentage values ($$\mu \in (-\infty,+\infty)$$)? If so, why would positive boundedness of a variable makes its estimation more reliable and lower estimation error?

• heuistically speaking, the volatility is a measure of variability of the observations from the population and the mean is a measure of the centrality of the population so it ( volatility ) doesn't move around as much. As the sample observations ( from the population ) arrive at every new observation period, the volatility is a measure of the spread of them so it is going to tend to change less than a measure of the center of them. The fst that volatiity is positive is unrelated to why it's "easier" to estimate. – mark leeds May 28 at 3:51
• for asset returns, i think volatility moves around more than their mean though – develarist May 28 at 9:48
• good point. even the term volatility is somewhat ambiguous because, with all these things, we're constructing estimates when we already know that there is no true underlying volatility parameter. Same with the true mean. But, if someone owned stock XXX and he-she asked me to predict either its return tomorrow or its volatility tomorrow and I had to get closer than someone else's prediction in order to win the prize, I'd choose volatility. – mark leeds May 28 at 22:35

To answer, the assertion that volatility is easier to predict than expected return requires clarification. The phrase "easier to predict" is particularly ambiguous.

To me this means that the estimation of volatility from a sample of returns is more robust than the estimation of expected return in the context of relative sampling error.

Suppose over a time period $$T$$ we observe asset prices $$S_0,S_1, \ldots, S_N$$ at uniformly spaced time intervals of length $$\delta t$$ where $$T = N \delta t$$. Assume that the log-return (over an interval of length $$\delta t$$) has a stable distribution and returns over non-overlapping intervals are independent. Let $$\mu$$ and $$\sigma$$ denote the annualized mean return and volatility, respectively.

The $$\delta t$$-period log-return has expected value $$\mu \delta t$$ and variance $$\sigma^2 \delta t$$, where the $$\delta t$$ scaling of the variance is a consequence of the independence. We now have an iid sample $$X_1,X_2,\ldots, X_N$$ where

$$X_j = \log \frac{S_j}{S_{j-1}}$$

and the estimators for expected retrun and volatility are

$$\hat{\mu}\delta t = \frac{1}{N}\sum_{j=1}^N X_j, \quad \hat{\sigma}^2\delta t = \frac{1}{N-1}\sum_{j=1}^N (X_j - \hat{\mu}\delta t)^2$$

Asymptotically, the sampling distributions for the estimators are

$$\hat{\mu}\delta t \sim \mathcal{N}(\mu \delta t, \sigma^2 \delta t/N),\quad \frac{(N-1) \hat{\sigma}^2 \delta t}{\sigma^2 \delta t} \sim \chi^2(N-1),$$ that is, normal and chi-square with $$N-1$$ degrees of freedom, respectively. The standard errors for the estimates of expected return and volatility are, respectively, $$\sigma\sqrt{\frac{\delta t}{N}}$$ and $$\frac{\sqrt{2} \sigma^2 \delta t}{\sqrt{N-1}}$$.

As expected, the absolute sampling error (given by standard error) for the both the expected return and the volatility diminish as $$1/\sqrt{N}$$ as the number of samples $$N$$ increases.

However, the relative errors tell a different story. The relative sampling error for the volatility is

$$\frac{\frac{\sqrt{2} \sigma^2 \delta t}{\sqrt{N-1}}}{\sigma^2 \delta t} = \sqrt{\frac{2}{N-1}}$$

This shows that the relative error improves simply by increasing the number of samples. Given a fixed time period $$T$$, we only need to sample returns at a higher frequency to improve the estimate of volatility. Sampling daily is more accurate than sampling monthly, sampling monthly is more accurate than sampling quarterly, etc.

On the other hand, the relative sampling error for the expected return is

$$\frac{\sigma \sqrt{\frac{\delta t}{N}}}{\mu \delta t} = \frac{\sigma}{\mu \sqrt{N \delta t}}= \frac{\sigma}{\mu \sqrt{T}}$$

The only way to get a better estimate for expected return is to increase the length of the period $$T$$ over which the samples are observed. For a fixed period $$T$$, say 3 years, the relative error cannot be improved by increasing the sampling frequency, regardless of how many additional samples are taken. In other words, in order to improve the accuracy of the estimated return by a factor of 5, we must increase the sampling period by a factor of 25 to 75 years -- clearly problematic.

The root cause of this phenomenon would seem to be the fact that return scales like $$\delta t$$ and volatility, with independent returns, scales like $$\sqrt{\delta t}$$ with respect to the measurement period $$\delta t$$.

• The TLDR summary: "Given a fixed time period T, we only need to sample returns at a higher frequency to improve the estimate of volatility. [Whereas t]he only way to get a better estimate for expected return is to increase the length of the period T over which the samples are observed." – noob2 May 28 at 21:51
• @noob2: What's TLDR ? thanks. Also, with that statement, are they assuming that volatility is iid so that in order to get the day's volatility from say the minute volatility, you can just multiply by the appropriate scale factor ? If so, that's quite an assumption. – mark leeds May 28 at 22:40
• This answer doesn't look like it addresses the original question regarding the mean having no lower bound, and volatility having a lower bound at 0. – develarist May 28 at 22:48
• @develarist: Because by that question you are speculating that lower bound is the reason for volatitlity to be "easier to predict" than expected return. First, you never made it clear what "easier to predict" means in a statistical sense. You say it is well known. What exactly is well known? Can you give me a reference? What is well known going back to work by Robert Merton is the so called Record Problem which defines more precisely that statement -- and this is what I am addressing here. – RRL May 28 at 23:15
• @markleeds: You are correct in that iid -- not to mention stationarity -- is quite an assumption. This is just a toy example merely to illustrate differences in the accuracy of estimating expected return versus volatility. I am under no illusion that asset return distributions are sufficiently stable to make estimation of expected return with historical data a practical or meaningful exercise. Empirical evidence does suggest that volatility is inherently more stable. – RRL May 29 at 0:09

The essential difference arises not from the lower bound on volatility, but rather the fact that volatility is mean-reverting and asset values are not.

To make this clearer, note that a period-$$T$$ return prediction $$\hat{r}=\hat{r}_T^{(0)}$$ at time $$t=0$$ for an asset with price $$P_0$$ is equivalent to a price prediction of $$P_T=P_0 e^{\hat{r} T}$$. And, of course prices are bounded below by zero just as volatility is. And yet, they are harder to predict than volatility.

The real difference is that any sane stochastic model for volatility has mean reverting terms, for example

$$d \sigma = \kappa (\sigma_0 - \sigma) dt + \eta \sigma^p dW$$

which for reasonable values of $$\kappa, \eta, p$$ cannot go below zero. A long-term average of $$\sigma$$ is then a good estimate of $$\sigma_0$$ and hence of long-term future volatility.

In contrast, reasonable stochastic models for $$P$$ have no such mean reversion, and the simplest ones like Black-Scholes can be proven to wander infinitely far from their initial values. Thus the returns themselves can be infinitely far from zero as well, making them much harder to predict than mean-reverting quantities.