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There seems to be some disagreement in the literature about this. Define predicability of a stationary series to be $\sigma^2_{t-1} / \sigma^2_t$

Finding mean reverting portfolios using canonical correlation analysis means minimizing predictability, while searching for portfolios with strong momentum can also be done using canonical correlation analysis, by maximizing predictability.

https://www.di.ens.fr/~aspremon/PDF/MeanRevVec.pdf

The traditional way to identify the optimal sparse mean-reverting portfolio is to find a portfolio vector subject to maximizing its predictability.

https://content.iospress.com/download/algorithmic-finance/af021?id=algorithmic-finance%2Faf021

The intuition behind this portfolio predictability is that the greater this ratio, the more $s_{t−1}$ dominates the noise, and therefore the more predictable $s_{t}$ becomes. Therefore, we will use this measure as a proxy for the portfolio’s mean reversion parameter $\lambda$ in (1). Maximizing this expression will yield the following optimization problem for finding the best portfolio vector $x_{opt}$

https://content.iospress.com/download/algorithmic-finance/af026?id=algorithmic-finance%2Faf026

Personally, I don't see any reason why a predictable time series should be necessarily trending or mean-reverting.

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  • $\begingroup$ If a stochastic process $\{x_t\}$ is stationary and ergodic then it must be mean reverting in the sense that if $x_t$ is unusually high or low, you can expect it to eventually decrease or increase respectively. A stationary ergodic process can't explode forever in some direction or get stuck. Over finite periods though, a stationary ergodic process can trend in some direction, like the tide flowing out for hours but later coming back in. $\endgroup$ – Matthew Gunn Aug 31 '18 at 14:58
  • $\begingroup$ @MatthewGunn thank you. The literature is talking about price processes. I believe there is no a priori assumption that the process is stationary, rather the goal is to find weights such that the portfolio becomes stationary or mean-reverting. I am familiar with time series terminology, mainly wanted to clarify the contradiction in the linked papers. $\endgroup$ – Edward Yu Aug 31 '18 at 18:11
  • $\begingroup$ Part of the issue may also be that "mean reversion" isn't a terribly well defined term. $\endgroup$ – Matthew Gunn Aug 31 '18 at 21:14
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The point of confusion may be in thinking that a predictable price process is synonymous with a mean-reverting process while using the definitions in these papers, it's actually the opposite! In the context of these papers, a random walk would be 100% predictable: the unpredictable component of a random walk (i.e. the period specific shock which has finite variation), comprises 0% of the process's total variation (which is infinite).

Some broader points of caution

Be cautious when an author gives an English language word a mathematical definition that may not perfectly align with the word's common use in English or a particular field.

Also, predictable is not the same as exploitable. You can use information on the relative strength of teams to predict Vegas, sports-betting lines. That's different though than whether the sports betting lines are exploitable!

Back to this case...

Consider an AR(1):

$$ x_t = b x_{t-1} + \epsilon_t$$

  • $b=1$ has no mean reversion in the sense that shocks are entirely persistent. It is also most predictable in the sense that as $b \rightarrow 1$, the fraction of the process's total variation that is forecastable also goes to 100 percent.
  • $b=0$ has entirely transitory shocks and in that sense is most mean reverting. It's always expected to move back to the unconditional mean! It is also the least predictable in the sense that 0 percent of the processes variation can be forecast.

With an AR(1) structure, shocks decay in a simple, exponential fashion. With higher order lags (which these papers don't do), you can instead get more complicated behavior such as cycles and, at some points, step ahead forecasts further away from the mean.

These papers assume prices are a stationary process (rather than containing a unit root) and that the price process takes a simple, 1 lag autoregressive structure. (I'll sidestep a whole discussion as to if and when that's useful or realistic.)

Box and Tiao Decomposition

Let $\{z_t\}$ be a stationary process. Define $\hat{z}_{t-1}$ as the expectation of $z_t$ based upon $t-1$ info: $$ \hat{z}_{t-1} = \mathbb{E}[ z_t \mid z_{t-1}, z_{t-2}, \ldots ]$$ Box and Tiao then decompose total variation $\sigma^2_z$ into a predictable component (the step ahead forecast) $\sigma^2_\hat{z}$ and an unpredictable component $\sigma^2_\epsilon$: $$ \underbrace{\mathbb{E} \left[ z_t^2 \right]}_{\sigma^2_z} = \underbrace{\mathbb{E}\left[ \hat{z}_{t-1}^2\right]}_{\sigma^2_{\hat{z}}} + \underbrace{\mathbb{E}[\epsilon^2_t]}_{\sigma^2_\epsilon}$$

They then define the predictability ratio $\lambda = \sigma^2_\hat{z} / \sigma^2_z$. If $\lambda = 0$, none of the total variation comes from variation in the step ahead forecast. If $\lambda \approx 1$, then nearly all of the total variation comes from the step ahead forecast.

Simple AR(1) case (what's effectively in these papers)

Let's assume we have a simple, mean zero AR(1):

$$ x_t = b x_{t-1} + \epsilon_t$$

Assume $-1 < b < 1$ so the process is stationary. The unconditional variance is $\sigma^2_x = \frac{1}{ 1 - b^2}\sigma^2_\epsilon$. The Box Tiao decomposition is:

$$ \underbrace{\frac{1}{ 1 - b^2}\sigma^2_e}_{\sigma^2_x} = \underbrace{\frac{b^2}{1 - b^2}\sigma^2_\epsilon}_{\sigma^2_\hat{x}} + \sigma^2_\epsilon $$

The predictability ratio is: $$ \lambda = b^2 $$

Discussion

Mean reversion isn't a precisely defined term.

Finding mean reverting portfolios using canonical correlation analysis means minimizing predictability...

$b=0$ has high mean reversion in either the sense: (1) the step ahead forecast is always the unconditional mean or (2) shocks are entirely transitory. $b = 0$ leads to $\lambda = 0$, minimum predictability.

... while searching for portfolios with strong momentum can also be done using canonical correlation analysis, by maximizing predictability.

When $b$ is close to 1, the process is close to a random walk. The author seems to be calling this momentum. I find that usage of "momentum" rather problematic.

The traditional way to identify the optimal sparse mean-reverting portfolio is to find a portfolio vector subject to maximizing its predictability.

Be aware that in the context of an AR(1) in prices, maximizing predictability implies finding a price process that decays towards its unconditional mean price as slowly as possible. A random walk in prices would have the most predictability (in prices).

The intuition behind this portfolio predictability is that the greater this ratio, the more $s_{t−1}$ dominates the noise, and therefore the more predictable $s_{t}$ becomes. Therefore, we will use this measure as a proxy for the portfolio’s mean reversion parameter $\lambda$ in (1). Maximizing this expression will yield the following optimization problem for finding the best portfolio vector $x_{opt}$.

As I discussed earlier $\lambda = b^2$ in the AR(1) context.

References

Box, G.E.P. and G.C. Tiao, "A canonical analysis of multiple time series," 1977, Biometrika

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In my opinion, the statement “absence of trend or mean reversion in a series does not necessarily prevents predictability” is absolutely correct in practice, if this is your question. Think about cointegration between two series: in that case we have two series of order 1 (like random walks) whose linear combination is a stationary time series. In that case, you take two series which are not necessarily supposed to be mean reverting or show stable deterministic trends in their conditional mean and you predict their future behavior based on their long-term equilibrium relationship (i.e. you find that the deviations from that relationship are just transitory and will sooner or later revert to their long term mean of 0, that is what Error Correction Models says, despite obvious simplifications). Ps: this is my opinion, but maybe I misunderstood the question in this case. To sum up, I would say that I agree with you.

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