If we assume the assets returns are stationary then the best forecast can only be the mean of the distribution.

But if we assume non-stationarity we are forecasting the mean parameter (assuming normal distribution) using either linear or non-linear models. And to emphasize we are forecasting the mean rather than some exact value of the return distribution domain.

Is my understanding correct?


2 Answers 2


This looks confused? I don't understand what you're saying in the second paragraph...

Comment 1: "Best" forecast depends on what you mean by "best."

Let $Y$ be a random variable and $\mathcal{F}$ be the information set. The "best" forecast depends on what the loss function is. If you're minimizing the expectation of squared loss:

\begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $x$)} & \mathbb{E}[(Y - x)^2 \mid \mathcal{F}] \end{array} \end{equation}

You have the solution that $x$ is the conditional expectation of $Y$ given information set $\mathcal{F}$ $$x^* = \mathbb{E}[Y \mid \mathcal{F}]$$

Of course you can have other loss functions. Consider minimizing the expected absolute error: \begin{equation} \begin{array}{*2{>{\displaystyle}r}} \mbox{minimize (over $x$)} & \mathbb{E}[|Y - x| \mid \mathcal{F}] \end{array} \end{equation} The solution here is that $x$ is the median of $Y$. Let $F^{-1}_Y$ be the quantile function for $Y$ conditional on information set $\mathcal{F}$.

$$ x^* = F_Y^{-1}(.5 \mid \mathcal{F})$$

Comment 2: The importance of stationarity

Let $\{ Y_t\}$ be a stochastic process. $Y_1$, $Y_2$, $Y_3$ etc... are all random variables.

Speaking with imprecise language:

  • Stationarity says that the unconditional distribution of $Y_1$ is the same as $Y_2$ is the same as $Y_3$ is the same as $Y_{1000}$ etc...
  • Ergodicity says that process doesn't get stuck somewhere.

Stationarity says that you can talk about a time invariant expectation $\mathbb{E}[Y]$. With ergodicity, a time-series mean $\frac{1}{T} \sum_{t=1}^T Y_t$ will estimate that time invariant expectation (by an ergodic theorem). With stationarity and ergdocity, averages over time converge to averages over space.

With a non-stationary series, that's not true! Example. Let $\{X_t\}$ be a stochastic process. Let $X_1$ be result of a die roll. Let $X_2$ be winning total of the Golden State Warriors vs. the LA Lakers. Let $X_3$ be the number of votes cast for BRexit. Let $X_4$ be the return on Apple stock December 10th. If I found a way to keep doing this, $X$ would be a non-stationary process. Talking about the expectation $\mathbb{E}[X]$ is non-sensical. There is no time-invariant expectation. And taking the average over time of $X$ does nothing useful at all.

(Note: Often times people have in mind a random walk when talking about a non-stationary process. In the case of a random walk, the first differences are stationary.)


If we assume the assets returns are stationary then the best forecast can only be the mean of the distribution.

This part is not accurate. Stationarity, even in its strongest sense, only implies that the unconditional distributions will be the same for every time index. Conditional distributions don't have to agree.

For instance, if the process $...,X_{-1},X_0,X_1,...$ is stationary, then $E(X_{t+1}) = E(X_t)$. However, $E(X_{t+1})$ doesn't have to agree with $E(X_{t+1}|X_t)$.

Of course, asset returns turn out to be very difficult to forecast, but this doesn't follow from any assumption of stationarity alone.

  • $\begingroup$ I am guessing the statement you quote is only valid if we assume a very naive view of the returns coming from a univariate distribution (i.e. being unconditional). However most likely they are coming from a multidimensional distribution (where they are conditional on other values) making the statement inaccurate. Hence we forecast the mean either to take account for that conditionality or/and to take into account non-stationarity. $\endgroup$ Mar 28, 2018 at 13:05
  • $\begingroup$ @a-l-verminburger Yes, if you in addition require the returns to be independent then that's correct. $\endgroup$
    – Winkelried
    Mar 28, 2018 at 14:15
  • $\begingroup$ Is independence not the same thing as stationarity? If it is stationary then at each subsequent moment in time I am sampling from the same distribution (with the same domain), e.g. tossing a coin. But if it is nonstationary (say a simple random walk) then the domain of the distribution I am sampling from depends on the realisation from the previous timestamp (i.e. is dependent on past a event) -- thinking of something like total earnings from tossing a coin. $\endgroup$ Mar 28, 2018 at 14:36
  • $\begingroup$ @Winkelfried OK, so when you say "require the returns to be independent" you mean non-conditional. As I wrote in the first comment this is likely a naive view of returns and agree with you that would be the only scenario when the quoted statement would work. Non-stationarity is a little broader concept. It may encompass something like conditionality on previous distributions or it may be a parameters / domain of at current time being a function of just time (unrelated to previous distributions). $\endgroup$ Mar 28, 2018 at 15:18
  • $\begingroup$ @a-l-verminburger Stationarity and independence are different notions. Neither of which implies the other. In a stationary process, the distribution (and also its domain) you sample from at a given time can depend on the previous timestamp. This is what I pointed out in the answer. A random walk is non-stationary but not because the distribution depends on the previous timestamp, but for other reasons, see: stats.stackexchange.com/questions/246357/… $\endgroup$
    – Winkelried
    Mar 28, 2018 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.