Stationarity, Conditional Forecasts, and Loss Functions

# Does forecasting asset returns by default assumes non-stationarity of asset returns?


# Does forecasting asset returns by default assumes non-stationarity of asset returns?












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?

## Answer by Matthew Gunn (score 2)

https://quant.stackexchange.com/a/37228

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.)

## Answer by Winkelried (score 1)

https://quant.stackexchange.com/a/39038

> 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.