# In mathematics, what is meant by "Mean Reversion"?

I am trying to understand the concept of "Mean Reversion" from a mathematical perspective. For example, suppose we are given a generic Stochastic Process - how do we know if this process is "mean reverting" or not?

Part 1: As an example, consider the following Stochastic Processes:

1. ARIMA(p, d, q) Process:

$$X_t = c + \sum_{i=1}^{p} \phi_i X_{t-i} + \sum_{i=1}^{q} \theta_i \epsilon_{t-i} + \epsilon_t$$

• p (order of the autoregressive part)
• d (degree of first differencing involved)
• q (order of the moving average part).
• $$\epsilon_t$$ is white noise with zero mean and constant variance.
1. Brownian Motion: A Weiner Process $$W_t$$ is a stochastic process that satisfies the following properties - a Browian Motion is the cumulative sums of the Weiner Process:
• $$W_0 = 0$$
• The increments are independent and normally distributed: $$W_t - W_s \sim N(0, t-s)$$ for $$0 \leq s < t$$.
1. Ornstein-Uhlenbeck Process: The Ornstein-Uhlenbeck process is a stochastic process that satisfies the following stochastic differential equation:

$$dX_t = \theta(\mu - X_t)dt + \sigma dW_t$$

• $$\theta > 0$$ is the rate of reversion to the mean
• $$\mu$$ is the mean value, $$\sigma$$ is the standard deviation
• $$W_t$$ is a Wiener process.

If I just look at the three equations, I would not be able to know which of these processes are mean reverting and which of them are not.

Part 2: Using the R programming language, we can plot these a single trajectory for these 3 processes:

library(ggplot2)

n <- 1000

# ARIMA(1,0,0) Process
set.seed(123)
arima_process <- arima.sim(n, model=list(ar=0.6))

# Brownian Motion
set.seed(123)
brownian_motion <- cumsum(rnorm(n))

# Ornstein-Uhlenbeck Process
set.seed(123)
theta <- 1; mu <- 0; sigma <- 1
ou_process <- numeric(n)
for (t in 2:n) {
ou_process[t] <- ou_process[t-1] + theta*(mu - ou_process[t-1]) + sigma*rnorm(1)
}

df <- data.frame(Time = 1:n,
ARIMA = arima_process,
Brownian = brownian_motion,
OU = ou_process)

ggplot(df, aes(Time)) +
geom_line(aes(y = ARIMA, color = "ARIMA")) +
geom_line(aes(y = Brownian, color = "Brownian Motion")) +
geom_line(aes(y = OU, color = "Ornstein-Uhlenbeck")) +
labs(x = "Time", y = "Value", color = "Process") +
theme_minimal()


Clearly, it looks like the Ornstein-Uhlenbeck Process and the ARIMA Process "hover" around the mean whereas the Brownian Motion does not. Does this mean the Brownian Motion is not mean reverting?

I also plotted multiple simulations for each process:

df_arima <- data.frame(Time = rep(1:n, n_simulations), Value = numeric(n * n_simulations))
df_brownian <- data.frame(Time = rep(1:n, n_simulations), Value = numeric(n * n_simulations))
df_ou <- data.frame(Time = rep(1:n, n_simulations), Value = numeric(n * n_simulations))

set.seed(123)
for (i in 1:n_simulations) {
# ARIMA(1,0,0) Process
arima_process <- arima.sim(n, model=list(ar=0.6))
df_arima$Value[(i-1)*n + 1:n*i] <- arima_process # Brownian Motion brownian_motion <- cumsum(rnorm(n)) df_brownian$Value[(i-1)*n + 1:n*i] <- brownian_motion

# Ornstein-Uhlenbeck Process
theta <- 1; mu <- 0; sigma <- 1
ou_process <- numeric(n)
for (t in 2:n) {
ou_process[t] <- ou_process[t-1] + theta*(mu - ou_process[t-1]) + sigma*rnorm(1)
}
df_ou$Value[(i-1)*n + 1:n*i] <- ou_process } # Plots p_arima <- ggplot(df_arima, aes(Time, Value)) + geom_line(alpha = 0.1, color = "blue") + labs(x = "Time", y = "Value", title = "(Multiple) ARIMA(1,0,0) Processes") + theme_minimal() p_brownian <- ggplot(df_brownian, aes(Time, Value)) + geom_line(alpha = 0.1, color = "red") + labs(x = "Time", y = "Value", title = "(Multiple) Brownian Motions") + theme_minimal() p_ou <- ggplot(df_ou, aes(Time, Value)) + geom_line(alpha = 0.1, color = "green") + labs(x = "Time", y = "Value", title = "(Multiple) Ornstein-Uhlenbeck Processes") + theme_minimal()  I might be hallucinating, but I can see some faint "bands" in the Brownian Motion simulations which might be illustrating the same concept of non-mean reversion? My Question: From a strict mathematical perspective, just by looking at the mathematical equation for a stochastic process - is it possible to determine if the process is mean-reverting or non-mean reverting? Thanks! • Note: Does "mean reversion" have anything to do with "memoryless property" or "stationarity"? • I think the bands are a graphical artifact. Are they still there if you zoom in? As for telling if a process is mean reverting at a glance, you can at least check that the coefficient of the$X_t$term of$dX_t$is negative. This is prevents X from getting too big or too negative. – Mark Commented Dec 31, 2023 at 8:54 ## 3 Answers Note: Does "mean reversion" have anything to do with "memoryless property" or "stationarity"? Mean reversion refers to the concept that given some observation of a random variable, the next event is more likely to be closer to the mean that this one was. In other words, given an extreme event, the next event is more likely to be less extreme. So, kind of to the memoryless property, in some sense, but generally not really. For stationarity yes! If it's variance or mean is not constant over time, then the memoryless property won't really happen in terms of values and not relative portions. Since stationarity demands that we have a constant mean and variance over time, and random variable/stoachastic process/time series that obeys stationarity also obeys the mean reversion principle. They are pretty closely tied together, and they may be synonymous assuming finite variance and mean, but I'm unsure. You ask why Brownian motion does not have mean reversion. We can see this very easily in the definition, where: $$W_t - W_s \thicksim \mathcal{N}(0, t-s)$$ The variance of our Brownian motion compared to the start is increasing over time. Mean reversion works because "as you move away from the mean, the proportion of the distribution that lies closer to the mean than you do increases continuously." but this doesn't happen when our variance is changing at a linear rate. Less amount of the distribution lies closer to the mean than it did $$1$$ weiner process ago (heuristically speaking), so it doesn't make sense that we should be more likely to fall within the same bounds. Basically, it means that over the long term, experimental probabilities revert to their theoretical "mean" probability or expected value. For example, let's say a basketball player is a 75% free-throw shooter. One game, he makes 72% of his free-throws, the next game he makes 87% of his free-throws, and the third game he makes 77% of his free-throws. The 3-game average is 78.67%. Assuming he stays at his same skill level, the more games he plays, the closer his average over all the games will approach 75%, regardless of whether he has a "hot streak" or a "cold streak". Same thing with recorded weather measurements in a given climate, etc. Mean reversion has nothing to do with the memoryless property but, if you prefer my definition of a mean-reverting process, then it is very related to stationarity, In the hedge fund world, people use the term "mean reverting" strategy which means that the strategy waits for the return ( or price ) of the stock to "come back" from its "unusual state". So, interpret that as you wish but, it basically means "Is there a long term mean that the process will eventually return to no matter how far process goes away from that long term mean ?". So, by the definition above (which is my definition and not necessarily everyone's), any stochastic process which has a long term ( and constant ) mean, ( along with a constant variance ) could be could be considered a mean reverting process. In your question, you mentioned the OU and Brownian Motion (BW) and the AR(1). The AR(1) should exhibit mean reversion ( to zero ) because the long term mean is zero and the process is stationary. BW cannot be mean-reverting because it doesn't have a stationary mean. In fact, the process itself doesn't have a long term mean. If you simulate a BW and it looks like it's mean reverting, then that's purely random and not a characteristic of BW. OU on the other hand is mean reverting. You may be able to see why it's mean reverting better in the discrete version of it which is an AR(1) with certain parameters. I'll see if I can find something and include it here. So, to try to answer your question, if you have an equation for a stochastic process and the process is stationary, one can technically categorize that process as mean reverting. OU is a specific mean reverting process ( continuous also ) where the mean that the process reverts to is specified in the model along with the speed of reversion. This is very convenient but one can achieve the same type of behavior using the discrete AR(1) counterpart of the OU. # ADDENDUM: 12-28-2023 #=============================================================== Below is a thread where the discrete AR(1) is derived from the OU. I didn't look at it closely and it seems to use a different parameterization than you used, but I just wanted to make the point that the AR(1) can do the same thing that the OU does in continuous time, as long as it's stationary. OU looks nicer though because the parameters have a very clear interpretation. How to approximate an AR(1) with an Ornstein-Uhlenbeck Process • thank you so much for your answer. Reading your comment :"BW cannot be mean-reverting because it doesn't have a stationary mean. In fact, the process itself doesn't have a long term mean." .Just by looking at the equation for a Brownian Motion - how can I know that the mean of the Brownian Motion is non-stationary? Is there some analysis I can do to see this? Thank you so much! Commented Dec 30, 2023 at 5:05 • Hi: Take an increment of Brownian motion so, say$(X_{t_{1}} - X_0)$. The variance of this increment is$\sigma^2 \times t_{1}$. Now, suppose we increase the same increment so consider say$(X_{t_{2}} - X_0)$where$t_{2} > t_{1}$. The variance of this increment is$\sigma^2 \times t_{2}$. So, as the increment gets bigger and bigger , the variance gets larger.and larger There's probably a more mathematical way of saying this but, as the variance increases without bound, the mean becomes non-existent. The process, when viewed as one long increment, is not stationary. Commented Dec 30, 2023 at 9:22 • @stats_noob : I just read what Robert Murray wrote below and I think we are kind of saying the same thing as far as why BM can't be stationary. Particular overlap is near the end of his statement. Commented Dec 30, 2023 at 9:28 • "BW cannot be mean-reverting because it doesn't have a stationary mean". What do you mean it doesn't have a 'stationary mean'? The mean of Brownian motion is zero for all$t\$, so it certainly exists. Commented Jan 19 at 20:21
• Hi Jose: brownian motion has a mean (E(W_t|F_t) = 0 but an long term equilibrium doesn't exist. By this I mean that, once the process goes away from zero, there is no tendency for the overall long term mean to go back towards zero. So, I shouldn't have said the mean doesn't exist but rather than brownian motion does not have an equilibrium. Maybe that's due to the increasing variance as time passes. Commented Jan 21 at 1:59