Solved – Why is a deterministic trend process not stationary

intuitionstationaritytime series

I am confused on why a simple trend process is not stationary. Consider the following process:

$Y_t = a + bt + \epsilon_t$

The variance is clearly constant. However, the mean $bt$ is dependent on $t$. When shifted in time, the mean only depends on the time interval and is independent of history. For example, $Y_{0,t}$ and $Y_{t,2t}$ would have the same mean and variance. So why is this process not stationary?

Secondly, if we now consider the following process:

$Y_t = a + \sqrt{t}\epsilon_t$

Assume $\epsilon_t$ is standard normal.

In this case, the mean is constant, however, the variance is dependent on $t$. However, in this case the variance is proportional to the time interval, which means $Y_{0,t}$ and $Y_{t,2t}$ would have the same mean and variance. So why is this process not stationary?

If you could explain it intuitively rather than definition/proof that would be helpful. My understanding of a stationary process is that the first two moments of the process (mean and variance) remain the same when shifted in time or space. I don't think I have the right understanding of stationary processes.

Best Answer

I think I nice way to get the intuition is to simulate 3 series for $t=0,...,500$ and plot them:

Autoregressive Stationary Series: $A_{t}=0.05+0.95A_{t-1}+u_{t}$
Random Walk with Drift: $R_{t}=0.05+1R_{t-1}+u_{t}$
Explosive Series: $E_{t}=0.05+1.05E_{t-1}+u_{t}$

where $u_{t}$ is just some white noise, like iid $N(0,1)$.

Look at $A$ and $R$:

enter image description here

The theoretical mean of $A$ is $1$ (red horizontal line) and its standard deviation is $3.2$. The graph will deviate from that mean over time, but not too far. $R$ will look qualitatively similar to $A$ early on, but begins to drift apart in the middle, but converges towards the end. In theory, the unconditional mean and variance of $R$ do not exist, and you can see that in the graph.

Now plot all 3 series on a graph with the same scale.

enter image description here

Can you see how $E$ just makes the other two look like a straight line? The slope parameter in $E$ exceeds $1$ by 0.05, the same amount that it falls short for $A$, but what a difference it makes! Here, the average makes no sense at all.

The other point is that $A$ and $R$ look like the sorts of things we see every day, but we are bad at guessing which ones are stationary, especially with fewer data.

This is shamelessly plagiarized from Econometric Methods by Jack Johnston and John DiNardo, which is sadly out of print.

Related Solutions

Solved – Is a model with a sine wave time-series stationary

Stationarity is a property of a stochastic process. A perfect sine wave is not a stochastic process. Hence, it can't be stationary or non-stationary. It doesn't have any random parts.

$$y_t=\sin (\phi t+\theta)$$

It's like asking whether a song is black or white. The music has no color, it has many other properties but color is not one of them.

Now, you could look at the problem differently. As you wrote the phase and frequency are unknown. So, if you look at the family of processes: $$y_t=\sin (\phi_i t+\theta_i)$$ Where $\phi_i,\theta_i$ come from some distribution, and you're to estimate $E[y_t]$, then it's a more interesting question. It's still not a stochastic process though.

The stochastic process represents an evolution of random variables. In the case of a perfect sine wave it's entirely defined by two random variables $\phi_i,\theta_i$. There's no evolution.

In other words there's got to be some kind of randomness and uncertainty introduced as time progresses in order for the process to be stochastic. In your case all the uncertainty is introduced at time 0.

Solved – Is a Markov chain with a limiting distribution a stationary process

Here's a simple example illustrating why the answer is no.

Let $$P = \begin{pmatrix} 0.5 & 0.5 \\ 0.5 & 0.5 \end{pmatrix}$$ be the transition matrix for a first-order Markov process $X_t$ with state space $\left\{0, 1\right\}$. The limiting distribution is $\pi = \left(0.5, 0.5\right)$. However, suppose you start the process at time zero with initial distribution $\mu = \left(1, 0\right)$, i.e. $X_0 = 0$ with probability one.

We then have $\mathbf{E}[X_0] = 0 \neq \mathbf{E}[X_1] = \frac{1}{2}$, meaning the moments of $X_t$ depend on $t$, which violates the definition of stationarity.

Here's some R code illustrating a similar example with

$$P = \begin{pmatrix} 0.98 & 0.02 \\ 0.02 & 0.98 \end{pmatrix}$$

p_stay <- 0.98
P <- matrix(1 - p_stay, 2, 2)
diag(P) <- p_stay
stopifnot(all(rowSums(P) == rep(1, nrow(P))))
mu <- c(1, 0)
pi <- matrix(0, 100, 2)
pi[1, ] <- mu
for(time in seq(2, nrow(pi))) {
    pi[time, ] <- pi[time - 1, ] %*% P
}
plot(seq(1, nrow(pi)), pi[, 1], type="l", xlab="time", ylab="Pr[X_t = 0]")
abline(h=0.5, lty=2)

The fact that $X_t$ is converging in distribution to some limit does not mean the process is stationary.

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Related Solutions

Solved – Is a model with a sine wave time-series stationary

Solved – Is a Markov chain with a limiting distribution a stationary process

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