# Is a model with a sine wave time-series stationary?

Today I was introduced to the definition of stationarity being that the marginal distribution of a process does not change over time, and the mean and variance remain constant over time. I questioned whether a process with a perfect sine-wave with an unknown starting point and unknown period would be stationary, as it is purely deterministic however the mean seems to change with time.

Asking my professor following class, his answer was that such a process would indeed be stationary, despite the mean changing. However on this answer I found on CV, the mean changing is cited as a reason for such a process being non-stationary. Now I'm a little bit confused about the definition of a stationary process, and still unsure about the stationarity of a sine wave.

Am I missing some point here? Or is there varying definitions for a stationary process?

• stationary means $y_t = y_{t-1} +\epsilon$ where $\mu(\epsilon) = 0$. If the process is deterministic, then $\epsilon = 0$ and $y_t = y_{t-1}$ on the nose. – aginensky Sep 17 '15 at 20:15
• @aginensky That's a new definition to me, but makes sense. So under this definition a sine wave time-series would be stationary, correct? – miradulo Sep 17 '15 at 20:25
• No, under this definition the sine wave is not stationary (either weakly or strongly). Weak stationarity requires that $E(X_t) = E(X_{t+\tau})$ where $\tau$ is a real number. If $X_t = sin(t)$, then $X_t$ is deterministic and $E(X_t) = sin(t)$. Since the expectation is not constant, the process is not stationary. – BLimkins Sep 17 '15 at 21:06
• @user44107 Now I'm more confused :P So in addition to meaning that $y_{t} = y_{t-1} + \epsilon$ where $\mu(\epsilon) = 0$, we require your expectation property. Thus making this process non-stationary? Under what definition of stationarity would my professor be correct then? He just loosely defined it as I did in my question. – miradulo Sep 17 '15 at 21:13
• He might have said that because often when people look for evidence of non-stationary, they look for trends or sustained deviations from the mean that might indicate deterministic trends or random walks. It is tempting to think of a sin wave as stationary because it is mean reverting, but it is not technically correct. I would recommend checking out chapter 1 of this book, which is free and easy to understand. It should clarify it for you: stat.pitt.edu/stoffer/tsa3 – BLimkins Sep 17 '15 at 21:25

## 2 Answers

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.

For the sine wave to be stationary it needs a random phase! As whuber points out, it is not enough that the phase is random, it must have a uniform distribution on the interval $[0,2\pi)$.

• I believe this is what was meant by the phrase "an unknown starting point" in the question. Note that having a random phase is insufficient for stationarity: there is a strong condition on its distribution. For instance, a uniform distribution of phase in $[0,2\pi)$ will work, but a Normal distribution of phase will not. – whuber Sep 20 '15 at 13:02