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By first glance of this time series; will you say it is stationary?

I can easily see some "seasonality" which means that this is not strictly stationary since the distribution will not be the same; higher variance around 1987 and 2008. But is it weakly stationary? The expected value at any timepoint will be zero, so I will say yes it is weakly stationary. Am I right? enter image description here

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    $\begingroup$ For "weak sense stationarity" we require that the first two moments (mean and variance) are constant over time, not just the mean $\endgroup$ – Alex C Jan 28 '18 at 6:09
  • $\begingroup$ Alecx C, "Weak stationarity" does not require that the variance is constant over time. That is one of them main differences between weak and strict. $\endgroup$ – Sanjay Jan 28 '18 at 17:04
  • $\begingroup$ Did you tried to decompose the series into Trend (without meaning) and seasonality (Wold decomposition)? $\endgroup$ – NunodeSousa Jan 28 '18 at 20:17
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    $\begingroup$ I thought weak stationary is what @AlexC mentioned and strong stationarity requires that the joint distribution remains the same over time? $\endgroup$ – Lipton Jan 28 '18 at 21:00
  • $\begingroup$ @AlexC is correct. $\endgroup$ – Matthew Gunn Feb 3 '18 at 17:50
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We can talk about whether a strictly stationary or weakly stationary process might usefully describe that data. My answer to both would be yes.

I also have issues with other text that people have written here.

A review of mathematical definitions:

  • A stochastic process $\{X_t\}$ is called strictly stationary if it's joint distribution function $F(X_{t}, X_{t+1}, \ldots, X_{t+k})$ does NOT depend on $t$.
  • A stochastic process $\{X_t\}$ is called weakly stationarity if it's first moment $\mathbb{E}[X_t]$ and second moments $\mathbb{E}[X_tX_{t+j}]$ do NOT depend on $t$.

Some intuition for the mathematical definitions

  • In the English language, an object is stationary if it does not move over time.
  • In time-series mathematics, a stochastic process is stationary if the joint probability distribution does not move over time.
    • We are NOT saying that the realizations of a stochastic process are constant over time. (i.e. we're not saying $X_t = X_{t+j}$.)
    • We are NOT saying that various conditional moments of a stochastic process are constant over time. Eg. we are not saying that our expectation of volatility given the past realizations of returns is constant. It's fine for $\mathbb{E}[X_t^2 \mid X_{t-1}] = f(X_{t-1})$.
    • What's constant over time in strict stationary is the joint distribution. What's constant in weak stationarity is the unconditional mean and auto-covariance function.

Stationarity and ergdocity are important time-series properties for a stochastic process. If the past and the future are drawn from the same distribution, we can learn about the distribution from the past and then use what we learned to say something about the future. Without stationarity, we're in a sense lost.

Pure mathematics vs. statistics

  • Given a mathematically well-defined stochastic process, we can say whether it satisfies either definition of stationarity (or not). That is pure mathematics.

  • Stationarity is NOT a mathematical property of data. Given some data, we can talk about whether a stationary process might have generated this data or whether the empirical data can be usefully described by a stationary process. But this isn't an exercise in pure mathematics. It's an exercise in statistics and judgement.

There's a famous quote of statistician George E. P. Box that "all models are wrong, but some are useful." When we build a model for data, we're almost certain to get the model wrong. Under Box's philosophy, what instead matters is whether the model is useful.

My big takeaway from visual inspection of that graph is that there's volatility clustering. An extremely simple, stationary model with stochastic volatility can also generate volatility clustering. For example something like:

$$ r_t = \mu + \sigma_t \epsilon_t$$ $$ \sigma_t = a + b \sigma_{t-1} + u_t $$

Does that capture a key feature of the data (i.e. some simple notion of volatility clustering)? Yes. Does my model have problems? Yes. Is it useful? Depends on the question.

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They probably can be modelled using a weakly stationary process.

To quote Section 1.2.1 from these lecture notes:

[Asset] returns [...] typically fluctuate around a constant level, suggesting a constant mean over time. [...] In fact most asset returns can be modeled as a stochastic process with at least time-invariant first two moments.

Mathematically, a time series $\{ Y_t \}$ is weakly stationary if, for all time indices $k,s,t$

  1. $\text{E}[Y_s] = E[Y_t]$, i.e. the first moment (the mean) is constant
  2. $\text{Cov}[Y_t, Y_{t+k}] = \text{Cov}[Y_s,Y_{s+k}]$, i.e. the second moment is constant

From a visual inspection of your series of asset returns,

  1. the mean/first moment does indeed appear to be constant
  2. the series clearly exhibits the phenomenon of volatility clustering, implying that it has a non-constant "conditional volatility" / exhibits heteroskedasticity - however we cannot make any visual judgements as to the behaviour of the unconditional volatility
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  • $\begingroup$ Your def. is correct but your application of the definition is flawed. Heteroscedasticity does not imply either strict or weak stationarity is violated! That volatility is autocorrelated does not imply that unconditional moments are non-constant. $\endgroup$ – Matthew Gunn Feb 3 '18 at 17:42
  • $\begingroup$ You're arguing that the conditional volatility can take different values. That does not imply unconditional volatility is time dependent. I can easily write a stationary model $r_t = \sigma_t \epsilon_t$ and $\sigma_t = 1 + .5 \sigma_{t-1} + u_t$ (where $\epsilon_t$ and $u_t$ IID standard normal). That model exhibits conditional volatility but is strict and wide-sense stationary. $\endgroup$ – Matthew Gunn Feb 3 '18 at 17:47
  • $\begingroup$ On the other hand, $\sigma_t = t^2 + u_t$ would be non-stationary. $\endgroup$ – Matthew Gunn Feb 3 '18 at 17:49
  • $\begingroup$ Important clarification here (previous 3 comments by Matthew Gunn). $\endgroup$ – Alex C Feb 3 '18 at 18:11
  • $\begingroup$ Thanks, I've edited my answer to clarify that important point. $\endgroup$ – A. G. Feb 4 '18 at 11:43
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Let us mentally walk through the implications of stationarity on the S&P 500. First, the likelihood function for $\log(p_{t+\Delta{t}})-\log(p_t)$ is the hyperbolic secant distribution, which has a mean and a variance, but no covariance. So on first blush, it should automatically be stationary, except this ignores a definitional fact.

The S&P 500 constantly has members being added and dropped. Presumably each separate company has its own unique log mean and log variance. So every time a firm is dropped and replaced it MUST change the long run future log mean from the long run past log mean. The alternative is to believe that all capital has the same mean log return.

There should be structural beaks every single quarter. Additionally, as a matter of policy the S & P 500 used to contain ADRs and so currency risk was present as well. Now it only contains American firms. I cannot imagine that such as series could be considered stationary, let alone have stable properties.

The only way you may seriously discuss it as stationary would be to adopt the subjective Bayesian approach where the mean and the variance are not considered points, but rather drawn from a distribution by nature. In that case, you could discuss them, but you should probably seek to defend that added and deleted firms were sufficiently similar that such a subjective statement is valid.

Sometimes you do not need a test, you just need to think about the logic of the implications.

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  • $\begingroup$ Question for you: if the imaginary portfolio get re-balanced properly, would the addition/removal of firms to the index be a huge deal at all? $\endgroup$ – Will Gu Jan 29 '18 at 20:57
  • $\begingroup$ @WillGu I cannot answer that. If the question is "is it stationary," then the answer must be no and it also cannot be fixed using standard techniques. I cannot answer it because I lack the necessary data to determine if you could ignore the problem by using a subjective Bayesian method. Objective Bayesian methods do what Frequentist methods do, which is assume that a parameter is a point. Subjective methods assume it is drawn from a density. It may be possible to escape the issue by presuming they are drawn from a highly similar class. $\endgroup$ – Dave Harris Jan 30 '18 at 1:30
  • $\begingroup$ I am not certain you could evade it intellectually by making these assumptions. It is sort of a cheat. You would really want to look at the two parallel series, the one from before rebalancing and the one after. You would have to do that for every quarter. $\endgroup$ – Dave Harris Jan 30 '18 at 1:33

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