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Log differenced returns, computed from stock prices, are known to be stationary. What about cumulative returns, are they also stationary? if not why not? Are there other properties, like non-i.i.d., that cumulative returns share with regular returns?

develarist
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3 Answers3

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Hi: Even if returns were stationary ( which is probably dependent on the time series one is considering ), cumulative returns, where $n$ is not fixed ( as it in say a rolling sum with a fixed window size or a non-overlapping sum with a fixed window size ) definitely can't be stationary. Consider a pure noise process.

$logret_t = log(P_{t}) - log(P_{t-1}) = \epsilon_t$

Now take the cumulative sum of $ret_t$ over $n$ periods. This sum = $\sum_{i=1}^{n} \epsilon_{i}$.

A) The expectation of this sum is zero but the variance is increasing because it's $\sigma^2 \times n $ so the variance of the cumulative return is not constant.

B) In the case, where $n$ is fixed and the cumulative return is not overlapping, then stationarity follows because the variance is $\sigma^2 n$ with $n$ fixed.

C) In the case, where $n$ is fixed and the cumulative return is overlapping ( a rolling sum say ), I don't think it's stationary but the definition of what the mean of the process is, is not clear to me so I'm not certain about this case.

mark leeds
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  • i don't think cumulative returns are calculated as the sum of returns – develarist Aug 19 '20 at 13:22
  • I was thinking of cumulative returns for the same $n$, e.g. the distribution of subsequent 5 day returns. – Bob Jansen Aug 19 '20 at 13:33
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    @develarist Also, log returns are: https://quantivity.wordpress.com/2011/02/21/why-log-returns/ – Bob Jansen Aug 19 '20 at 14:01
  • @Bob Jansen: Thanks. I should have said log returns. Develarist: Is the second cumulative return overlapping with the first or non-overlapping ? If you assume returns are stationary and you calculate the non-overlapping cumulative returns, then you could assume stationary in that specific case. – mark leeds Aug 19 '20 at 14:38
  • Note that, if you use fixed $n$ as Bob described and your returns were overlapping, then you wouldn't have stationarity because your cumulative return at time $t+1$ would depend on the cumulative return at time $t$. So, the variance would be constant ( say $n \times \sigma^2$) but the mean would always be changing because of the dependence on the previous return. – mark leeds Aug 20 '20 at 00:35
  • Cumulative returns are stationary for fixed $n$ if the return process is stationary. – fes Aug 21 '20 at 06:57
  • @fesman: I agree, if you include non-overlapping (i.e: not rolling ) cumulative returns. – mark leeds Aug 21 '20 at 13:15
  • @markleeds In grad school, I was taught that prices are typically integrated of order 1 and that returns are (covariance) stationary. These observations were called ''stylized facts''. Of course, I can always test for a unit root but your answer suggests that my profs were out-dated and newer empirical facts point to integrated returns and that the returns of returns are stationary? – Alex Aug 21 '20 at 15:22
  • @markleeds Thanks for clarifying. – fes Aug 21 '20 at 16:38
  • @alex: I think returns being stationary is possible. I'm not sure where I implied that they couldn't be but I didn't mean to. I think it could go either way but the RW model says that they are, Overlapping cumulative returns is more tricky because you then have past returns in the model for current cumulative return. That's why I said that non-overlapping cumulative would insure stationarity in the cumulative returns fixed n case. I could be wrong on overlapping but I'm not sure how to define the mean in that overlapping fixed $n$ cumulative case. – mark leeds Aug 21 '20 at 18:57
  • @Alex: The confusion might due to me differentiating between returns and cumulative returns. I'm assuming that there is some process for the returns ( say RW ) and then, the cumulative return is the sum of those returns. So, the original return in the RW DGP is stationary. But, then, when you sum them to get cumulative return, I think the overlapping fixed $n$ case may not be stationary but the non-overlapping fixed $n$ case is still stationary. Finally, the not fixed $n$ case where $n$ just gets larger and larger ( initially what I thought develarist was referring to ) is not stationary. – mark leeds Aug 21 '20 at 19:06
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    @Alex: I see now where I said: to quote: "even if we assume returns are stationary (which is usually not true)". That was probably too strong a statement so my apologies. At the same time, to say that they are stationary might be a little strong also. If one assumes normality, it's probably true. But, of course, there's all that literature on returns not being normal. So, my answer would be: "test, test, test" !!!! :). And again, my apologies for the strong and confusing statement. Note that time-frame is crucial to this discussion also and I never specified a time frame !!!! – mark leeds Aug 21 '20 at 22:49
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Stock prices are definitely not stationary as tomorrows closing price is strongly influenced by today's closing price and prices tend to change. Returns can be potentially stationary and are therefore a much better target for analysis. They don't need to be stationary and I don't believe they are. For example, historical returns exhibit heteroscedasticity. This answer explains more.

Cumulative returns don't need to be stationary either as their building blocks don't need to be. However, you could argue that if you aggregate returns some of the short term noise averages out and become more homoscedastic and thus become more stationary. On the other hand, it seems that the mean return varies over longer periods. The expected return over longer periods seems to vary though as the world economy moves through different stages. A varying mean again makes returns non-stationary.

Bob Jansen
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    Not sure homoscedasticity (of returns) invalidates stationarity??? Sure, variance is not then equal across the sample; but if its fluctuations are independent of time; then not sure this makes them non-stationary. Ditto time-varying means if these are not dependent on (location in, as opposed to distance in) time. But agree with everything else!!! – demully Aug 19 '20 at 15:12
  • @demully I think I just wrote it down in a confusing way. – Bob Jansen Aug 19 '20 at 15:49
  • I find the first sentence a bit confusing: it seems to associate stationarity with independence. For example the standard AR(1) / OU short rate model implies a stationary yet highly autocorrelated short rate process. – fes Aug 20 '20 at 08:30
  • @fesman: I think what Bob meant is that the random walk model says that today's price is a FUNCTION of yesterday's price in that $log(P_{t} = log(P_{t-1}) + \epsilon_t$ so the mean of today's price can be thought of as yesterday's price so the mean of the process is not constant. So, in that sense, today's price is "influenced" by yesterday's price but not in the standard sense of what the term influenced usually means. – mark leeds Aug 20 '20 at 21:05
  • @Bob Jansen: Of course, correct me if above is not what you meant but I have a feeling you're using the term "influenced" to mean that yesterday's price is the mean of the distribution of today's price.. – mark leeds Aug 20 '20 at 21:06
  • @markleeds $log(P_t)=\lambda log(P_{t-1})+\epsilon_t$, where $\epsilon_{t}$ is Gaussian white noise is stationary for $-1<\lambda<1$ and here $log(P_t)$ still depends on $log(P_{t-1})$. The problem with the random walk $\lambda=1$ is that unconditional variance is increasing over time. – fes Aug 21 '20 at 06:38
  • I agree. that's why I was saying that Bob's use of the term "influence" was implying the mean rather than correlated. – mark leeds Aug 21 '20 at 13:17
  • and I also agree that the unconditional varance is increasing over time. But, also, the mean of the $ log(P_t) $ process in the RW is $log(P_{t-1})$ which is changing over time. So, to me, there are 2 problems with the random walk as far as stationarity goes. – mark leeds Aug 21 '20 at 13:20
  • Correction to what I said immediately above. I guess one could say that there is no unconditional mean in the RW because $log(P_{t-1})$ is a conditional mean so I think you're right about the only problem being the unconditional variance. The unconditional mean really doesn't exist so it shouldn't be spoken about. My mistake. Thanks for clarification. – mark leeds Aug 21 '20 at 13:23
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In a nutshell...

  • It's always prudent and conservative to assume that prices are non-stationary.

  • But it's not actually as obvious this is true as it intuitively sounds. Intuitively, any random walk or any trend will always lead to a non-stationary process... but you'll practically struggle to prove the unit root (ie a random walk versus a slowly-correcting autoregressive process) more than intuition suggests. And if your "trend" is merely a co-integrated process with inflation/GDP etc., then the real price could indeed be stationary in nature ;-)

  • Simple example: take the gold price going back to the Roman times, and Dickey-Fuller it against prior gold prices, time, and the wages of a commander of ~100 troops in any army (the Roman century, the modern infantry company etc.). I bet you'd struggle to disprove the null of neither a significant historical "trend", nor any statistical "unit root" there...

  • Back in the real world, log-differencing prices to generate log-returns obviously removes any/much doubt here. These might exhibit some time-variation in means and in variance/volatility; but that is not prima facies evidence of non-stationarity. These might have sustained periods of high and of low, and of extremes of both (ie homoskedasticity); but unless you believe (and can demonstrate) that these tend to significantly correlate with respect to time, this is "omitted variable bias" in your model; not "non-stationary behavior". For the uncertainty about price behaviour above, looking at returns removes most "reasonable doubt" from the problem.

  • So to "cumulative returns" - these come in two basic forms. You can have trailing returns from say 1 Jan this year; and you can have rolling hourly/daily/weekly/monthly returns.

  • Put simply - the former are non-stationary. If I peg my starting point, then the variation of cumulative return will grow with respect to time, ie non-stationary. But if I look at rolling weekly, monthly, quarterly, or annual returns, these will remain stationary. Obviously, my last-12m return today will be very, very, highly correlated to the same yesterday or tomorrow. That doesn't matter - because the dropping off of the date last year generates an autoregressive process (that is stationary). My 12m returns today vs yesterday will always be very highly correlated... but this does not make them non-stationary. What matters here is not "distance in time" but "location in time". So if 12m today vs 12m yesterday get consistently more or less correlated as time goes by, then they're non-stationary.

  • So as a general rule of thumb, it's OK to assume that returns (over most reasonable investible frequencies) are indeed stationary.

  • Where this gets tricky is eg in bonds. At any given bond yield, any given change in yields will have a different "convexity effect". IE if yields structurally rise and fall over time, then the mean and the variance of associated bond returns will be affected; non-stationary. Likewise, however, economic regimes that might produce different PE (ie reciprocal earnings yield) regimes in stocks will have similar impacts. So stock returns might, likewise, be "trending" but nevertheless a "stationary" process. Because the returns are biased derivatives of a stationary risk-premium process :-(

Short answer:

  • It's impossible to say (let alone prove) most of the time.
  • It's always conservative, and never very wrong, to assume prices are non-stationary.
  • It's usually sound to assume that returns are stationary.
  • Rolling-cumulative returns as above; cumulative returns from a fixed starting point, non-stationary.
demully
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