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Question: How do I calculate the effective sample size of a time series under overlapping observations?

By overlapping observations of $x_t$ where the overlap is of length $k$ I mean a case where $x_t=\sum_{\tau=t-k+1}^t \xi_\tau$ where $\xi_\tau$ is some random process. Hence, $x_t$ and $x_{t-\kappa}$ measure partly the same thing for $\kappa<k$; they "overlap". An example would be measuring monthly financial returns every day. The monthly return $x_t$ of today overlaps with the monthly return of yesterday $x_{t-1}$ to a large degree: given a month with 30 trading days, 29 daily returns $\xi_{t-29},\dots,\xi_{t-1}$ constitute both $x_t$ and $x_{t-1}$, while only $\xi_{t}$ and $\xi_{t-30}$ make $x_t$ and $x_{t-1}$ differ. (How many trading days a month has depends on the market.)

Richard Hardy
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  • This is a well known problem for MCMC diagnostics. The MCMC output does not consist of overlapping observations, but your problem isn’t overlap per se but rather dependence. So my advice would be to estimate the effective sample size as you would for MCMC output, ie use the long run variance as the key ingredient. The coda package in R has functions for it. – hejseb Dec 11 '19 at 21:45
  • @hejseb, thank you! This is helpful. Also, a rather obvious point that I did not think of when asking the question: the effective sample size depends on the estimation target (or related things). – Richard Hardy Dec 12 '19 at 10:38
  • Related threads: "$R^2$ and adjusted $R^2$ in presence of overlapping observations" and "Effective sample size: does it depend on the model?". – Richard Hardy Dec 12 '19 at 10:39
  • @hejseb, I checked out the coda package, and it seems to base the estimation of the effective number of observations on comparing the long-run variance estimated adjusting for autocorrelation (via fitting an AR model) with the naive estimate – as simple as that. So the answer would be to take $n\frac{ \hat\sigma^2 }{ \tilde\sigma^2 }$ with $n$ being the nominal number of overlapping observations, $\hat\sigma^2$ the naive estimate and $\tilde\sigma^2$ the one adjusted for autocorrelation. – Richard Hardy Mar 22 '23 at 18:46

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