We obviously try to avoid doing statistics on overlapping returns, but here is an interesting little tidbit that has turned up.If we had for example a two-day overlapping return series, so that returns come from data on day 1 to day 3, then day 2 to day 4, etc, the autocorrelation is of the order of 50%.If we were to separate the data into non-overlapping series (odd days and even days for example) we could view the two series as having a joint distribution (with Pearson correlation of about 50%) and therefore being joined by a copula.So the copula is created purely from the marginal distribution, as that is the only information available.If we begin with data that is Gaussian distributed, the copula "looks" like a Gaussian copula (numerically, at first glance at least).If we start with Student t distributed data, would we end up with what looks like a Student t copula?So we seem to be able to map a marginal distribution to a copula. Does the opposite direction mapping exist, whereby we take for example the Clayton copula and map to a unique marginal distribution?What is this "overlap copula"? I can't see a quick proof.Interested to know if someone can see right through this one, or if this is a well-known result.Thanks

Any speculation from the copula experts out there?

If I am asking a dumb question feel free to just say "dude this is obvious - it is the XYZ copula, just Google XYZ copula to find a paper."This just seems pretty interesting as it is a case where a copula is uniquely generated from a marginal distribution.Thanks

I don't think it is a dumb question. Basically you are saying what is the class of copulas that correspond to (X,Y), where X = Z1+Z2, Y=Z2+Z3, where Z1,Z2,Z3 are independent (these are your daily returns) and possible identically distributednot that I know everything written on copulas but I have not seen an answer to this. could be an interesting Master/PhD/paper topic...Vladimir

THe marginal distributions won't determine the copula. The copula will depend on (1) how you align the two series and (2) the cyclical pattern of volatilities.Let r(n) denote the 2-day return starting on day n. Then for issue (1) you could pair r(2n) with either r(2n-1) or r(2n+1). If volatility is not stationary, and especially if it is cyclic, these two different alignments will give different copulas.To see this, imagine that the factor has high volatility on even days and low vol on odd days. If r(2n) is paired with r(2n-1) the high vol days will match between the two marginal distributions, so there will be high correlation between the two series. But if r(2n) is paired with r(2n+1) the high vol day in one element of a pair will be different from the high vol day in the other element. So there will be a low correlation between the series.It is conceivable that the marginal distributions will impose some constraints on the copula, within this particular framework, but they won't fully determine it. EDIT: If we assume the daily returns are IID, the copula for overlapping returns [$]X,Y[$] will be fully determined, and can be written as [$]C(u_x,u_y)=\int_{s=-\infty}^\infty p_{m-h}(s)\cdot F_h\big(F^{-1}_m(u_x)-s\big) \cdot F_h\big(F^{-1}_m(u_y)-s\big)ds[$]where [$]F_n[$] is the cdf and [$]p_n[$] the pdf of the return on a sequence of [$]n[$] consecutive days, and the return periods are [$]m[$] days long, sampled at rests of [$]h[$] days (we assume [$]h\vert m[$]). Note that [$]m-h[$] is the length of the overlapping pieces and [$]h[$] is the length of the non-overlapping piece at either end of the overlapping piece.The integration variable [$]s[$] is the return on the overlapping piece. If we write [$]u_x\equiv F_m(x),u_y\equiv F_m(y)[$] then Prob[$](X<x\wedge Y<y)=C(u_x,u_y)[$].

Vladimir and AndrewK thanks for responding.Andrewk: yes the basic assumption was that of iid data. Interesting expression for [$]C(u_x, u_y)[$]. To make sure I am following the notation, let's take the case of iid Gaussian distributed returns, where we are taking one-day returns and creating overlapping two day returns, and then splitting them into an even day series and an odd day series.Can you then perform the integration? I was guessing it would end up being the bivariate normal copula with correlation 50%.[$]C(u,v)=\Phi_{\rho}\left(\Phi^{-1}\left(u\right),\Phi^{-1}\left(v\right)\right)[$]I was thinking one could generate autocorrelated time series by starting with a copula, such as the above bivariate normal, and some assumed marginals, and was hoping this line of thinking would lead me to the reverse problem, which would then lead to the copula given the one-day marginals. No need for that now it seems!Thanks

QuoteOriginally posted by: erstwhileI was guessing it would end up being the bivariate normal copula with correlation 50%.Yes it does end up being that. With a bit of mucky algebra (I think it's easier to work with conditional distributions for the Gaussian rather than integrating the general formula) one can demonstrate that, for Normal IID daily returns, the joint distribution of the returns for two m-day periods that overlap by (m-h) days is a bivariate Gaussian with correlation (1-h/m). In your case m=2 and h=1, giving the 50% correl. So the copula for the overlapping returns is indeed Gaussian, with the expected correlation.Going in the reverse direction, consider a bivariate Gaussian with parameters [$]\mu_1=\mu_2=0[$], [$]\sigma_1=\sigma2=\sigma'[$] and correlation [$]\rho[$]. The copula from this will be the same as that for returns from overlapping periods, as specified previously, with [$]h=(1-\rho)m[$] and [$]\sigma=\frac{\sigma'}{\sqrt{m}}[$] where daily returns are [$]\sim N(0,\sigma^2)[$] and [$]m[$] can be freely chosen as whatever we want. I expect the result can be extended to multivariate Gaussians, with a bit of matrix manipulation. I don't think it will work for t distributions because, unlike Gaussians, the sum of t-distributed random variables is not t-distributed. The fact that sums of Normal RVs are normal is used repeatedly in deriving the above.EDIT: Hmmm. Latex code processor doesn't seem to be working on this site. I wonder what's going on there.

Excellent - thanks for this