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Let $\Phi$ be the CDF of the normal distribution, and let $u,v,s\sim\mathrm{Unif}[0,1]$ be iid uniform variables, then $X_1:= \Phi^{-1}(u),Y_1:= \Phi^{-1}(v)$ will be independent normal variables, therefore $Z_1:=(X_1+Y_1)/\sqrt{2}$ will follow a normal Gaussian. Now if we shift $u,v$ by $s$ and define $X_2:=\Phi^{-1}(u+s - \lfloor u+s\rfloor ),Y_2:=\Phi^{-1}(v+s - \lfloor v+s\rfloor )$, where $\lfloor \cdot \rfloor $ stands for floor, $Z_2:=(X_2+Y_2)/\sqrt{2}$ will all analogously follow the normal Gaussian distribution. My question is, is $Z_1$ independent of $Z_2$?

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As $s \to 0+$, $Z_2 \to Z_1$. So they should certainly not be independent if $s$ is sufficiently small. I would guess that they are dependent for all $s$, but it'll be a bit messy to prove.

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  • $\begingroup$ actually by numerical testing, $s$ near $0$ seems to be the only place that they become dependent, but seem to be independent almost everywhere else. $\endgroup$
    – kvphxga
    Jul 24 '20 at 16:54
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    $\begingroup$ Consider, for some $\alpha$ and $\beta$, $g(s) = \mathbb P(Z_1 < \alpha, Z_2 < \beta) - \mathbb P(Z_1 < \alpha) \mathbb P(Z_2 < \beta)$. I would think that this is real-analytic as a function of $s$. If this is the case, and $g(s) \ne 0$ for some $s$, then $g(s) \ne 0$ with at most countably many exceptions. So they should be dependent almost everywhere. $\endgroup$ Jul 24 '20 at 17:42
  • $\begingroup$ I did montecarlo experiments: I generated $10^7$ $(Z_1,Z_2)$ pairs this way, and tabulated them in a $20\times20$ histogram grid, and plotted the data and calculated the obvious Pearson $\chi^2$ number. The histogram did not look flat, and I got off-the-chart chisquare values. (I used drand48 and the GSL cdf routines.) $\endgroup$ Jul 25 '20 at 13:29

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