# Are Normal variables constructed by CDF inverse of uniform variables indepdent?

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 $$Z_2$$?

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.
• 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. Jul 24 '20 at 16:54
• 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. Jul 24 '20 at 17:42
• 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.) Jul 25 '20 at 13:29