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Let $X$ be a normally distributed variable with mean $0$ and standard deviation $1$. I will consider its fractional part $$\overline{X} = X - \lfloor X \rfloor = X \, \bmod \, 1.$$

I have done some numerical testing and it seems likely that $\overline{X}$ is uniformly distributed on $[0,1].$ To be specific I computed $$\sum_{k=-200}^{200} \Big( \Phi(k+b) - \Phi(k+a) \Big)$$ for a few values of $b \ge a$ in $[0,1]$ and the result is consistently very close to $b-a$.

Here is the code in Sage: I first define

def Phi(x):
    return (1/2 + erf(x / sqrt(2)) / 2).n()

then a few examples of these computations:

s = 0
for k in range(-200,200):
    s = s + Phi(k+3/5) - Phi(k + 2/5)
print s.n()
0.199999998998919

and

s=0
for k in range(-200,200):
    s = s + Phi(k+4/9) - Phi(k + 2/9)
print s.n()
0.222222221674844

Question: is $\overline{X}$ in fact uniformly distributed? From the examples I've done I am confident that it is but I am not sure how to prove it.

user404188
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1 Answers1

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$\overline{X}$ is not actually uniformly distributed, but it is ridiculously close to being so.

After some numerical experiments that also convinced me, I tried looking for sources online to confirm this, and found this article. Figure 1 on page 3 plots the PDF of $\overline{X}$ from 0 to 1; it is very close to uniform, but apparently it actually varies between 1.000000004 and 0.999999996.

To be honest, even now I half suspect there's some mistake in the article, because coincidences like this don't just happen. But there's a proof and everything.

There's also an intuitive explanation. To quote the article:

A better explanation relates to a well-known quick and dirty way of generating normal variates on a computer by simply summing 12 uniforms and subtracting 6. Since the variance of a uniform is 1/12, by the central limit theorem, this procedure should have a density that is very close to a standard normal. But the distribution of the fractional part of the sum of any number of uniforms is exactly uniform. The conclusion can only be that the fractional part of a standard normal must be very close to uniform.

Misha Lavrov
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    The actual PDF of $\overline{X}$ is $f(x) = 1 + 2 \sum_{k=1}^\infty e^{-2(\pi k)^2}\cos 2\pi kx$. If we just take the first term of this infinite sum (by far the most significant), we get $1 + 2e^{-2\pi^2} \cos 2\pi x$, and $2e^{-2\pi^2} \approx 5.35 \cdot 10^{-9}$. – Misha Lavrov Mar 17 '17 at 01:18
  • I guess it was too good to be true after all. Thanks! – user404188 Mar 17 '17 at 01:20