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I'm trying to prove that if

$X_n$ iid normal
$S_n = \sum_1^n X_i$
$U_n=S_n-\lfloor S_n\rfloor$

then $U_n$ is asymptotically uniform in distribution. I've got no idea how to approach this, and it's a past exam question, so I should be able to do it inside 30 mins. My only ideas for how to approach it were

(1) possibly use an ergodic theorem (since the floor function looked inviting).
(2) work with characteristic functions and use Levy's Continuity Theorem (but I can't see how to work out the characteristic functions of the $U_n$).

Has anyone got an idea how to proceed? A hint rather than a full solution would be ideal, since I'd quite like to have something to work out myself, once I've got an idea how to start!

Many thanks.

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1 Answer 1

up vote 1 down vote accepted

Since you asked for hints:

  • It would be worth finding the maximum (absolute) gradient of the density function of a normal distribution with variance $\sigma^2$.

  • That can be used to put an upper bound on the difference between the maximum and minimum density of the distribution of the fractional part of a normal distribution

  • Combine this with the variance of the sum of $n$ iid normal distributed random variables being $n$ times their individual variance

  • Let $n$ increase without limit and see the upper bound reduce towards zero.

Alternative approach added later

  • The aim is to put an upper bound on the difference between the maximum and minimum density of the distribution of the fractional part of a normal distribution with variance $\sigma^2$. The density of the distribution of the fractional part can be found by adding densities of different parts of the original distribution corresponding to the same fraction.
  • For the original distribution above the mean the density is decreasing, so the difference cannot be more than the maximum density: $h(z_1) - h(z_2) = \sum_n f(n+z_1) - \sum_n f(n+z_2) $ $= f(z_1) - \sum_n \left(f(n+z_2) - f(n+1+z_1)\right) \le f(z_1) \le \frac{1}{\sqrt{2 \pi}\sigma }$
  • Similarly below the mean. So the difference on the fractional density $g(z_1) - g(z_2)$ cannot be more than twice this, i.e. $\frac{2}{\sqrt{2 \pi}\sigma }$.
  • This bound reduces towards $0$ as the variance increases, i.e. as more terms are included in the sum.
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Thanks very much for the hints Henry! I can't quite see how to use my bound on the maximum of the gradient to get a bound on the density of the distribution of the fractional part. So far I have that if $g$ the distribution of the fractional part $g(z)=\sum_n f(n+z)$ where $f$ is the distribution of $N(0,\sigma^2)$. Is this right? If so, where do I go next? Thanks! –  Edward Hughes May 28 '12 at 20:33
    
@hgbreton The absolute gradient is less than or equal to $\frac{1}{\sigma^2 \sqrt{2e\pi}}$, so the difference between the maximum and minimum density of the fractional part of a normally distributed random variable is also less than or equal to this. But I will add a different approach to my reply –  Henry May 28 '12 at 21:47

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