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I asked a question on the conditions when $S_n-\lfloor S_n \rfloor \sim U([0,1))$ where

$$S_n = \sum\limits_{i=1}^n X_i$$

and the $X_i$ are i.i.d. here. The accepted answer states the conditions that lead to this along with a proof provided by H. H. Rugh. The proof is above my level, but I want to close the gap. So, I decided to ask follow-up questions around the things I don't grasp.

In the answer, he defines for each $m\in {\Bbb Z}$: $\gamma_m = {\Bbb E} \left( e^{2\pi i m X_1} \right)$ and it is claimed that the following two statements are equivalent:

  1. The law of $S_n \ {\rm mod}\ 1$ converge in distribution to $\ { U}([0,1))$
  2. $|\gamma_m|<1$ for every $m\in {\Bbb Z}^*$.

A proof is provided for the equivalence (to understand it, I need to close a lot of gaps), but even before getting into it I don't understand how $|\gamma_m| = |E(\cos(2\pi m X_1)+i \sin(2\pi m X_1))|$ can possibly be anything but $1$.

Isn't it true that for any realization of $X_1$, $|\cos(2\pi m X_1)+i \sin(2\pi m X_1)|=1$? And if so, the expected value over $X_1$ should also be $1$.

What am I missing?

EDIT: Once I wrote this down here, I noticed that we can write -

$$|\gamma_m| = |E(\cos(2\pi m X_1)) + i E(\sin(2\pi m X_1))|$$

This gives me a sense that it might not be $1$ necessarily, but still lost as to why it has to be $<1$.

Rohit Pandey
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    Your edit is right. It is true that $E[|e^{2 \pi i m X_1}|] = E[1] = 1$, but $|\gamma_m| = |E[e^{2 \pi i m X_{1}}]| \le E[|e^{2 \pi i m X_1}|]=1$ (using the property $|E[Z]| \le E[|Z|]$. So in general, $|\gamma_m| \le 1$, not necessarily with equality. To prove the strict inequality requires the extra assumptions in condition (1) and the context of the problem. – angryavian Jul 05 '21 at 01:14
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    $Z:=e^{2i\pi nX_1}\in\mathbb{S}^1$, so $E[Z]$ is in the convex hull of $\mathbb{S}^1$, i.e. $\overline{B(0;1)}$. In fact, unless $X_1$ is degenerate, $E[Z]\in\operatorname{Int}(\overline{B(0;1)})=B(0;1)={(x,y)\in\mathbb{R}^2: x^2+y^2<1}$. – Mittens Jul 05 '21 at 03:06

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\begin{align} \big|E\big(e^{2\pi i mX}\big)\big|&=\Bigg|\int_\mathbb{R} e^{2\pi i mx}dF_X(x)\Bigg|\\ &\le\int_\mathbb{R}\big|e^{2\pi i mx}\big|dF_X(x)\\ &=1\ . \end{align} The inequality is not necessarily strict, however, but if $\ \big|E\big(e^{2\pi i mX}\big)\big|=1\ $ then $\ E\big(e^{2\pi i mX}\big)=e^{2\pi i\theta}\ $ for some $\ \theta\in[0,1)\ $, and this can be true only if $\ e^{2\pi i mX}=e^{2\pi i\theta}\ $ almost surely—or, equivalently $\ mX-\theta\ $ is almost surely an integer. To see why this is true, note that $\ E\big(e^{2\pi i mX}\big)=e^{2\pi i\theta}\ $ if and only if $\ E\big(1-e^{2\pi i m(X-\theta)}\big)=0\ $. But if this is true, then \begin{align} 0&=\big|E\big(1-e^{2\pi i (mX-\theta)}\big)\big|^2\\ &=\Bigg(\int_\mathbb{R}\big(1-\cos2\pi(mx-\theta)\big)dF_X(x)\Bigg)^2\\ &\hspace{4em}+\Bigg(\int_\mathbb{R}\sin2\pi(mx-\theta)dF_X(x)\Bigg)^2\ , \end{align} and this can be zero only if $\ \int_\mathbb{R}\big(1-\cos2\pi(mx-\theta)\big)dF_X(x)\ $ and $\ \int_\mathbb{R}\sin2\pi(mx-\theta)dF_X(x)\ $ are both zero. So we have \begin{align} 0&=\int_\mathbb{R}\big(1-\cos2\pi(mx-\theta)\big)dF_X(x)\\ &\ge\epsilon P\big(\cos2\pi(mX-\theta)<1-\epsilon\big) \end{align} for any $\ \epsilon>0\ $, which implies that $$ P\big(\cos2\pi(mX-\theta)<1-\epsilon\big)=0 $$ for any $\ \epsilon>0\ $, so $\ \cos2\pi(mX-\theta)=1\ $ with probability $1$, or, equivalently, $\ mX-\theta\ $ is an integer with probability $1$.

lonza leggiera
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