If $X$ is Gaussian, prove that $X-\lfloor X \rfloor \sim U(0,1)$ as its variance becomes large

Question

I have a normal distributed random variable, $X$ with mean $\mu$ and standard deviation, $\sigma$. I don't believe it matters, but this distribution was obtained as a result of summing a large number of independent, identically distributed random numbers with finite variance (hence invoking the central limit theorem).

It seems intuitive that $X - \lfloor X \rfloor$ should become closer and closer to a uniform random number between $(0,1)$ as the variance of $X$ increases. And in the limit, it should become a uniform random number. Is there a proof for this claim or a refutation of it?

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Context: this is going to help "complete" the accepted answer here: As the variance of a random variable grows, the conditional distribution of it residing in an interval of length $1$ becomes uniform. Larger picture, I'm trying to prove Blackwell's theorem from renewal theory. See here for details: Going "well into the lifetime" of a renewal process means the time until the next event will be uniform conditional on inter-arrival?

Answer 1

Here I confirm the claim about the given weak convergence to $U[0,1]$.
Moreover, I also present a upper bound for the convergence rate, which shows the underlying convergence is extremely rapid.

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For simplicity, WLOG $\mu=0$
Let $(X_{\sigma} , \sigma \in \mathbb{R}_+)$ be the respective sequence of random variables, $f_{\sigma}$ be the density function of $\{ X_{\sigma} \}$
For any positive number $\sigma $ and integer $n$, we have: $$\int_{0}^1 f_{\sigma}(t)e^{-2i\pi n t} dt= \mathbb{E}(e^{-2i\pi n X_{\sigma}}) = e^{-2\pi^2 n^2 \sigma^2}$$ Because $\sum_{n \in \mathbb{Z}} \left| e^{-2\pi^2 n^2 \sigma^2}\right|^2<\infty$, so according to Riesz-Fischer theorem, we have that $$f_{\sigma} \in L^2([0,1])$$ And this gives ,by Perseval's identity: $$\int_{0}^1 |f_{\sigma}(t)-1|^2dt=\sum_{n \in \mathbb{Z}} \left| e^{-2\pi^2 n^2 \sigma^2}-\mathbb{1}_{\{n=0\}}\right|^2=\sum_{n \ge 1}2e^{-4\pi^2n^2\sigma^2}\xrightarrow[]{\sigma \rightarrow \infty} 0$$ Hence, $$\{ X_{\sigma} \} \xrightarrow[\sigma \rightarrow \infty]{\text{(d)}} \mathcal{U}([0,1])$$ And in particular,

For any function bounded measurable function $g$, we imply the rate of convergence

$$\left| \mathbb{E}( g( \{X_{\sigma}\}))-\int_0^1 g(x)dx \right| \le \|g\|_{\infty}e^{-2\pi^2 \sigma^2}\sqrt{\frac{2}{1-e^{-4\pi^2 \sigma^2}}}$$

The rate of convergence is the exponential of minus $\sigma^2$, hence the rate is extremely rapid. $\square$

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Side note Using density argument (replace $X$ in the following statement with any random variable whose density function is in $\mathcal{C}^{2}_{c}$), we can deduce an even more generalized result

Lemma Let $X$ be any random variable with density, $f_{\sigma}$ be the density function of the random variable $\{\sigma X\}$ , then $$\lim_{\sigma \rightarrow \infty} \int_{0}^1 |f_{\sigma}(t)-1|dt = 0$$ In particular, $$ \{ \sigma X\} \xrightarrow[\sigma \rightarrow \infty]{\text{(d)}} \mathcal{U}([0,1])$$

Answer 2

Here we show that for any random variable $X$ whose distribution has density with respect to Lebesgue measure $\lambda$ (i.e. there is $\phi\in L^+_1(\mathbb{R},\lambda)$ such that $P[X\in A]=\int_A\phi(x)\,dx$ for any $A\in\mathscr{B}(\mathbb{R})$), we have that $$\begin{align}\{\sigma X+\mu\}\Longrightarrow U(0,1),\quad \text{as} \quad\sigma\rightarrow\infty,\tag{0}\label{zero}\end{align}$$ where $\mu\in\mathbb{R}$ is fixed, $\{x\}=x-\lfloor x\rfloor$, $U(0,1)$ is the uniform distribution on the interval $(0,1)$, and $\Longrightarrow$ stands for weak convergence of probability measures. In particular, if $N(\mu;\sigma)$ denotes the normal distribution with mean $\mu$ and variance $\sigma>0$, then $N(\mu;\sigma)\Longrightarrow U(0,1)$ as $\sigma\rightarrow\infty$, for if $X\sim N(0,1)$, then $\sigma X+\mu\sim N(\mu;\sigma)$. Incidentally, \eqref{zero} also answers the question in here by considering $1-\{x\}$ in place of $\{x\}$ an noticing that $1-U(0,1)\stackrel{law}{=}U(0,1)$.

The approach we follow is based on a simple extension of Fejér's formula which we state here and prove after showing \eqref{zero}. We conclude this answer with an observation of the apparent uniformity that the transformation $\sigma X$ has as $\sigma\rightarrow\infty$.

Theorem: Let $g$ be a bounded measurable $T$-periodic function, $\sigma_n\xrightarrow{n\rightarrow\infty}\infty$, and $\alpha_n$ any sequence in $\mathbb{R}$. For any $\phi\in L_1(\mathbb{R},\lambda)$, where $\lambda$ is Lebesgue's measure, $$\begin{align} \lim_{n\rightarrow\infty}\int_{\mathbb{R}} \phi(x)g(\sigma_nx+\alpha_n)\,dx=\Big(\frac{1}{T}\int^T_0 g(x)\,dx\Big)\Big(\int_{\mathbb{R}} \phi(x)\,dx\Big)\tag{1}\label{one} \end{align}$$

Let $\mu\in\mathbb{R}$ and let $\sigma_n>0$ be a sequence such that $\sigma_n\xrightarrow{n\rightarrow\infty}\infty$. Let $\phi$ be the density function of the distribution of $X$. The function $\{x\}:=x-[x]$ is bounded measurable and $1$-periodic. For any $f\in\mathcal{C}_b(\mathbb{R})$, $x\mapsto f(\{x\})$ is measurable, bounded, and $1$-periodic. By Fejér's formula \eqref{one} $$\begin{align} \mathbb{E}\Big[f\big(\{\sigma_n X+\mu\}\big)\Big]&=\int_{\mathbb{R}} f(\{\sigma_n x+\mu\})\phi(x)\,dx\xrightarrow{n\rightarrow\infty}\Big(\int^1_0f(\{x\})\,dx\Big)\Big(\int\phi(x)\,dx\Big)\\ &=\int^1_0f(x)\,dx=\mathbb{E}[f(U)]\tag{2}\label{two} \end{align}$$ This proves the weak convergence along any sequence $\sigma_n\xrightarrow{n\rightarrow\infty}\infty$, and hence that $\{\sigma X+\mu\}\Longrightarrow U(0,1)$ as $\sigma\rightarrow\infty$.

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Proof of Fejér's formula: We first consider functions of the form $\phi(x)=\mathbb{1}_{(a,b]}(x)$. Since $$\sigma_nb+\alpha_n=\sigma_na+\alpha_n+\frac{\sigma_n(b-a)}{T}T= \sigma_na+\alpha_n+\left\lfloor\frac{\sigma_n(b-a)}{T} \right\rfloor T+ r_nT$$ where $r_n=\big\{\frac{\sigma_n(b-a)}{T}\big\}$, we have that $$\begin{align} \int_{\mathbb{R}} g(\sigma_nx+\alpha_n)\phi(x)\,dx&=\frac{1}{\sigma_n}\int^{\sigma_nb+\alpha_n}_{\sigma_na+\alpha_n}g(x)\,dx\\ &=\left\lfloor\frac{\sigma_n(b-a)}{T}\right\rfloor\frac{1}{\sigma_n}\int^T_0 g(x)\,dx +\frac{1}{\sigma_n}\int_{I_n}g(x)\,dx\tag{3}\label{three} \end{align}$$ where $I_n$ is an interval of length $r_nT$. The first term in \eqref{three} converges to $\frac{b-a}{T}\int^T_0g=\Big(\frac{1}{T}\int^T_0g\Big)\int \phi$; whereas the second term in \eqref{three} converges to $0$, for $\frac{1}{\sigma_n}\Big|\int_{I_n}g\Big|\leq \frac{T}{\sigma_n}\|g\|_u\xrightarrow{n\rightarrow\infty}0$. This proves \eqref{one} for finite intervals, and by linearity, \eqref{one} extends to all step functions. As step functions are dense in $L_1(\mathbb{R},\lambda)$, we conclude that \eqref{one} holds for all $\phi\in L_1$.

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Observation: The transformation $\sigma X$, where $X$ is a random variable with density (no integrability assumed), flattens out locally the density of $\sigma X$ as $\sigma\rightarrow\infty$ giving the appearance of uniformity. To be more precise, suppose the density $\phi$ of $X$ is continuous at $0$, and $\phi(0)>0$. Let $A$ be a Borel set with $0<\lambda(A)<\infty$ and $P(X\in A)>0$, and consider the conditional distribution $$ P^A_\sigma(dx):=P[\sigma X\in dx|\sigma X\in A]$$ Then, by dominated convergence, we have that for any $f\in\mathcal{C}_b(\mathbb{R})$ $$\begin{align} E[f(\sigma X)|\sigma X\in A]&=\frac{\int \mathbb{1}_{A}(\sigma x) f(\sigma x)\phi(x)\,dx}{\int\mathbb{1}_A(\sigma x)\phi(x)\,dx}\\ &=\frac{\int \mathbb{1}_{A}(x) f(x)\phi\big(\tfrac{x}{\sigma}\big)\,dx}{\int\mathbb{1}_A(x)\phi\big(\tfrac{x}{\sigma}\big)\,dx}\xrightarrow{\sigma\rightarrow\infty}\frac{1}{\lambda(A)}\int_A\,f(x)\,dx \end{align}$$ Therefore, $P^A_\sigma\stackrel{\sigma\rightarrow\infty}{\Longrightarrow}\frac{1}{\lambda(A)}\mathbb{1}_A(x)\,dx$, that is, $P^A_\sigma$ converges weakly to the uniform distribution over the set $A$. This in particular, holds for $X\sim N(\mu;1)$, $\mu\in\mathbb{R}$, in which case $\sigma X\sim N(\mu,\sigma)$.

Answer 3

You are wrong. It is of crucial importance how you obtain this approximate gaussian distribution. You are looking at the fine structure of the distribution so the CLT is of no help here. A counter-example: Let $X_k$ be an i.i.d. sequence of integer valued random variables with variance $0<\sigma^2<+\infty$. Then $S_n=X_1+\cdots X_n$ will satisfy the CLT: $(S_n-{\Bbb E}(S_n))/\sqrt{n} \sim {\cal N}(0,\sigma^2)$ but $S_n-\lfloor S_n\rfloor$ is identically zero.

Thus you may not necessarily have a normal distributed variable in the sense you would like to have. In the generality stated the claim does not hold. Other posts deal with various ways of considering the limit. In view of your description, I believe the relevant problem you want to address is under what conditions on the distribution of an i.i.d. sequence $X_k$, does $S_n \ {\rm mod}\ 1$ converge in distribution to $\ { U}([0,1))$. For this we have the following complete characterization:

Theorem: Let $(X_k)_k$ be a sequence of i.i.d. real valued random variables and define for each $m\in {\Bbb Z}$: $\gamma_m = {\Bbb E} \left( e^{2\pi i m X_1} \right)$. Set $S_n=X_1+\cdots X_n$. Then the following are equivalent:

1. The law of $S_n \ {\rm mod}\ 1$ converge in distribution to $\ { U}([0,1))$
2. $|\gamma_m|<1$ for every $m\in {\Bbb Z}^*$.
3. For every $m\in {\Bbb Z}^*$, $\theta\in [0,1)$: $\ {\rm supp} (m X_1) \not\subset \theta + {\Bbb Z}$

Proof: By the i.i.d condition $${\Bbb E} \left( e^{2\pi i m S_n} \right) = {\Bbb E} \left( e^{2\pi i m X_1} \right)^n = \gamma_m^n$$ Thus if $g$ is a 1-periodic trigonometric polynomial, then ${\Bbb E}(g(S_n))\to \int_0^1 g$ whenever $|\gamma_m|<1$ for every non-zero $m$. Conversely if for some non-zero $m$, $\gamma_m=e^{i\theta}$ then the convergence does not take place for $g=\exp(2\pi i m x)$ (like in the above counter-example). As trigonometric polynomials are dense in the 1-periodic continuous functions we get that $1\Leftrightarrow 2$. To see that 2 and 3 are equivalent, simply note that for non-zero $m$ $${\Bbb E} \left( e^{2\pi i m X_1} \right) = e^{2 \pi i \theta}$$ iff $mX_1 \in \theta +{\Bbb Z}$ almost surely.//

Note that the above holds without further assumptions on $X_1$. It need not be integrable, so in particular, the usual CLT need not even apply. I imagine that the above result is a well-known theorem to specialists.

Answer 4

Sketch of proof with heat kernel: Recall the heat kernel on $\mathbb{R}$ is $$ K(t,x,y)=\frac1{\sqrt{4\pi t}}\exp\left(-\frac{(x-y)^2}{4t}\right) $$ which is also the pdf of $N(y,2t)$ (as a function of $x$) from the Brownian motion interpretation of the heat kernel. Also recall the correspondence between heat kernels on $M$ and $M/G$ (for a Riemannian manifold $M$ and a discrete group $G\leq\operatorname{Isom}(M)$ acting freely properly discontinuously on $M$, action on the right so I don't have to do the quotient on the left): $$ K^{M/G}(t,xG,yG)=\sum_{g\in G}K^M(t,x,yg). $$ Thus, for $S^1=\mathbb{R}/\mathbb{Z}$, $$ K^{S^1}(t,x+\mathbb{Z},y+\mathbb{Z})=\sum_{m\in\mathbb{Z}}K^{\mathbb{R}}(t,x,y+m) =\sum_{m\in\mathbb{Z}}\frac1{\sqrt{4\pi t}}\exp\left(-\frac{(x-y-m)^2}{4t}\right) $$

On the other hand, we have Fourier series on the compact $S^1$: an orthonormal basis of eigenfunctions of the Laplacian on $S^1$ is $e^{i2\pi n\theta}$ with eigenvalue $4\pi^2n^2$ for each $n\in\mathbb{Z}$. Thus the heat kernel on $S^1$ has eigenfunction expansion $$ K^{S^1}(t,x+\mathbb{Z},y+\mathbb{Z})=\sum_{n\in\mathbb{Z}}\exp(-4\pi^2n^2t)\overline{\exp(i2\pi nx)}\exp(i2\pi ny) $$ This $L^2$-expansion converges locally uniformly on $(0,\infty)\times S^1\times S^1$ and hence also valid for $C((0,\infty)\times S^1\times S^1)$. Equating the two expressions $$ \sum_{n\in\mathbb{Z}}\exp(-4\pi^2n^2t)\exp(-i2\pi nx)\exp(i2\pi ny) =\sum_{m\in\mathbb{Z}}\frac1{\sqrt{4\pi t}}\exp\left(-\frac{(x-y-m)^2}{4t}\right) $$ and letting $t\to\infty$ gives $$ \left\lvert\sum_{m\in\mathbb{Z}}\frac1{\sqrt{4\pi t}}\exp\left(-\frac{(x-y-m)^2}{4t}\right)-1\right\rvert\leq\sum_{n\in\mathbb{Z}-\{0\}}\exp(-4\pi n^2 t)\to 0\text{ uniformly} $$ Hence $B_t\pmod 1$ converges (e.g. in distribution) to $U(0,1)$.