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

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?

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?

• One-liner: WLOG $\mu=0$, take the heat kernel on $S^1$ and let $t\to\infty$. Commented May 31, 2021 at 22:33
• What's $S^1$? What's the heat kernel? Hitting Google now for the latter. Commented May 31, 2021 at 22:50
• This is very exciting. Please, please post an answer with more details :) Commented May 31, 2021 at 22:51

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.

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. $$\square$$

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, $$\{ \sigma X\} \xrightarrow[\sigma \rightarrow \infty]{\text{(d)}} \mathcal{U}([0,1])$$

• Sorry, I'm missing a lot of context due to my illiteracy in some of these techniques. What is $L^2$? And what is the difference between $f$ and $f(t)$? Is $f(t)$ just one between $t \in [0,1]$ and $0$ otherwise? Commented Jun 3, 2021 at 4:40
• Yes, you get the function $f$ right. For $L^2$, by definition, $g \in L^2([0,1])$ if $$\int_0^1 |g(t)|^2dt <\infty$$ Commented Jun 3, 2021 at 4:44
• There is something odd about wlog'ing $\mu = 0$. It is easy to establish that the distribution of $X - \lfloor X \rfloor$ is the same as $X - n - \lfloor X - n \rfloor$ where $n \in \mathbb{Z}$, but that is not sufficient to give the required WLOG symmetry. In the case where $\mu \in \mathbb{Z}$, the distribution of $X - \lfloor X \rfloor$ is clearly symmetric around 0.5, which is not the case otherwise. Commented Jun 4, 2021 at 10:50
• What is to say, in fact, I don't need $\mu=0$, even $\mu \ne 0$ and vary with $n$, the Parseval's identity stays the same. I just made $\mu =0$ to simplify the presentation. Commented Jun 4, 2021 at 10:55
• @RohitPandey First question: The characteristic function of Gaussian variable Commented Jun 6, 2021 at 14:11

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 $$\sigma\rightarrow\infty$$.

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 $$\phi\in L_1$$.

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 $$\sigma X\sim N(\mu,\sigma)$$.

• Sorry, what does this mean: $f\in\mathcal{C}_b(\mathbb{R})$? Commented Jun 6, 2021 at 10:03
• @RohitPandey: $f$ bounded and continuous. Weak convergence of measures $\mu_n\Rightarrow\mu$ means that for any such $f$, the numeric sequence $\int f\,d\mu_n$ converges to $\int f\,d\mu$. A consequence of this is that for any Borel set $A$ whose boundary $\partial A$ has measure $0$ under $\mu$ ($\mu(\partial A)=0$), the numeric sequence $\mu_n(A)$ converges to $\mu(A)$. Commented Jun 6, 2021 at 10:43
• In the $\Big(\frac{1}{T}\int^T_0 g(x)\,dx\Big)\Big(\int_{\mathbb{R}} \phi(x)\,dx\Big)$ part, won't $\Big(\int_{\mathbb{R}} \phi(x)\,dx\Big)$ always be $1$? Commented Jun 6, 2021 at 23:17
• Got it, thanks! Also, proving $\mathbb{E}\Big[f\big(\{\sigma_n X+\mu\}\big)\Big] =\mathbb{E}[f(U)]$ for $f\in\mathcal{C}_b(\mathbb{R})$ is sufficient to demonstrate weak convergence of ${\sigma_n X+\mu}$ to $U$? Commented Jun 6, 2021 at 23:40
• (+1) Great answer. I did learn something from your answer. It's really interesting that the my accepted answer has lower scores than yours. Perhaps, OP was also curious about the convergence rate. Commented Jun 14, 2021 at 4:42

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.

• Sorry, what is the set ${\Bbb Z}^*$? Commented Jun 6, 2021 at 9:29
• And what does "$X_1$ has a density w.r.t. Lebesgue" mean? Commented Jun 6, 2021 at 9:31
• ${\Bbb Z}^* = {\Bbb Z}\setminus \{0\}$ and density w.r.t. Lebesque means that the law of $X_1$ may be given as $dP=f(x)\; dx$ with $f\geq 0$ and $L^1$ function with integral 1. (But the statement is much more general than that). Commented Jun 6, 2021 at 9:36
• @RohitPandey When $g(t) = \sum_{m} c_m \exp(2\pi i m t)$ (finite sum) then ${\Bbb E}(g(S_n)) = \sum_{m} c_m \gamma_m^n \to c_0 \gamma_0 = c_0 = \int_0^1 g$ as $n\to +\infty$. Commented Jun 7, 2021 at 11:51
• @RohitPandey Indeed, thanks for spotting. I have corrected. Commented Jul 5, 2021 at 8:07

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 $$U(0,1)$$.

• Not the one-liner you said it would be! Commented Jun 3, 2021 at 22:14
• @JeanL. It is a one-liner if you know enough about heat kernel: Take limit $t\to\infty$ in $K^{S^1}(t,[x],[\mu])$. Commented Jun 3, 2021 at 23:22