Prove that $\lim_{m \to \infty} (\lceil m U \rceil -mU ) \sim U(0,1)$

Question

Title says it all. If you take a really large uniform random number and the distance from it to the next integer, you should get a uniform random number between $0$ and $1$ ($U$ everywhere in the equations is a uniform random number between $0$ and $1$).

$$\lim_{m \to \infty} (\lceil m U \rceil -mU ) \sim U(0,1)$$

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For context, this identity is useful in proving the conjecture here: Need help validating a proof that for any point process with MTBF $t$, the events in an interval sized $u$ will be $\frac{u}{t}$, which says that if I take a large uniform random number and start an observation period there for any point process, the average number of events inside an interval size $u$ will be $\frac{u}{t}$ where $t$ is the mean time between events for the point process. Basically, you can rescale your time so that $t$ becomes $1$ unit and the result here shows that the time from the start of the interval to the next event will be uniform between $0$ and $t$.

Answer 1

$U(0,1)$ is standard notation for a uniform random variable so I will keep the name of the random variable (number) $V$ instead of $U$. Suppose $m\not\in\Bbb Z$.

Then $Y_m=mV\sim U(0,m)$ and $Z_m=\lceil Y_m\rceil-Y_m\in[0,1)$. The distribution function of $Z_m$,$$\begin{align*}F_{Z_m}(x)&=P(\lceil Y_m\rceil-Y_m\le x)\\&=\begin{cases}0,&x\le0\\P(Y_m\in[k-x,k],k\in\Bbb N_{\le\lfloor m\rfloor}),&0<x\le \lceil m\rceil -m\\P(Y_m\in[k-x,k],k\in\Bbb N_{\le\lfloor m\rfloor})+P(\lceil m\rceil-x\le Y_m\le m),&\lceil m\rceil -m<x<1\\1,&x\ge1 \end{cases}\end{align*}$$

Note that$$\begin{align*}&P(Y_m\in[k-x,k],k\in\Bbb N_{\le\lfloor m\rfloor})+P(\lceil m\rceil-x\le Y_m\le m)\\&=\sum_{k=1}^{\lfloor m\rfloor}\int_{k-x}^k\frac{dy}m+\frac1m\int_{\lceil m\rceil -x}^mdy\\&=\frac{x\lfloor m\rfloor}m+\underbrace{\frac1m\int_{\lceil m\rceil -x}^mdy}_{\le\frac1m}\end{align*}$$

so the underlined expression vanishes as $m\to\infty$ and the asymptotic behaviour of the second and third cases is identical. Taking the limit as $m\to\infty$,

$$\begin{align*} \lim_{m\to\infty}F_{Z_m}(x)&=\begin{cases}0,&x\le0\\\lim_{m\to\infty}\frac{x(m-\{m\})}m,&0<x<1\\1,&x\ge1 \end{cases}\\ &=\begin{cases}0,&x\le0\\x,&0<x<1\\1,&x\ge1 \end{cases}\\ &\sim U(0,1) \end{align*}$$where $\{\cdot\}$ is the fractional part function.

Answer 2

Adding a visualization of cases 2 and 3 from the first conditional equation of @ShubhamJohri's answer (for my own future reference; this helps see why the equation is true). Pasting the equation here:

$$\begin{align*}F_{Z_m}(x)&=P(\lceil Y_m\rceil-Y_m\le x)\\&=\begin{cases}0,&x\le0\\P(Y_m\in[k-x,k],k\in\Bbb N_{\le\lfloor m\rfloor}),&0<x\le \lceil m\rceil -m\\P(Y_m\in[k-x,k],k\in\Bbb N_{\le\lfloor m\rfloor})+P(\lceil m\rceil-x\le Y_m\le m),&\lceil m\rceil -m<x<1\\1,&x\ge1 \end{cases}\end{align*}$$

In case-2, we have to ignore the purple area. But in case-3, some of the purple area turns green and adds an additional term. The orange regions are common to both cases.

Answer 3

Adding a proof that this works for $m \in \Bbb Z$ even if it isn't large. For some integer $0 \leq i \leq m$, if $V \in \left(\frac{i}{m},\frac{i+1}{m}\right)$ we have:

$$Y_m = (i+1) -m\left(\frac{U+i}{m}\right) = 1-U \sim U$$

Since this holds for all $i$, we get that $Y_m \sim U$.

Answer 4

Here is another proof that relaying in a rather humble result from Fourier analysis: Fejer's formula.

Notice that $\lceil x\rceil -x=1-\{x\}$, where $\{x\}=x-\lfloor x\rfloor$. Suppose $f\in \mathcal{C}[0,1]$, then the function $\phi(x)=f(1-\{x\})$ is a bounded $1$-periodic (i.e. periodic with period $1$).

Féjer's result states that for any $g\in L_1([0,1])$ $$ \int^1_0 g(x)\phi(nx))\,dx\xrightarrow{n\rightarrow\infty}\Big(\int^1_0\phi(x)\,dx\Big)\Big(\int^1_0 g(x)\,dx\Big) $$ In particular, for $g\equiv1$ we get that $$E[f(1-\{nU\})] = \int^1_0 f(1-\{nu\})\,du\xrightarrow{n\rightarrow\infty}\int^1_0 f(1-u)\,du=E[f(U)]$$ since $1-U\stackrel{law}{=}U$.