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

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 $$1$$).

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

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 $$t$$.

• Okay, nice result. What is $U$ in the limit? Commented May 15, 2021 at 5:24
• $U$ everywhere is just a uniform random number between $0$ and $1$. Commented May 15, 2021 at 5:28
• Why only in the limit? Commented May 15, 2021 at 5:34
• Does it hold for all real $m$? I see it holds when $m$ is an integer, not obvious for a general real number. Commented May 15, 2021 at 5:34
• Regarding the meta post : Good enough questions stay. You improved yours, and that's all that matters. Thank you for the prompt response and wish you a good day ahead with good questions ,better answers and great conversations. Commented May 17, 2021 at 21:57

$$U(0,1)$$ 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

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,&0where $$\{\cdot\}$$ is the fractional part function.

• When $m\in\Bbb Z$ our work is considerably reduced as the second case of $F_{Z_m}(x)$ applies when $0<x<1$. Commented May 15, 2021 at 6:22

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

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.

• That's an image that's incredibly hard to see. Commented May 17, 2021 at 18:09
• It expands on the answer. I'd gladly post it as a comment but can I add images to comments? Commented May 17, 2021 at 20:09
• @amWhy - changed the background of the image to white. Hopefully this makes it easier to see. Commented May 17, 2021 at 20:18
• Thanks, @RohitPandey ! Commented May 18, 2021 at 0:34

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 $$Y_m \sim U$$.

• This should have been part of you question. Commented May 17, 2021 at 18:10
• I can edit and add it there now, let me know.. I came up with it much after I'd asked the question and an answer had been posted+accepted. What do I do in such cases (where I come up with some insight after people have posted answers)? If I edit the question, it might change the nature of it and make the answers look strange since they assumed the original question. Though admittedly, this wasn't the case here. Commented May 17, 2021 at 18:27
• What is $I$? Do you mean $\Bbb Z$? Commented May 17, 2021 at 19:08
• Oh, its the set of integers.. should I use another notation? Commented May 17, 2021 at 19:09
• @RohitPandey Fixed. Commented May 17, 2021 at 19:10

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 $$1-U\stackrel{law}{=}U$$.

• What is $\mathcal{C}[0,1]$? Commented May 21, 2021 at 21:12
• The space of continuous functions in the interval $[0,1]$. Commented May 21, 2021 at 21:23
• @RohitPandey: What I am doing (as others did through by directly using the CDF of $Y_m:=\lceil mU\rceil - mU$) is to prove that $Y_m$ converges weakly to $U$ based on the definition of weak convergence: that is, that for any bounded continuous function $f$ on $[0,1]$ (this is where $Y_m$ and $U$ take values) one has that $E[f(Y_m)]\xrightarrow{m\rightarrow\infty} E[f(U)]$. Commented May 21, 2021 at 21:32
• I see, that seems to be a stronger result! I'm not familiar with these techniques, so need some time to digest this. Very interesting and thanks for sharing! Commented May 21, 2021 at 21:34
• Finally, what I called Féjer formula does not appear as such in the literature. I learned of this result from a book of problems by Claude George, Exerceises in Integration, Springer-Verlag1, 1984, pp. 115-116, and who attributes it to Féjer. Commented May 21, 2021 at 21:51