Is there a constant $c>0$ such that for $X,Y$ two iid variables supported by $[0,1]$, $$ \liminf_\epsilon \epsilon^{-1}P(|X-Y|<\epsilon)\geqslant c $$ I can prove the result if they have a density, of if they have atoms, but not in the general case.


2 Answers 2


If $\epsilon \geqslant \tfrac{1}{n}$, then $$ \mathbb{P}(|X-Y|<\epsilon) \geqslant \sum_{i=1}^n \mathbb{P}(X, Y \in [\tfrac{i-1}{n}, \tfrac{i}{n}]) = \sum_{i=1}^n (\mathbb{P}(X \in [\tfrac{i-1}{n}, \tfrac{i}{n}]))^2 . $$ It follows that $$ \mathbb{P}(|X-Y|<\epsilon) \geqslant \frac{1}{n} \biggl(\sum_{i=1}^n \mathbb{P}(X \in [\tfrac{i-1}{n}, \tfrac{i}{n}])\biggr)^2 = \frac{1}{n} \, . $$ If we choose $n$ so that additionally $\epsilon < \frac{1}{n-1}$, then we obtain $$ \mathbb{P}(|X-Y|<\epsilon) > \frac{\epsilon}{\epsilon + 1} \, . $$ This leads to the desired result with $c = 1$.

  • 2
    $\begingroup$ Interestingly, while the estimate given above is sharp at least when $\epsilon = \tfrac{1}{n}$ (consider a uniform distribution over $\frac{i}{n - 1}$, $i = 0, 1, \ldots, n - 1$), my guess is that the optimal value of $c$ is in fact $2$. However, I fail to see a proof. $\endgroup$ Commented Mar 10, 2020 at 21:44

Mateusz Kwaśnicki's guess, that the optimal value of $c$ is 2, is correct! In fact:

Theorem: Suppose $X,Y$ are i.i.d. random variables on $\mathbb{R}$. Then the limit $\lim_{\varepsilon \rightarrow 0} \varepsilon^{-1} \mathbb{P}[|X - Y| < \varepsilon]$ exists. It is $+\infty$ unless $X$ has a density function $f$ satisfying $||f||_2^2 = \int f^2\ \text{d}x < \infty$, in which case the limit is equal to $2 ||f||_2^2$.

Restricting to distributions on $[0,1]$, Cauchy-Schwarz says $$ 1 = \int_0^1 f\ \text{d}x \le \left ( \int_0^1 f^2\ \text{d}x \right )^{1/2} \left ( \int_0^1 1\ \text{d}x \right )^{1/2} = ||f||_2, $$ with equality iff $f = 1$. That is, the constant $c = 2$ is valid for every distribution, and it is sharp only for the uniform distribution.

Here are a few examples where the limit is infinite:

Example 1: If $X$ has an atom then $\mathbb{P}[|X-Y| < \varepsilon]$ is bounded away from 0. So the limit in question diverges like $\varepsilon^{-1}$.

Example 2: The Cantor ternary function is the CDF of a probability distribution on $[0,1]$ which is nonatomic but not absolutely continuous. A random sample from this distribution is given by $\sum_{k \ge 1} c_k \tfrac{2}{3^k}$ where $\{c_k\}$ are i.i.d. Bernoulli random variables. (Thought of in terms of the "repeatedly take out the middle third" construction of the Cantor set, the $c_k$'s are the choices of left versus right third.) If $\varepsilon = 3^{-n}$, then $|X-Y| < \varepsilon$ iff the first $n$ $c_k$'s agree. Thus $$ \mathbb{P}[|X-Y| < \varepsilon] = 2^{-n} = \varepsilon^{\log_3 2}. $$ So the limit in question diverges like $\varepsilon^{-1+\log_3 2} = \varepsilon^{-0.37}$.

Example 3: The power law $f(x) = \tfrac{1}{2\sqrt{x}}$ is an example of a density function which is not in $L^2$. In this case we can evaluate $\mathbb{P}[|X-Y| < \varepsilon]$ exactly (well, if you believe in inverse hyperbolic trig functions). The asymptotic behavior is $$ \mathbb{P}[|X-Y| < \varepsilon] = \tfrac{1}{2} (-\log \varepsilon) \varepsilon + (\tfrac{1}{2} + \log 2) \varepsilon + O(\varepsilon^2). $$ So the limit in question diverges logarithmically.

Mateusz's slick argument worked on the block diagonal, a sum of $n$ squares of width $1/n$. This covers about half of the area of the strip $|X-Y| < \tfrac{1}{n}$, which is why the resulting constant is half of optimal. There's probably a hands-on way to extend it, but you start using words like "convolution" and I found it easier to reason in Fourier space. This question is asking for a bound on the density of $X-Y$ at 0. This random variable has nonnegative Fourier transform (characteristic function, if you prefer), and that condition alone is enough to guarantee positive density (shameless self-citation: Lemma 3.1 in this paper). In general, when it makes sense, the density at 0 of a random variable is equal to the integral of its Fourier transform. So our job is to make that concrete and translate it back to a statement about PDFs.

Convention: The Fourier transform (characteristic function) of a random variable $X$ is the function $t \mapsto \mathbb{E}[e^{2 \pi \mathrm{i} X t}]$. The Fourier transform of a function $f$ is $t \mapsto \hat f(t) = \int e^{2 \pi \mathrm{i} x t} f(x)\ \text{d}x$. If $X$ is absolutely continuous (has a density function), then its Fourier transform is equal to the Fourier transform of its density function. With this convention the Plancherel identity reads $\int f \bar g\ \text{d}x = \int \hat f \overline{\hat g}\ \text{d}t$ (no normalization constant).

Lemma: Let $\psi$ denote the Fourier transform of $X$. Then $\psi \in L^2$ iff $X$ is absolutely continuous and its density function $f$ is in $L^2$. Moreover, if this holds, then $||\psi||_2 = ||f||_2$.

Proof of Lemma: The Fourier transform is an isometry of the space $L^2$. The ($\Leftarrow$) direction is immediate: if $X$ has a density function in $L^2$, then certainly its Fourier transform is in $L^2$. For ($\Rightarrow$), if $\psi$ is in $L^2$ then it is the Fourier transform of some function $f \in L^2$. But then $f$ defines the same distribution as $X$, so it is indeed the density function. $\square$

(This is probably a basic result in some probability textbook?)

Proof of Theorem: Let $\psi$ denote the Fourier transform of $X$. Then the Fourier transform of $X-Y$ is $t \mapsto \psi(t) \psi(-t) = \psi(t) \overline{\psi(t)} = |\psi(t)|^2$. Let $T_\varepsilon$ denote the "triangle filter" $T_\varepsilon(x) = \varepsilon^{-1} (1 - |x|/\varepsilon)$ for $|x| \le \varepsilon$ and $T_\varepsilon(x) = 0$ otherwise. It is a standard calculation that $\hat T_\varepsilon(t) = \operatorname{sinc}^2(\pi t \varepsilon)$. (Here $\operatorname{sinc} x = \tfrac{\sin x}{x}$.) This is integrable, so we can compute the expected value of $T_\varepsilon$ using the Fourier transform. We find $$ \varepsilon^{-1} \mathbb{P}[|X-Y| < \varepsilon] \ge \mathbb{E}[T_\varepsilon(X-Y)] = \int |\psi(t)|^2 \operatorname{sinc}^2(\pi t \varepsilon)\ \text{d}t. $$ The integrand is nonnegative and, as $\varepsilon \to 0$, it converges (pointwise) to its (pointwise) supremum $|\psi(t)|^2$. So, by an appropriate incantation of the monotone convergence theorem (the one in this comment applies exactly), the integral converges to $\int |\psi(t)|^2\ \text{d}t = ||\psi||_2^2$. If this is infinite, then also $\lim_{\varepsilon \to 0} \varepsilon^{-1} \mathbb{P}[|X-Y| < \varepsilon] = +\infty$.

Suppose now that $||\psi||_2^2 < \infty$, i.e., $\psi \in L^2$. Consider the "box filter" $B_\varepsilon$ defined by $B_\varepsilon(x) = (2 \varepsilon)^{-1}$ for $|x| < \varepsilon$ and $B_\varepsilon(x) = 0$ otherwise. This has $\hat B_\varepsilon(t) = \operatorname{sinc}(2 \pi t \varepsilon)$. This is bounded and $|\psi|^2$ is integrable. So we can again compute expected values in Fourier space: $$ (2\varepsilon)^{-1} \mathbb{P}[|X-Y| < \varepsilon] = \mathbb{E}[B_\varepsilon(X-Y)] = \int |\psi(t)|^2 \operatorname{sinc}(2 \pi t \varepsilon)\ \text{d}t. $$ The integrand converges pointwise to $|\psi(t)|^2$. We don't have nonnegativity this time, but now we know integrability of the bounding function $|\psi|^2$. So we can apply the dominated convergence theorem, concluding that $$ \lim_{\varepsilon \to 0} (2 \varepsilon)^{-1} \mathbb{P}[|X-Y| < \varepsilon] = \int |\psi(t)|^2\ \text{d}t = ||\psi||_2^2. $$ The lemma completes the proof. $\square$

  • $\begingroup$ Wow, this is an excellent answer! I learned a lot $\endgroup$ Commented Apr 15, 2020 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.