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If what you ask is to find the distribution of $Z=\min|x_i-y_j|$ where $x_i, i=1,\dots,m$ and $y_j, j=1,\dots,n$ are independent random points on $[0,1]$ (scaling to $L$ is trivial), then one can writ …

Since I was requested to elaborate, here goes. First, let's look at the automorphism of the unit circle induced by this mapping (written in the least revealing way). With $z=e^{it}$, as usual, we have …

There is a good reason why you cannot get anything for the standard Levy-Prokhorov distance in high dimensions. Let's consider the uniform distribution on the sphere of radius $R$ in $\mathbb R^3$ and …

This should really be a comment, but it just takes too much space to put in the comment box. I was quite puzzled by Ryan's remark that Feige's problem (with just some constant) is hard while it is a t …

For these matters, I usually rely on the following trickery: $$ 2\cosh t-2=\int_0^t (e^s-e^{-s})\,ds\le \int_0^t 2s e^s\,ds \\ \le\int_0^t 2se^{as^2+\frac 1{4a}}\,ds=\frac 1a e^{\frac 1{4a}}[e^{at^2}- …

You understand that it may be a good time to quit when you start forgetting what used to be your favorite tricks. Let $u=\frac ab>0$. Let $F_s$ be the CDF corresponding to $a'=u(1+s),b'=1+s$, so $F_s' …

Note that $1\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}+3\cdot\begin{bmatrix}1\\2\\4\end{bmatrix}=3\cdot\begin{bmatrix}1\\1\\1\end{bmatrix}+1\cdot\begin{bmatrix}1\\3\\9\end{bmatrix}=\begin{bmatrix}4\\6\\ …

Here is a positive answer to Q1. Step 1: Without loss of generality, each $X_i$ is symmetrically distributed. Indeed, consider an independent copy $X_i'$ of the family $X_i$ with the same joint distri …

There is a simple but often efficient trick to facilitate such computations. Let $H_k$ be the normalized Hermite polynomials and let $\Phi,\Psi$ be any two functions square integrable with respect to …

Yes, with essentially the same proof. In the scalar case, just notice that $e^{cX_n+bZ_n}$ is a supermartingale as long as $p(e^{c+b}-1)\le b$. In the vector case you will need to find positive vector …

"How do you see this?" It is quite simple, actually. For a positive random variable $Z$, we have $E[e^{-\beta Z}]=\beta\int_0^\infty e^{-\beta t}P[Z\le t]\,dt$. Thus, if $P[Z\le t]\asymp t^q$ for $0<t …

I am not quite sure if the previous long discussion has already resulted in a full proof of anything but here is the crude bound that shows that the expectation in question is infinite. Let $X_i$ be i …

How was your formula derived? You can do most of the work in your head without touching pen or paper. First, project to a random plane. Then the area will drop twice on average (Archimedes), so the an …

In my opinion, that other question has been answered by michael completely in the very first comment. However, since the question arose again, let me just spell the details of michael's answer out. W …

Here is the lower bound of $0.49$ for all $\alpha\ge 1$. Note that $\min_\pi F(\pi)$ is a non-decreasing function of $\alpha$, so it is enough to consider $\alpha=1$. Also, the truth is about $0.55$ f …