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I think the following geometric argument is interesting and maybe sufficient to answer "why" at an intuitive level (?). When we take the powers of $x$ in the complex plane, the absolute value scales g …

Here's a step that seems nice enough to point out. It still leaves a parameter to pick, and I'm not sure it's ever better than applying Bernstein, but it does something different. We can get a probab …

Throw $m$ balls into $n$ bins independently, each ball selecting a bin from the distribution $A \in \Delta_n$. This question is about lower-bounding the max-loaded bin. Background. In this MO answer I …

If $X \sim Normal(0,1)$, then we have the tail bound: $$ (*) \qquad\Pr[X > t] \leq \mathcal{O}\left(\frac{e^{-t^2/2}}{t}\right) .$$ Now for general variables $X$, a nice condition is that $X$ be su …

The Good-Turing estimator addresses a very similar problem. The model is that there is an unknown number of types $n$ and a probability distribution $p \in \Delta_n$. Each time you pick up a new she …

I'm looking for "tail-bound-like" inequalities that look like this (I state a specific setting but more general settings are interesting): Let $D$ be a distribution on a set of "nice" functions $g …

Don't forget that far out in the left tail, the Binomial CDF is multiplicatively approximated by the PMF, because terms grow geometrically. Example. For $t \leq \frac{np}{2}$, we claim $\Pr[X = t] \le …

No, it looks like many different sufficiently symmetric distributions with enough concentration at $0$ will have $\arg\min_c E[|X-c|^p] = 0$ for all $p > 0$. Concrete family of examples: if $$ X = \be …

Maybe this is a stretch, but laws of large numbers, and more precise concentration of measure bounds, can be proven by the "isoperimetric" approach which can be called geometric and was pioneered by T …

For the expectation, linearity of expectation should help a lot. There are ${n\choose 2}$ pairs of chords, and if each chord is drawn i.i.d. then each pair has some probability of intersection of $p$, …

This is a bit different and doesn't address the question, but hopefully close enough to be useful: we know some things about $\mathbb{E} L(Z,Y)$ for other loss functions $L$. If and only if $L$ is a …

Turns out that Gneiting and Raftery give an example in Section 4.2 of the continuous ranked probability score (CRPS), which is strictly proper for $\mathcal{P}$ equal to the Borel probability measures …

Adding to Iosif Pinelis' answer, there are two points here. First, as he says, the fact that we have a martingale rather than i.i.d. variables doesn't change much as proofs generally extend. So, secon …

Here's a sketch and a link for how I prove it. Let $$ f(x) = - \left( \frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5}\right) \phi(x) .$$ Now show (lemma): $\frac{df}{dx} = \left(1 + \frac{15}{x^6}\righ …

A self-contained proof. Step 1: $\mathbb{E} \|\hat{P}_n - P\|_2^2 \leq \frac{1}{n}$. Step 2: McDiarmid's inequality. Let $X_i$ be the number of samples of $i \in \{1,\dots,k\}$. Then $X_i \sim \te …