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Replacing $b$ by $b/\|a\|_2$, we may assume WLOG that $\|a\|_2=1$. Let $R$ be a rotation matrix that maps $a$ to $v=(1,0,\ldots,0)^T$. Then $a^Tx=R(a)^T R(X)=v^T R(X)$, which has the same distribution …

One can avoid the $n$-dependence. Let $H$ be a Hilbert space, endowed with the norm $\|\cdot\|$. Given $f:\mathbb{S}^{d-1}\to H$ which is $L$-Lipschitz, the goal is to prove a concentration inequalit …

If $f_\theta(x)$ is a Lipschitz function of $X$ then standard concentration inequalities for Lipschitz functions (e.g. Mcdiarmid's inequality, see for instance https://people.eecs.berkeley.edu/~bartl …

The implication goes the other way. The "standard" inequality you quote, usually called McDiarmid's inequality is derived from the second inequality that you ask about. See e.g. http://empslocal.ex.a …

I will answer the question as stated though I am not sure you wanted to square the norm of $P_n-P$. Let $e_1,\ldots,e_k$ be the standard basis of ${\bf R}^k$. Write $\mu:=\sum_{j=1}^k p_j e_j$. Let $d …

The function you propose is related to the L'evy concentration function, studied by Kolmogorov, Rogozin, Esseen and others. See the special volume [1] https://link.springer.com/chapter/10.1007/97 …

The exponent 2 on $\epsilon$ cannot be improved by passing to the $L^1$ norm. Consider $X_i$ i.i.d. uniform in $[0,1]$. Let $C$ be a constant, and denote $\varepsilon_n= C/\sqrt{n}$. Let $A_n $ be the …

Here is a simple counter example to the original question with $B=y=1$, which also gives an alternative proof of Theorem 2 from Iosif Pinelis' answer. Let $\{R_i\}_{1 \le i \ge n}$ be independent Ra …

Such an estimate cannot hold in general. Here are two different counterexamples: Consider a complete bipartite graph with $n/2$ nodes on each side and $m=n^2/4$ edges. Take $q$ bounded, e.g. $q=2$. …

Let $\Phi(r)=P(X_1>r)$. Then for $s<t$ we have $$P(X_{(1)}<t, \; X_{(2)}<s)=\Phi(s)^n+n(\Phi(t)-\Phi(s))\cdot \Phi(s)^{n-1}$$ and from this one can obtain the exact distribution of $X_{(1)}- X_{(2)}$. …

Observe that $X_n=X_{n-1}(1+Z_n)$ where $\{Z_k\}_{k \ge 1}$ are i.i.d. standard normal. Hence to analyze the asymptotic distribution of $|X_n|$, pass to logarithms, to get $$\log(|X_n|)= \log(|X_1|) + …

Let $b$ denote the LHS. Expanding $e^{\lambda X}$ in a power series you can deduce that $$E(e^{\lambda X}) \ge \sum_{k \ge 0} \frac {b \lambda^k \delta^k}{k!}=b e^{\lambda \delta} \,.$$

If $f$ is the sum and all the components are identical then the inequality fails for $t=n$.

There are better bounds than the Azuma-Hoeffding bound for the tail probability of a Martingale . A good starting point is the classical inequality of Freedman (1975) [1]. But in the union bound for y …

You can use the Bernstein inequality along a discrete skeleton and then pass to the limit using continuity.