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$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such th …

Any vector in $R^n$ whose coordinates sum to $0$ is a scalar multiple of the difference $x-y$ between probability vectors $x$ and $y$. So, your condition on $A$ and $B$ can be restated as $$\mathbf1^\ …

(This is to address the previous comment by the OP. Since the previous answer is already long, this is being done in a separate answer.) The answer to the mentioned comment is the following: there is …

do we at least have $Y_n=\Theta(\frac{\log n}{n})$ almost surely? The answer to this is no. It is not even true that $Y_n=\Theta(\frac{h(n)}{n})$ almost surely (a.s.) for any (deterministic or rando …

I am now trying to show that $H(\mu) \leq 1$ for $|\mu| \geq c$. As $H$ is an even function, it is sufficient to show that $H(\mu) \leq 1$ for $\mu \geq c$. This is not true. E.g., suppose that $c=1 …

$\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}$As I understand the problem, it is to compute $$e_{k_1,\dots,k_n}:=EM_nH_{k_1}(X_1)\cdots H_{k_n}(X_n),$$ where $n\ge1$ is an odd integer, the $H_k$ …

Suppose that $(X_t)_{t\in\Bbb R}$ is a Gaussian process stationary in the wide sense, so that $m(t):=EX_t=m$ and $Cov\,(X_s,X_t)=g(s-t)$ for some real $m$, some real-valued function $g$, and all real …

$\newcommand\ka\kappa$Intuition for this result is as follows. The condition on $h'$ implies that $h(y)=(1+o(1))y$ (as $y\to\infty$). So, for the tail function $T$ given by $T(y):=P(Y\ge y)$ we have $ …

Counterexample: $g_i(x)=x-1/2$ for all $i$ and all $x\in[0,1]$.

This conjecture is not true. E.g., let $P(X_i=q)=p=1-P(X_i=-p)$, where $p\in(1/2,1)$ and $q:=1-p$, so that $P(X_i>0)>1/2$, $EX_i=0$, and $Var\,X_i=pq$ for each $i$. Then for any real $c>0$ and $S_n:=X …

For any random variable $X$ with $E\max(X,0)<\infty$, you can determine the distribution of $X$ if you know the values of $$g(c):=E\max(X,c)$$ for all real $c$. Indeed, take any real $c$ and any real …

After cancellations and a bit of algebra, for $t:=\gamma>1$, we get $$\begin{aligned} f(n_0)&=\prod_{j=0}^{n_0-1}\Big(1-\frac{m}{n-j}\Big) \\ &\le\exp\Big(-m\sum_{j=0}^{n_0-1}\frac1{n-j}\Big) \\ &\l …

$\newcommand\si\sigma$The answer is no. E.g., suppose that $P(X_i=1)=2/e=1-(X_i=0)$ for $i=0,1$. Then $X_0$ and $X_1$ are sub-Gaussian with parameter $\si=1/\sqrt2$, so that we can take $\si_0=\si_1=1 …

This statement is false in general. E.g., suppose that $C=I=C[0,1]$. Then the function $1_{[0,1/2]}$ is bounded and $\sigma(C)$-measurable but not in $I$.

Let $S$ be the survival time. Then $$P(S\ge s)=\binom{n-s}a\Big/\binom na$$ for $s=0,1,\dots$. So, $$ES=-1+\sum_{s=0}^\infty P(S\ge s)=\frac{n+1}{a+1}-1.$$ So, $n=3$, $a=1$, $n^*=8$, $a^*=3$ will do. …