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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 …

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 …

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. …

$\newcommand\vpi\varphi\newcommand\de\delta\newcommand\ep\varepsilon\newcommand\R{\Bbb R}$Continuous compactly supported functions are dense in $L^1$. So, the problem reduces to the following: Given …

$\newcommand\la\lambda\newcommand\be\beta$Yes, this is true. Indeed, on the event $A_\la:=\{|K-\la|\le\la^{5/8}\}$, $$Z=Z_\la:=\frac{\la+\be}{\sqrt\la}\, \ln\Big(1+\frac{K-\la}{\la+\be}\Big) \\ =\fra …

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$. So, $$I_n^*=n/2$$ for even $n$. It also follows …

$\newcommand\om\omega$This is of course not true in general (and almost never true). Indeed, $I:=I_p$ is a continuous function, at least when $\mu$ and $\nu$ are compactly supported. Then the equality …

$\newcommand\si\sigma$This is not true. For instance, suppose that $\mu_1=\mu_2=\mu_3=0$ and $(\si_1,\si_2,\si_3)=(1,2,6).$ Then $d(P, Q) + d(Q, R) - d(P, R)=-263/1764<0$.

$\newcommand\fm{f_{\max}}\newcommand\si\sigma\newcommand\Si\Sigma\newcommand\R{\Bbb R}$Let $f$ be the p.d.f. of the normal distribution $N(\mu,\si^2I_n)$ over $\R^n$, where $\si>0$ is a real number an …

Counterexamples, with $d=1$: $$f(x)=\frac{\ln2}{x\ln^2 x}\,1(0<x<1/2)$$ for all real $x$ -- for your statement 1; $$f(x)=\frac{\ln2}{x\ln^2 x}\,1(x>2)$$ for all real $x$ -- for your statement 2.

We have \begin{equation*} F(x_0)=\infty \tag{1}\label{1} \end{equation*} for any nonzero $x_0$. Indeed, by spherical symmetry, without loss of generality \begin{equation*} x_0=(2a,0,\dots,0) \ …

Not if $C>1$. (The case $C=1$ is trivial, and $C$ cannot be $<1$.) Indeed, let $H(x):=P(Y\ge x)$ and $G(x):=\min(1,CH(x))$ for real $x$. Then $G(x)=P(X\ge x)$ for some random variable $X$ and all real …

The answer is yes. Indeed, for any real $t\ge0$, \begin{equation*} P(XY\ge t)\le P(X\ge\sqrt t)+P(Y\ge\sqrt t) \le 2P(Z\ge\sqrt t)=:G(t); \tag{10}\label{10} \end{equation*} note that $G$ is a …

First this fact was proved by Hoteling himself -- see formula (25). See also e.g. formula (28.40) in The Advanced Theory of Statistics, Volume 2, 1946 by Kendall.

Note that $X-Y=\sum_1^n X_i$, where the $X_i$'s are i.i.d. random variables with $P(X_i=1)=(1-p)p=P(X_i=-1)$ and $P(X_i=0)=1-2(1-p)p$. So, assuming that $0<p<1$, by the local limit theorem -- see e.g. …