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$\newcommand\th\theta$$T$ is a sufficient statistic and thus a random variable. So, the integral $\int f(T)\Phi(T,\theta)dT$ cannot possibly have a meaning. The conclusions that Rao is trying to reach …

$\newcommand\R{\Bbb R}\newcommand{\1}{\mathbf{1}}\newcommand{\xx}{\mathbf{x}}\newcommand{\yy}{\mathbf{y}}\newcommand{\uu}{\mathbf{u}}\newcommand{\vv}{\mathbf{v}}\newcommand{\0}{\mathbf{0}}$ Your quest …

The answer is $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$. This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du} …

Your question is Find the minimum of $f(x,y)=(x-2)^2+y$ subject to $y-x^3\ge0$, $y+x^3\le0$ and $y\ge0$. The restrictions $y-x^3\ge0$, $y+x^3\le0$, and $y\ge0$ can be rewritten as $0\le y\le-x^3$ or …

$\newcommand\R{\mathbb R}\newcommand\ep\epsilon\newcommand\bad{\text{bad}}$One would think that the answer is "of course no, descent does not have to monotonic". However, the no stationary points cond …

For real $u,v$, let \begin{equation*} Q(u,v):=P(Y_1>u\sqrt2,Y_2>v\sqrt2)=\int_u^\infty dz\,\varphi(z)\Big(1-\Phi\Big(\frac{2v-z}{\sqrt3}\Big)\Big), \end{equation*} where $\varphi$ and $\Phi$ are t …

$\newcommand\R{\mathbb R}$Let $S:=\mathbf S$. Let $X$ be any random vector in $\R^n$ with a spherical symmetric distribution (say such that $P(X=0)=0$). Then $|X|$ is independent of $U:=X/|X|$ and $U$ …

Let $n=1$, $f(t)=t^2 + |t|^{7/2}\sin(1/|t|)$ for $t\ne0$, $f(0):=0$. Then $f'(0)=0$ and $f''(0)=2>0$, so that $0$ is a strict local minimum of $f$. However, $f''(t)\sim-|t|^{-1/2}\sin(1/|t|)$ as $t\to …

$\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$First, one should not denote a random vector in $\R^n$ (which is not actually a vector in $\R^n$ but a function with values in $\R^n$) and a true, n …

$\newcommand{\R}{\mathbb R}$Your conjecture is true. Indeed, \begin{equation*} \begin{aligned} f(x)&=F(a,u):=a\frac{\sin u}u+4\cos u, \\ g(x)&=G(a,u):=a\frac{\cot(u/2)}{u^2}\,h(u)-4\frac{\sin …

The answer is no. E.g., let $f(x):=|x-1|^{3/2}$. Then $$ F(s)=\begin{cases} F_1(s) &\text{ if } 0<s\le1/9, \\ F_2(s) &\text{ if } 1/9\le s<1, \end{cases} $$ where $$F_1(s):=1 - 3 s - 2s^{3/2},$$ $ …

The answer to your question is: No, in general $F$ is not differentiable everywhere on $(0,\infty)$. First, to simplify the notations a bit, consider the change of variables $x=e^u$, $y=e^v$, $s=e^t …

Let $M:=M(x,y,z)$ be the $8\times8$ matrix in question. Let $m(x):=M(x,0,0)$. We have $$\det m(x)= -256 e^{i x/2} x^2 \cos (2 x) \big((x^2+1)^2 \cos (2 x)-(x^2-1)^2\big).$$ So, $|\det m(x)|$ will be o …