Search Results

Your objective function is already convex. This follows because the Hessian matrix $$\frac2{y^3}\,\begin{bmatrix} y^2&-xy\\-xy&x^2 \end{bmatrix}$$ of the map $\mathbb R\times(0,\infty)\ni(x,y)\mapsto …

Let $S:=\mathcal{S}$ and $a\cdot b:=\langle a,b \rangle$. Without loss of generality $x=0$ (or replace $S$ by $S-x$). That $(s^*,i^*)$ is a solution to the max-min optimization problem means the follo …

The answer is no. By considering random variables $X$ taking only one real value $c$ and then letting $c\to\infty$, we see that the infimum of your set is $-\infty$ and hence not attained.

Let us provide an explicit (albeit complicated) expression for $EM_k$, where \begin{equation*} M_k:=\min_{1\le i\le k}c^TX_i \end{equation*} and $k$ is any natural number. Without loss of generality …

As in the comment by leo monsaingeon, let $$p_*(x):=e^{h(x)}/c,$$ where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $c:=\int_0^1 e^{h(x)}\,dx$, so that $p$ is a pdf on $(0,1)$ with mean $1/2$, and $$V(p)=\int_ …

$\newcommand\th\theta$ It is not true in general that the Cramér--Rao lower bound is a solution to a meaningful optimisation problem: "Under some regularity conditions, the Cramér-Rao lower bound is …

$\newcommand\C{\mathcal C}$If $\C$ is convex, then the projection onto the convex set $X\C$ is uniquely defined and, moreover, $1$-Lipschitz, so that $\|X\hat w-Xw^\ast\|\le\|v-Ev\|$ and hence $\|\hat …

$\newcommand\si\sigma\newcommand\Si\Sigma$If $x$ is centered, then $u\cdot x\sim N(0,\si_u^2)$, where $\si_u^2:=u\cdot\Si u$ and $\Si$ is the covariance matrix of $x$. So, $$\frac{E(u\cdot x)^4}{E(u\c …

For real $a$ and $x\in(0,1)$, let \begin{equation*} p_a(x):=e^{ax+h(x)}/C(a), \end{equation*} where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $C(a):=\int_0^1 e^{ax+h(x)}\,dx$, so that $p_a$ is a pdf on $ …

Note that the constant function $f=c$ satisfies your conditions for any real $c\le-1$. So, $$\inf\int_0^1 f=-\infty.$$ Consider now the same problem but with the additional condition $f\ge0$. Then, …

The upper bound is not correct. E.g., let $n=101$, $x_1=\dots=x_{100}=1/1000$, $x_{101}=9/10$. Then $$\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}>7>2.$$ More …

We need to show that $$\sum x_j^2 y_j^2\le1/2+\Big(\sum x_j y_j\Big)^2\tag{0}\label{0}$$ given that $\sum x_j^2=1$ and $\sum y_j^2=1$. Consider the maximization of $\sum x_j^2 y_j^2$ for a fixed value …

Replacing $x_k^*$ by $x_k$ and noting that $|x_k^*|=|x_k|$, we see that the problem is to show that $$|s|^2-\sum_{k=1}^n|x_k|^2|y_k|^2\ge-1/2$$ for $x,y\in\mathbb{C}^n$ with $\|x\|_2=\|y\|_2=1$, where …

In fact, $x_1$ can be $\pi$-periodic. Indeed, let $m:=\mu$, $$a(t,m):=m\sin2t,\quad b(t,m):=m\cos2t-c_m,$$ where $c_m$ is the unique solution to the equation $$\int_0^\pi e^{-A(s,m)}(m\cos2s-c_m)\,ds …

The answer is no. Indeed, let $n=1$, $T=1$, and $\Lambda(t,u,v)\equiv v^2$, so that \begin{equation} I(x)=\int_0^1 x'(t)^2\,dt. \end{equation} Let $x_0:=0$ and, for each real $b\ge1$ and all $t\in[ …