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Recall that $E[X] = \int_0^\infty (1-F(x))\,dx$. So this shows that something like $F(x) = 1-1/(x \log x)$ for large $x$ is a counterexample. This does however show that $\lim_{x \to \infty} x^p (1-F …

I suppose here that $E$ is separable. Let $\epsilon > 0$ and fix a sequence $\epsilon_i > 0$ such that $\sum_{i=1}^\infty \epsilon_i < \epsilon$. Let $B_r$ denote the closed ball of radius $r$. Set $ …

Yes, it's attained. Note that the desired expression can be written as $E[(f 1_B)(\mathbf{x})]$. Then it's clear that we get the maximum by taking $B = \{f \ge 0\}$, so that $f 1_B = f^+$, the posit …

I assume $X_1 \sim f(x_1)$ means that that the distribution of $X_1$ has density function $f$. Note first that your $\kappa$ can only be $2$, otherwise the integral of $p$ will not equal $1$. Then $ …

Yes, both statements are true. For (a), note that for every Borel set $A$ of $C([0,t],X)$, the set $\{u \in C([0,T], X) : u|_{[0,t]} \in A\}$ is a Borel set in $C([0,T],X)$. This follows immediately …

In fact, every reasonable function can be made into an example by adding an appropriate constant. I'll write $Z$ for a standard Gaussian random variable. Recall the Gaussian Poincaré inequality: …

First of all, I suppose you mean $\kappa$ to be defined as $$\kappa(x,\;\cdot\;):=\bigotimes_{n\in\mathbb N}\mathcal N_{x_n,\:\sigma^2}$$ with $x_n$ on the right side instead of $x$, where $x = (x_1, …

The limit is not $u(\int_0^1 x p(x)\,dx)$ but rather $\int_0^1 u(x) p(x)\,dx$. I prefer probabilistic notation, so let $X_n \sim p_n$ and $X \sim p$. We are then supposing that $X_n \Rightarrow X$ i …

If it were subgaussian then we would have $E[e^{\lambda X Y}] < \infty$ for all $\lambda$. However, by conditioning and using independence we find $$E[e^{\lambda X Y}] = E[E[e^{\lambda X Y} \mid X]] …

The claimed inequality is not true. The simplest possible counterexample works: let $x,y \in \mathbb{R}^n$ with $|x-y| = \epsilon$, and take $\mu = \delta_x$, $\nu = \delta_y$. Then $W_1(\mu,\nu) = …

Yes, they are normally distributed. This is the Lévy-Cramér theorem.

No. Note that your hypothesis $F_Y = g(F_X)$ will be satisfied whenever $F_X, F_Y$ are both strictly increasing (simply take $g = F_Y \circ F_X^{-1}$). So let's take $X \sim N(0,1)$, $Y \sim N(1,1)$ …

Let $f(x) = x$, $P = \delta_0$, and $Q = (1-2\epsilon)\delta_0 + \epsilon \delta_c + \epsilon \delta_{-c}$. Then $\mathbb{E}_P[f] = \mathbb{E}_Q[f] = 0$ and $\operatorname{Var}_P(f) = 0$. By taking …

Note that $E[\varepsilon_n] = E[E[\varepsilon_n \mid X]] = 0$. Let $\sigma^2$ denote the variance of $\varepsilon_n$. Let $Y_n = X + \varepsilon_1 + \dots + \varepsilon_n$. Note that $\frac{\vareps …

It's true; we can use the following rather trivial construction. Notation: for measurable $A \subset S^2$, let $F_A : \mathcal{P}(S)^2 \to [0,1]$ be the "evaluation map" $F_A(\alpha, \beta) = (\alpha …