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No. It's easy to construct a sequence $Y_n$ with $Y_n \to 0$ a.s. but $E Y_n \to +\infty$. (You can even have $E Y_n \equiv +\infty$ if you wish.) Now let $X$ be a fixed Poisson random variable and $ …

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 …

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 …

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 $X \sim N(0,1)$, $Y \sim N(1,1)$ …

How about this: the set of finite mixtures of (non-degenerate) Gaussians is weakly dense in the space of probability measures on $\mathbb{R}$. Proof: the set of finite mixtures of constants is certai …

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: …

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, 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 …

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

Converting Yemon Choi's comment to an answer: If $E[X g(X)] \le 0$, we are done because the left side of your proposed inequality is nonnegative. So assume $E[X g(X)] \ge 0$. First, by the triangle …

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 …

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]] …

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 …

No, this is not true. Let $X \sim N(0,1)$ and define $Y_n$ to be independent of $X$, such that $Y_n = \sqrt{n}$ with probability $1/n$ and 0 otherwise. Set $X_n = X+ Y_n$. Since $Y_n \to 0$ in $L^1 …

I think the following example says no. Consider the state space $\{0,1\}$. Let $U$ be a uniform random variable on $[0,1]$ and let $X_t = 1$ if $t=U$ and $X_t = 0$ otherwise. Note that $X_t$ is a.s …