Search Results

For the question at hand, we don't need the full strength of the RPZ theorem cited by Sam Hopkins; just the Kolmogorov three-series theorem. Since $\sum a_n$ converges (conditionally), we have $a_n \ …

No. For a simple counterexample, let's work in discrete time. Consider the following gambling strategy: start with \$0 and bet \$1 on a fair coin flip. If you win, you take your dollar and go home. …

I think Iosif's Fatou lemma argument can be fixed, as follows. Assume without loss of generality that $X_0 = 0$. Suppose to the contrary that $X_t \to +\infty$ a.s. Then it must be that $\inf_{t \ge …

Yes. First, one can show the $\sigma$-algebra in question is equal to the Borel $\sigma$-algebra induced by the weak topology. (We could do without this fact, but I find it makes the proof simpler.) …

I think you can adapt the proof in these notes of mine, Theorem 4.44. The first step in the proof of my notes is to pick, for each $n$, a finite-rank projection $P_n$ such that for all finite-rank pro …

Yes, under mild assumptions. If the state space $E$ is Polish (including $E = \mathbb{R}^n$ in particular), then the space $\mathcal{P}(E)$ of Borel probability measures on $E$, with the weak topology …

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 $ …

It looks fine to me. If we let $I_k = [k t^\ast_M, (k+1) t^\ast_M]$ be the relevant subintervals of $[0,T]$, then the supremum of $|u_n|$ over $[0,T]$ must be almost attained along some sequence of po …

No, not even for sequences of countably additive measures. Take $X = \mathbb{N} = \{0,1,2,\dots\}$ with its discrete $\sigma$-algebra, and let $\mu_n$ put mass $1/n$ at the point $n$ and mass $1-1/n$ …

Note that your argument contains an implicit assumption that $\kappa_t \mu \in \mathcal{S}^1$ for every $\mu \in \mathcal{S}^1$ (otherwise the Banach fixed point theorem does not apply). I will also …

No. Take $E=[0,1]$ with Lebesgue measure. Let $f_n = n 1_{[0, 1/n]}$, so that $\int_E |f_n|\,d\mu = 1$ for every $n$, and $B_p = [0, 1/p]$. Note that $|f_n| \le p$ on $E \setminus B_p = (1/p, 1]$ fo …

Yes. By definition of the inf, we can find a sequence of integrable functions $\phi_n$ such that $\psi \le \phi_n$ for every $n$, and $\int \phi_n \to \int^* \psi$. Set $\phi = \liminf \phi_n$, which …

Not in general. The standard counterexample is to let $U \sim U(0,1)$, $X_n = n 1_{\{U \le 1/n\}}$, and $X=0$. There are several basic theorems giving sufficient conditions for this to hold, e.g. mo …

Let $\Omega = \mathbb{N}$ and let $\mathcal{F}$ be the field consisting of all finite and cofinite subsets of $\mathbb{N}$. Let $\mu_n = \delta_{2n}$ be a point mass at the integer $2n$, and let $\mu …