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NOT AND ANSWER, ONLY TOO WEAK INEQUALITIES AT THIS TIME You should replace all $n_i$ by $X_i^{(n)}$ in your question, and your proof. I did not read the proof, but the answer is trivially yes. Note th …

I give a crude lower bound, which does not use the distribution of $Z$, but only that $|Z|=\sqrt{d}$. It relies on the triangle inequality for the angular distance on the unit sphere. I assume $c_1$ a …

By the law of Large numbers, $X_t/t \to b$ almost surely as $t \to +\infty$, hence $X_t \to +\infty$ almost surely as $t \to +\infty$. Therefore $X_\infty = +\infty$ almost surely under $P$ and also u …

Partial answer The question may be whether the process $f$ is progressively measurable under the assumptions. If yes, we can conclude as follows. Fix $t \ge 0$. For each $\omega$, $f(\cdot,\omega)$ is …

Let $\epsilon>0$ and $n \ge 1$. Then $$\bigcap_{k=1}^\infty\{|S_{n+k}-S_n|<\epsilon = \bigcap_{j \geq n}\{|S_j-S_n|<\epsilon\} \subset \bigcap_{j,k\geq n}\{|S_j-S_k|<2\epsilon\} .$$ Hence, taking comp …

A sufficient condition is unimodality of $f$, namely the existence of some $c \in~]a,b[$, such that $f$ non-decreasing on $[a,c]$ and $f$ non-increasing on $[a,c]$. If this property holds, then by Fat …

I think that I have a counterexample. Let $(X,Y)$ be a Brownian motion in $\mathbb{R}^2$. Then $M = \int_0^\cdot X_s \mathrm{d}Y_s$ is a martingale, in the natural filtration of $(X,Y)$, in its own fi …

Almost surely, we can write for every $t \ge 0$, $M_t=M_0+\beta_{\langle M,M \rangle_t}$, where $\beta$ is some Brownian motion. By Kahane's theorem, almost surely, for every $s \ge 0$, $\limsup_{\del …

As I mentionned in my comment, there is an ambiguity in the statement of you question. Anyway, if $B$ is a standard one dimensional Brownian motion, if $\lambda$ is a real number, then $(B_t-\lambda t …

I assume that $(a_n)_{n \ge 1}$ are random variables taking values on a finite subset $B$, and that $\nu_l(b) \le P[a_n = b|a_1,\ldots,a_{n-1}] \le \nu_u(B)$ almost surely for every $n \ge 1$ and $b \ …

Is $Y$ independent of $(X_1,\ldots,X_n)$? If yes, write it. If yes, it suffices to prove that for any fixed unit vector $u$, $\langle u,Y \rangle \le c/\sqrt{d}$ with high probability. By rotational i …

Partial answer. Continuous (local) martingales are time-changed Brownian motion. A way to obtain funny local martingales which also are Markov processes is to start from a Brownian motion $B$ and its …

The answer to the first question is positive, and the lower bound is achieved, since the set of all probability measures on $[0,1]^k$ with uniform marginals is compact and since the integral of the bo …

Partial answer For every $a \in [0,1]$, $(\mu_n([0,a]))_{n \ge 0}$ is still a non-negative martingale, hence it converges almost surely to some random variable $L_a$ with values in $[0,+\infty]$. One …

I hope I did not make a mistake, but I think it works. The convergence $$\exp\Big(\int_\mathbb{R} (e^{izx}-1)d\eta_n(x) \Big) \to \exp\Big(\int_\mathbb{R} (e^{izx}-1)d\nu(x)\Big)$$ yields the converge …