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Sum of $X_k$ with $\mathbb{P}(X_k=\pm 1) = 1/2\pm 1/(2\sqrt{k})$

Let $\{X_k\}$ be a sequence of mutually independent random variables with \begin{align} \mathbb{P}(X_k = 1) & = \frac{1}{2} + \frac{1}{2\sqrt{k}}, \\ \mathbb{P}(X_k = -1) & = \frac{1}{2} - \...
Nuno's user avatar
  • 269
0 votes
0 answers
71 views

Conditions for existence of a semi-martingale representing a system of probability measures

Let $(\nu_t)_{t \in [0,1]}$ be Borel probability measures on a stochastic basis $(\Omega,\mathcal{F},(\mathcal{F}_{t \in [0,1]})_t,\mathbb{P})$. Does there exist a semi-martingale $(X_t)_{t\in[0,1]}$ ...
ABIM's user avatar
  • 5,405
2 votes
1 answer
250 views

Compactness of cadlag martingales w.r.t. to the point-wise topology

Given a sequence of cadlag (right-continuous with left limits) martingales $X^n=(X^n_t)_{0\le t\le 1}$, we may use the well known criteria to determine whether it is weakly convergent, i.e. subtract a ...
CodeGolf's user avatar
  • 1,835
5 votes
1 answer
652 views

Proof of Pinelis (1992) - Banach space inequalities

I am reading Pinelis "An approach to inequalities for the distributions of infinite -dimensional martingales" and cannot follow his proof of Theorem 3: Let $(f_n)$ be a martingale in a separable ...
Manuel Schmidt's user avatar