The martingales tag has no wiki summary.

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### Examples of a continuous martingale with $E[\sup\limits_{0\leq s\leq t} |M_s|]=\infty$?

A local martingale is a martingale iff it is in the class DL.
The condition: for every $t\in[0,\infty)$
$$E[\sup\limits_{0\leq s\leq t} |M_s|]<\infty\tag1$$
guarantees a local martingale $M$ is ...

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### explicit characterization of the stochastic integrand

Let $V$ be a cadlag positive supermartingale with the following decomposition:
$$V_t=V_0+\int_0^tH_sdX_s-K_t$$
where $X$ is a cadlag local martingale and $K$ is an adapted increasing process with ...

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### Question about the stochastic integral of martingales

Let $M=(M_t)_{t\ge 0}$ be a continuous martingale defined on some filtered probability space taking values in $R$. Let $H=(H_t)_{t\ge 0}$ be some bounded progressively measurable process, i.e.
...

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214 views

### Supremum in a Markov chain model

A Markov chain $X$ with finite state space $\{1,2,\cdots,N\}$ is defined on a probability space $(\Omega, P, \mathcal{F})$ equiped with filtration $\{\mathcal{F}_t\}$. And we assume that we can reach ...

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### Maximal inequalities for certain functions of a martingale difference sequence

Suppose $\xi_1,\ldots \xi_T$ is a martingale difference sequence. Then,
1) For any $a\in \mathbb{R}^{+}$, can we say something about the sequence $\xi_1^2\mathbb{1}(\xi_1\geq a),\ldots, ...

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148 views

### Savings property: A transformation which turns nonnegative martingales into uniformly integrable ones

Background
I work in a subfield of computability theory called algorithmic randomness. We have been using martingales as long as probability theory (going back to work of von Mises). However, since ...

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208 views

### Doob's inequality for martingale “convolution”

Let $(X_t, t \in \mathbb{N})$ be a martingale, and let $a \leq b \leq T \in \mathbb{N}$ be constants. Is there something like Doob's inequality for $\mathbb{E} \sup_{a \leq t \leq b} X_t(X_T-X_t)$, ...

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### A generalization of Jensen's Inequality

Jensen's inequality is well known as
$$E\big[f(X)\big]\le f\big(E[X]\big)$$
where $X$ is a integrable random variable and $f: R\to R$ is a bounded concave function, see also ...

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### A note on Doob's theorem

I have faced the following problem, regarding to the Martingale Theory. Because this area far from my area I don't know whether this problem is in literature or this can be simple question for ...

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### Reference request: Stochastic integration and martingale theory on the whole real line

I'm looking for a thorough treatment of stochastic integration and/or martingale theory on the whole real line, i.e. a way to construct a Brownian motion $(B_s)_{s \in \mathbb{R}}$ (if a two-sided BM ...

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124 views

### compactness of a probability set

I have a question about the compactness of a set of martingale measures. Let $\Omega=\mathcal{C}[0,1]$ be the space of continuous functions on $[0,1]$ and $\mathcal{M}_{\Omega}$ be the family of ...

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205 views

### Some constants in Martingale Stein inequality

Dear all,
the following is a special case of Stein inequalities for martingales.
$\textbf{Theorem}$ Let $(\Omega, \mathbb{P})$ be a (standard) probability space equipped with a filtration of ...

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111 views

### Hitting time of two dimensional continuous martingale

Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which $\mathcal{F}_t$ is filtration satisfying general conditions. $W_{t}=\left(W_{t}^{1},W_{t}^{2}\right)^{T}$ is a two dimensional Brownian ...

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86 views

### A result on absolute mean of a stopped supermartingale

The reason of posting the following problem here is that I heard that it is a result from some paper.
Let $(X_n, \mathscr{F_n}), n \geq 0$ be a super martingale and $T$ an $\{F_n\}$-stopping time ...

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### Pointwise convergence of ergodic averages of unconventional conditional expectations

Let $(X_i,Y_i)_{i\in\mathbb{Z}}$ be a finite-valued stationary process whose $\sigma$-algebra of tail events is trivial. Let $\mathcal{F}_n^m$ be the $\sigma$-algebra generated by $X_n,\dots,X_m$ ...

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### weaker version of the martingale convergence theorem

Let $\mathcal{A}_n$ be a sequence of finite sigma-algebras, let $\mathcal{B}_{q,p}= \sigma(\mathcal{A}_n, q \geq n \geq p )$. Moreover, we suppose $\mathcal{A}_k \subset \mathcal{B}_{\infty,p}$ for ...

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89 views

### a generalization of Monge-Kantorovich Problem

I am thinking about the martingale version of Monge-Kantorovich Problem.
Let $\mu(x)$ and $\nu(y)$ denote two density laws on $\mathbb{R}$, and define $M(\mu,\nu)$ the set of densities $f(x,y)$ on ...

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61 views

### question related to Tanaka Formulae

Supposse $X=(X_t)$ is a cadlag martingale taking values in $\mathbb{R}$. If $f:\mathbb{R}\to\mathbb{R}$ is a convex function, then we have Tanaka Formulae. Now let $g: ...

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### a question about the modification of a supermartingale

Let $\mathbf{D}\subset\mathbf{D}([0,1],\mathbb{R}_+)$ denote the space of positive cadlag functions $\mathbf{x}$ defined on $[0,1]$ with $\mathbf{x}(0)=1$. Define the canonical process
...

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89 views

### question about Doob-Meyer decomposition

Given a filtered probability space and let $X$ be a cadlag local martingale defined on this space. Let $V$ be a cadlag supermartingale and assume we know the following decomposition:
...

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152 views

### asymptotic variance of sample autocorrelation of two iid random variables

I am trying to prove that the variance of the sample lag-1 autocorrelation
$$\hat{\rho}=\frac{\sum_{t=1}^n(x_t-\bar{x})(x_{t-1}-\bar{x})}{\sum_{t=1}^n(x_{t-1}-\bar{x})^2}$$
for an i.i.d. R.V is ...

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### Supermartingale inequality on a particular event

Say, I have a supermartingale $Y_t$ with respect to the filtration $F_t$. Let $T$ and $S$ two stopping times greater than $t>0$ such that on the event $A$, $T>S$, then since $Y_t$ is a ...

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213 views

### What conditions on a filtration guarantee that a (sub)martingale has a continuous modification?

There is a theorem as follows:
Theorem. Let $\mathcal{F}_t$ be a filtration which is right-continuous and complete. Assume $M_t$ is a submartingale adapted to $\mathcal{F}_t$ such that $t \mapsto ...

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### When are the limits of Martingales are Martingales?

Suppose I have a sequence of continuous time random variables $X_n(t)$ where $t \in [0,1]$, adapted to a filtration $F_t$, that are martingales with respect to this filtration and that $\sup_n ...

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299 views

### Change of Time in Stochastic Integral

Hi everyone,
Let's be given $I(0,t)$ a Stochastic Integral with respect to a local martingale $ M_t$ of the form :
$I(0,t)=\int_0^t h(s_-)dM_s$ with $h\in L(M)$ (for example $h$ is an adapted ...

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### Tail inequality for orthomartingales/martingale difference random fields

It is known that if $(S_i= \sum_{j \leqslant i }X_i, \mathcal F_i)$ is a martingale,
then for each
$
\beta>1$, $\delta\in (0,\beta-1)$ and $\lambda>0$, and each integer $N \geqslant 1$, the ...

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### Question about Skorokhod embedding problem

Let $B=(B_t)_{t\ge 0}$ be a standard Brownian motion on some probability space. Now for every centered probability distribution $\mu$ on $R$, i.e. $\int_{R}|x|d\mu(x)<+\infty$ and ...

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### Zeros of non-lipschitz functions (when noisy estimates are available only)

Given noisy (martingale difference) of a Lipschitz continuous function $f$ it is known how to compute zeros of it. It is the stochastic approximation approach (by Borkar, Kushner and Yin etc.). Is ...

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### a question about Dambis, Dubins-Schwarz Theorem

Let $M=(M_t)_{0\le t\le 1}$ be a continous $\mathbb{F}=\{\mathcal{F}_t\}_{0\le t\le 1}$-martingale s.t. $M_0=0$. Now my question is whether there exists a Brownin motion $B$ s.t.
...

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### Sufficient condition for local martingale property of stochastic integral

Is the following correct and/or a (simple) known result?
Let $X$ be a local martingale and $H$ an integrand for $X$, such that the stochastic integral $\int H\cdot dX\ge x$ for some random variable. ...

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### Is stopped brownian motion not a martingale ?

In page 45 of the book "Financial Derivatives In Theory and Practice by P.J.Hunt and J.E.Kennedy, it seems to me that the author says the stopped Brownian Motion is not a martingale as follows.
...