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Joint measurability of strongly continuous vector-valued function

The main result in "The measurability of a stochastic process of second order and its linear space," Proc. Amer. Math. Soc. 47 (1975) 467–475, by S. Cambanis, implies a given selection $(\...
John Dawkins's user avatar
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1 vote

Exercise on Girsanov's theorem

There is no difference with the case your drift is non-deterministic. Your calculation shows that $X$ is a $Q$-Brownian motion; since $Q$ and $P$ are equivalent and $Q(\{X_t > M\}) > 0$ it ...
Jose Avilez's user avatar
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1 vote

Proving The Fundamental Theorem of Markov Chains

One reference which has this proof is Introduction to Probability for Computing by Harchol-Balter, specifically Theorem 25.12 [PDF link].
Ziv's user avatar
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1 vote
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Holding points are absorbing

This is a consequence of strong Markov property. Assume $0\leq\lambda(x)<\infty$. Define $T_{0}=\inf\{t>0,X_{t}\neq X_{0}\}$ is the first exit time, and $T_{1}=T_{0}\circ \theta_{T_{0}}+T_{0}$ ...
Boey C's user avatar
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2 votes

Walds identity and difference of stopping times

Consider the following (it technically doesn't match your requirements because $N_2 \leq N_1$, but I hope that doesn't matter much): $X_1, X_2, \ldots$ is a an i.i.d. sequence of Bernoulli random ...
Brian Moehring's user avatar
2 votes
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Question about the integrand space of stochastic integral wrt martinagles

Yes, $\int_a^b |f(t)|^2d\langle M \rangle_t$ is a Lebesgue-Stieltjes integral for each $\omega$. Since $\langle M \rangle_t$ is a non-decreasing process, it has finite variation, and so $\int_a^b |f(...
user6247850's user avatar
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0 votes
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A Gaussian process and a Rademacher proecss are sub-Gaussian

Gaussian Process. Let $Z=(Z_1,\cdots ,Z_d), t=(t_1,\cdots ,t_d), s=(s_1,\cdots, s_d)$. By definition, $Z_i \sim N(0,1)$ are independent Gaussian variables (Jointly normal random variables that are ...
Kaira's user avatar
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1 vote

What to call a sequence of Bernoulli trials with different probabilities?

It is Poisson Binomial Distribution : In Probability theory and Statistics, the Poisson Binomial Distribution is the Discrete Probability Distribution of a sum of independent Bernoulli trials that are ...
Prem's user avatar
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Sub-Gaussian $X_t$, prove $\mathbb{E}\left[\sup_{t\in T}X_t \right] \leq 2 \mathbb{E}\left[\sup_{\rho(t,s)\leq \delta}(X_t-X_s) \right]+J(\delta,T)$

For this problem, bounding the expectation of the supremum is crucial. The following inequality can help in this case. Proposition. Let $\{Z_i\}_{i=1}^{N}$ be $\sigma^2$-sub-Gaussian random variables. ...
Kaira's user avatar
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0 votes

Help with Integration by Parts for a Markov Chain

OP here, I think I was able to figure out the problem. Part 1: Let's start with the Q Matrix: $$ Q = \begin{bmatrix} B & A=-BI \\ 0 & 0 \end{bmatrix} $$ Within this matrix, we are interested ...
stats_noob's user avatar
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1 vote

What to call a sequence of Bernoulli trials with different probabilities?

If $p$ is fixed and all the experiments are Bernoulli trials then we can call it a Binomial Distribution. If $p$ is not fixed, then it is more useful to consider the Bernoulli trials as separate ...
Red Five's user avatar
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1 vote

Transition probability density function for a non-trivial diffusion process.

Here we will only focus on finding the eigenfunctions of the infinitesimal generator. In other words we want to solve the following ODE below: \begin{equation} \left(\mu z^{\beta_1} \frac{d}{d z} + \...
Przemo's user avatar
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1 vote

Solve SDE $dX_t = X_tW_tdt + dW_t,$

To solve problems like these, rather than trying to guess the correct process to introduce, it may be easier to just let $Y_t := e^{Z_t}X_t$ where $dZ_t = \alpha_t dt + \beta_t dW_t$ for some ...
user6247850's user avatar
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0 votes

1D Brownian motion: stopping time and stopping state not independent

Given $X_{T}=-3$, it means that state 1 has not been hit by time $T$. Hence, by the reflection principle, $$P(T\le t|X_{T}=-3, X_0=0)=1-P(hitting\; 1\; by\; time\; t)\\ =1-2\int_1^\infty \frac{1}{\...
toronto hrb's user avatar
0 votes

Exercise on irreducible positive recurrent markov chain

1. Let $N_n(A)$ be the number of visits to $A$ by time $n$; that is, $N_n=\sum_{j=1}^n 1_A(X_j)$. What can you say about $N_n(A)/n$ as $n\to\infty$? 2. Observe that, for each $n$ and $k$, $T_k^A\le n$...
John Dawkins's user avatar
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d-dimensional random walk: scaling of probability of self-crossing

Well, on every step of a random walk, it can go backwards and cross itself immediately, so the probability of not crossing itself decays at least as $\left(1-\frac1{2d}\right)^N$.
van der Wolf's user avatar
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The almost sure event in the law of the iterated logarithm for the Brownian motion: what it looks like

Because $\sup_{0<s\le t}W_s(\omega)/h(s)$ is monotone in $t$, you have $$ \{\omega: \limsup_{t\to 0}W_t(\omega)/h(t)\le 1 \} = \cap_{\epsilon\in\Bbb Q_+}\cup_{\delta\in\Bbb Q_+}\cap_{s\in\Bbb Q\cap(...
John Dawkins's user avatar
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4 votes
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Computing the quadratic covariation $\langle B,B^2\rangle_t$ of a Brownian motion and its square

See first this discussion on how to obtain the quadratic variation of two Itô processes. By Itô's formula we obtain that $B_{t}^{2}$ follows the dynamics $$ B_{t}^{2} = \int_{0}^{t}ds + \int_{0}^{t}...
minginator's user avatar
1 vote
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Ergodic series converge to the expectation?

Great counter-example by Michael. Essentially this boils down to the fact that if $(X_n)_{n\in\mathbb N}$ and $(Y_n)_{n\in\mathbb N}$ are two sequences of random variables satisyfing the Strong Law of ...
Maximilian Janisch's user avatar
0 votes

Markov process characterization and application of monotone class theorem

It's my first time answering questions in stochastic processes, so please correct me if I'm wrong. Following @TheBridge's comment, we can use the Functional Monotone Class Lemma (Theorem 2 in ...
isomorphicdude's user avatar
2 votes

Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

Define $A_t := \int_0^t e^{\lambda s} dB_s$. You are asking if $X_t := e^{-\lambda t} A_t$ is a martingale. By Ito's formula, we have \begin{align*} dX_t &= e^{-\lambda t} dA_t -\lambda e^{-\...
user6247850's user avatar
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5 votes
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Is the stochastic integral process $\left(\int_{0}^{t}e^{-\lambda\left(t-s\right)}\mathrm{d}B_s\right)_{t\geq 0}$ a martingale?

The criterion you mentioned is certainly a sufficient condition, but not a necessary condition. So, let's go back to the basic. Let $M_t = \int_{0}^{t} e^{-\lambda (t-s)} \, \mathrm{d}B_s$. Then for $...
Sangchul Lee's user avatar
2 votes

Ergodic series converge to the expectation?

Consider the counter-example: $U \sim Unif[0,1]$. $\{Y_n\}_{n=1}^{\infty}$ i.i.d. $Bern(1/2)$ (and independent of $U$) For all positive integers $n$ define $$X_n=\left\{\begin{array}{cc} U & \...
Michael's user avatar
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0 votes

Creating a martingale given

You are correct, the filtration by construction makes the process adapted. You also need to check the integrability condition of the definition. To check the martingale property, you need to use the ...
Oscar's user avatar
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1 vote

Properties of a transient state in a Markov Chain

As you have pointed out, $\sum\limits_{n=1}^{\infty}P_{jj}^{n}<\infty$, and that $\mathbb{P}_{ij}(s)=\mathbb{F}_{ij}(s)\mathbb{P}_{jj}(s)$, what you need follows directly from Abel's theorem, since ...
Sparkle-Lin's user avatar
1 vote
Accepted

In this case does convergence of marginal distribution imply joint convergence in distribution

Clearly no. Consider $(S_n)_{n\geq 1} = (Z_n)_{n \geq 1}$ where $Z_n \sim \mathcal{N}(0, 1)$, a standard normal. Let $Z$ also denote a generic standard normal random variable. Consider $f(x)=x$ and $g(...
pbb's user avatar
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0 votes

Upgrade dominance in distribution to almost sure in a new probability space

It sounds like you are trying to keep the same $(\Omega, \mathcal{F})$, so the functions $X_i:\Omega\rightarrow\mathbb{R}$ and $Z_i:\Omega\rightarrow\mathbb{R}$ are always the same, but just change $P:...
Michael's user avatar
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2 votes

The almost sure event in the law of the iterated logarithm for the Brownian motion: what it looks like

The event you have captured in more explicit terms is not $\{\limsup_{t\to 0}W_t/h(t)=1\}$, but rather $\{\limsup_{t\to 0}|W_t/h(t)-1|=0\}$. The placement of the absolute value is crucial!
John Dawkins's user avatar
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4 votes

Proof of a martingale condition regarding martingale transform

$$ \begin{align} C_n(\mathbb E[X_n\vert\mathcal F_{n-1}]-X_{n-1})&=\mathbb E[C_n(X_n-X_{n-1})\vert\mathcal F_{n-1}]\\ &=\mathbb E[(C\bullet X)_n-(C\bullet X)_{n-1}\vert\mathcal F_{n-1}]\\ &...
Will's user avatar
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1 vote

Natural $\sigma$-algebra on $C[0,\infty[$: Prove equivalence of definitions

Note that under the given topology, the projections $\pi_{t}$ are continous and thus measurable in respect to the Borel measure. As for the other direction, this topology is metrizable and separable, ...
FZan's user avatar
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1 vote

Example of a process that yields a non-martingale

Easy example: $C_k = 1_{X_k > X_{k-1}}$. This is adapted because $X$ is adapted, but $\sum_{k=1}^n C_k (X_k-X_{k-1}) = \sum_{k=1}^n (X_k-X_{k-1})^+$ is non-decreasing and hence cannot be a ...
user6247850's user avatar
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1 vote
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Converse of a martingale transform theorem

Under the stated conditions, it is not true as you can easily construct a counterexample since you basically left the process $X$ to be unrestricted. For example, assume for simplicity that $C_{n}(\...
minginator's user avatar
2 votes
Accepted

Understanding stochastic integration with semimartingale as integrator: The locally bounded variation part.

Indeed, for each $\omega$ you get a different measure $\mu_g$. So the integral can be defined path-by-path: i.e. for a fixed $\omega$ $$\left( \int_a^b f(s) dA_s \right)(\omega) = \int_a^b f(\omega, s)...
Jose Avilez's user avatar
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0 votes

Convergence Rate of "Infinite Monkey"-type-probability

I found a solution to the problem that is sufficient for my use case. This is not the exact convergence rate. If somebody still manages to find it in time I'll still give the bounty of course. However,...
Joseph Expo's user avatar
1 vote

Using Ito's Lemma to take a stochastic integral

Please avoid to ask too many questions at the same time in future posts. Let's tackle the side questions first. First side question. I mean, why not ? If you have a stochastic variable $X_t$ itself ...
Abezhiko's user avatar
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2 votes
Accepted

How can I show that the first exit time by a planar Brownian motion is a.s. finite, i.e. $\mathbb{P}_z(\tau_D<\infty)=1$?

My first question is, where do i need that $W_t$ is a complex Brownian motion, I mean why can't I only work with $B_t$ instead of $W$? You could do it with just using neighborhood-recurrence for 2d-...
Thomas Kojar's user avatar
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2 votes
Accepted

Is the quadratic variation process also a martingale?

The quadratic variation process is non-decreasing, so it is not a martingale (unless it is constant at $0$). The flaw in your argument is that $M_t - \langle M\rangle_t$ is not a martingale, rather, $...
user6247850's user avatar
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1 vote

Verifying Ito isometry for simple stochastic processes

A simple process $\Phi = (\Phi_t)_{t\geq 0}\in\mathcal{E}$ is a process of the form $$\Phi_t = \sum_{i=0}^{n-1}U_i1_{(t_i,t_{i+1}]}(t)$$ where $0=t_0<\dots<t_n<\infty$ and $U_i\in b\mathcal{F}...
Wilfred Montoya's user avatar
0 votes

The distribution of the first hitting time for the Constant Elasticity of Variance process.

We are going to prove that the Laplace transform $(3)$ approaches the correct limit when $\beta \rightarrow 1_+$. We denote ${\mathfrak N}:= 1/(2(-1+\beta)$ and $\theta := 2 \mu/\sigma^2$ and we have: ...
Przemo's user avatar
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0 votes

Markov Property of a Ito Process

Rather than bringing in $B_{t+h}-B_h$, consider using $B_{t+h}-B_t$: Because $X^x_{t+h} = X^x_t\cdot\exp(ch+\alpha(B_{t+h}-B_t))$, and $B_{t+h}-B_t$ is independent of $\mathcal F_t$ with the same ...
John Dawkins's user avatar
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Talagrand's functional and Dudley's sum (Vershynin 8.5.2)

I'm writing with respect to the recent version of Vershynin, i.e. it is $1+\log k$ instead of $1+\log n$. Choosing $T_k$ to be the first $2^{2^k}$ vectors will not work as suggested by the OP (or as ...
MikeG's user avatar
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1 vote
Accepted

How to prove $\sup_{t\geq 0} \mathbb{E}(M_t^2)<\infty$ implies $\mathbb{E}(\sup_{t\geq 0}M_t^2)<\infty$?

Let $f_{N}:=\sup_{N\geq t\geq 0}M_{t}^{2}$, then by (liminf)-Fatou's lemma we get $$E[\liminf_{N}f_{N}]\leq \liminf_{N}E[f_{N}]\leq 4\liminf_{N}E[M_{N}^{2}].$$ Since the sequence $f_{N}$ is increasing,...
Thomas Kojar's user avatar
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deriving covariance of SDE from fokker-planck

It holds $$ \begin{aligned} \frac{\partial \phi}{\partial t}&=-\frac{\partial m_u}{\partial t}(t)m_v(t)-m_u(t)\frac{\partial m_v}{\partial t}(t)\\ \frac{\partial \phi}{\partial x_u}&=x_v\\ \...
user408858's user avatar
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1 vote
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Optional Stopping Theorem and Stopped $\sigma$-fields

For real $x$ you have $$ \{X_T\le x\}\cap\{S\le n\} = \cup_{k=0}^n\{X_k\le x, T=k, S\le n\}. $$ which is clearly $\mathcal F_n$-measurable, for each $n$. It follows that $X_T$ is $\mathcal F_S$-...
John Dawkins's user avatar
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1 vote
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Is $\inf\left\{t\in\left[0,1\right]\vert t+B^2_t=1\right\}$ a stopping time?

(Edited) Useful theorems Assuming nothing about the filtration: If $X=(X_t)_{t\geq 0}$ is a continuous $\mathbb{R}^d$-valued $(\mathcal{F}_t)_{t\geq 0}$-adapted process, then $T_A:=\inf\left\{t\geq ...
Wilfred Montoya's user avatar
1 vote
Accepted

Calculate $\mathbb{E}(\exp (- \lambda T_x ))$ for $\lambda > 0$ where $X$ is a Brownian Motion with drift

We have $$E[e^{\theta B_t+\theta ct-\lambda t}]=e^{\theta^{2}t/2+\theta c t-\lambda t}$$ and so we indeed need $\theta^{2}/2+\theta c-\lambda=0\Rightarrow \theta=-c\pm \sqrt{c^{2}+2\lambda}$ to get ...
Thomas Kojar's user avatar
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0 votes

Rate of increase of maximum process of Brownian Motion

This indeed just follows from reflection principle and Borel-Cantelli $$\sum_n P[M_{t_{n}}\frac{1}{g(t_{n})}\geq \epsilon]=\sum_n P[|B_{t_{n}}|\geq g(t_{n})\epsilon]\leq c \sum_{n}exp(-\frac{(g(t_{n}))...
Thomas Kojar's user avatar
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Equivalent definition of Markov property

The converse cannot hold, take a process with trajectories that are equal to a random constant $Z$, so $X_s \equiv Z, s \in T$. Then knowing $X_t$ for some $t \in T$, you know $X_s$ for all times $s \...
Moritz Schauer's user avatar
1 vote
Accepted

Decaying inequality in expectation implies almost sure convergence to zero?

Assuming finiteness as mentioned in the comments $E[X_{1}]$. By the decreasing we get $\lim_{n}E[X_{n}]=0$. By Fatou's lemma for $X:=\liminf_{n}X_{n}\geq 0$ we get $$E[X]\leq \liminf_{n}E[X_{n}]=0.$$ ...
Thomas Kojar's user avatar
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1 vote

How to deduce an expression of a specific conditional expression

I am not entirely sure if this proof is complete and correct, but I would like to share what I have attempted. Let $z\in\mathbb{R}$, then it holds $$ \begin{aligned} \int_{\mathbb{R}}-\mathbb{1}_{\{z\...
user408858's user avatar
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