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2 votes

Showing a basic market admits no arbitrage

The idea is actually really simple... presumably the difficulty is in cutting through the abstract formalism. We have an asset whose price is $S_0^1$ and tomorrow it can either go to $\beta S_0^1$ or ...
spaceisdarkgreen's user avatar
0 votes

Stochastic Integral with respect to Compensated Poisson Process

Proof First, the compensated process of Poisson prcocess is a strict martingale. To see this, note that $$\begin{aligned}E(N_t|\mathcal{F}_s)&=E[N_t-N_s+N_s|\mathcal{F}_s]\\&=E[N_t-N_s]+N_s\\&...
Mingzhou Liu's user avatar
1 vote
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If $X$ is a continuous supermartingale, why is, for every $N$, $(X_s)_{s\leq N}$ uniformly integrable?

I have doubts about this assertion. (Where did you read it?) Example. Let $(B_t)$ be a standard 1-dimensional Brownian motion (with $B_0=0$). Let $Z_t:=\exp(B_t-t/2)$ be the corresponding geometric ...
John Dawkins's user avatar
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3 votes
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Derive the 0-1 law from a functional equation of a martingale limit

For $z>0$ and $0<x<1$ you have $$ \lim_{t\to+\infty}\Bbb E(e^{-tM(z)})=\lim_{t\to+\infty}\Bbb E(e^{-tzx^{2z-1}M(z)})=\lim_{t\to+\infty}\Bbb E(e^{-tz(1-x)^{2z-1}M(z)})=\Bbb P(M(z)=0). $$ ...
John Dawkins's user avatar
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1 vote

Book recommendation for stochastic integral wrt local martingales

I would suggest, that you consider "Brownian Motion, Martingales, and Stochastic Calculus " by Le Gall. He does not use the Reisz representation theorem to define the stochastic integral, ...
user123234's user avatar
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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

Book recommendation for stochastic integral wrt local martingales

I'm a big fan of "Continuous Martingales and Brownian Motion" by Revuz and Yor. They define the stochastic integral using the Reisz representation theorem, then show it agrees with the ...
user6247850's user avatar
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4 votes

Uniform Integrability and Proving martingale of Poisson product process

$E(M_{n})=n!E(\prod_{k=1}^{n}X_{k})=n!\prod_{k=1}^{n}E(X_{k})=n!\prod_{k=1}^{n}\frac{1}{k}=\frac{n!}{n!}=1$. The expectation of a martingale cannot vary with $n$, it has to be constant. $X_{k}$'s are ...
Mr.Gandalf Sauron's user avatar
0 votes

Exercise 4.3.11 of Durrett's Probability: Theory and Examples

Suggestion: Let $\pi:=P(\lim_nZ_n/\mu^n=0)$. Show that $\pi=\varphi(\pi)$ and then appeal to Theorem 4.3.12.
John Dawkins's user avatar
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1 vote
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maximum between martingale and constant

Consider standard Brownian motion $(B_t)$. This is a martingale. If $B_t\vee 1$ is a martingale, then $E(B_t \vee 1)=E(B_0\vee 1)=1$. But then $ E[(B_t \vee 1)-1]=0$ which implies $B_t \le 1$ a.s.. ...
geetha290krm's user avatar
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0 votes

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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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

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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0 votes

Optional Stopping Theorem for martingales bounded except at the stopping time

My friend pointed out a simple counterexample: $X_0 = 0$, $$X_{n+1} = \begin{cases} X_n & \text{with probability $1/2$} \\ X_n - 2^n & \text{with probability $1/4$} \\ X_n + 2^n & \text{...
lily's user avatar
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1 vote
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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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0 votes

$\mathcal{F}_n$ measurable

In general no. $$ \begin{align*} &\{T\in\{(m,k)\mid m>n\}\}\\ =&\left(\bigcap_{m\le n}\bigcap_{0\le\ell\le K}\{S_{m,\ell}\le\lambda\}\right)\cap\bigcup_{m\ge n+1}\left(\left(\bigcap_{0\le\...
Will's user avatar
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1 vote

Definition of Left-Closable Martingale

To build on Jacob Maibach's answer on the positive integers being open on the right, later on in Resnick's book he discusses reversed/backwards martingales: given a DECREASING family of $\sigma$-...
picklechu's user avatar
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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2 votes
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Definition of Left-Closable Martingale

I haven't run into this terminology before, but here is my guess. The term closed probably refers to the set of indices $n$ where the martingale is defined. That is, consider the domain $$ \{ n : X_{n}...
Jacob Maibach's user avatar
0 votes
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Is $(e^{i \lambda B_t + \frac{1}{2}\lambda^2t})_{t\geq 0}$ a martingale?

$\mathbb{E}\left[e^{c\left(B_t - B_s\right)}\right] = e^{\frac{1}{2}c^2\left(t-s\right)}$ does hold for $c\in\mathbb{C}$. The proofs are correct.
Wilfred Montoya's user avatar
1 vote
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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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1 vote
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Expected value of the square of a stopping time

$\def\={\mathrel{\phantom=}}$Your calculation writes\begin{gather*} E( (a^2 - S_a) I_{\{ B_{S_a} = a \}} ) + E( ((-a)^2 - S_a) I_{\{ B_{S_a} = -a \}} )\\ = a^2 - E( S_a I_{\{ B_{S_a} = a \}} ) + (-a)^...
Ѕᴀᴀᴅ's user avatar
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1 vote

Expectation of the indicator function

$\{k\}$ is not independent of $\mathcal F_n$. Since $\mathcal F_n$ is generated by a partition, $$ \mathbb E\left[X_{n+1}\mid\mathcal F_n\right]=\sum_{k=1}^n \mathbb E\left[X_{n+1}\mathbf{1}_{\{k\}}\...
Davide Giraudo's user avatar
1 vote
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Expected value of the exponential of a stopping time

All good. Perhaps, you can add more details on the "by symmetry" Independence of $T$ and $B_T$ i.e. we have $B_t\stackrel{d}{=}-B_{t}$ and $S_{a}$ is only a function of $|B_{t}|$, which is ...
Thomas Kojar's user avatar
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0 votes
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Stirling approximation of the probability that the stopping time is finite

Consider $x=0$. Using Stirling's approximation you find $$P(S_{2m}=0)=\binom{2m}{m}2^{-2m}\sim \frac{1}{\sqrt{\pi m}},$$ which yields $$\sum_{m=0}^\infty P(S_{m}=0)= \infty.$$ Define $$\tau_{0}^{(m)}:=...
user408858's user avatar
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