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50 votes

Example of filtration in probability theory

Another simple example. The natural filtration when we model tossing a die twice in a row. Here is how it works. Let $X_1$ be the outcome of the first toss. So the values of $X_1$ are in the set $\...
GEdgar's user avatar
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35 votes

Martingale theory: Collection of examples and counterexamples

convergence results: pointwise convergence of martingale $M_n$ does not imply $\sup_n \mathbb{E}(M_n^+)<\infty$ (this means that the converse of the martingal convergence theorem does not hold ...
saz's user avatar
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29 votes
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Example of filtration in probability theory

Let me first state an interpretation for the meaning of a filtration: A filtration $\mathcal F_t$ contains any information that could be possibly asked and answered for the considered random process ...
davidhigh's user avatar
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20 votes

Example of filtration in probability theory

Take the following simple model: a stochastic process $X$ that starts at some value $0$. From that value, it can jump at time $1$ to either the value $a$, either the different value $b$. And at time $...
Raskolnikov's user avatar
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19 votes
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Martingale / local martingale : some confusion

Before trying to understand the difference between martingales and local martingales on a technical level, it pays to have an intuitive understanding of the difference: that is what I will attempt to ...
pre-kidney's user avatar
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17 votes
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Expected stopping time

The linear map $T(x):=(x_1,x_1+x_2,\ldots,x_1+\ldots+x_n)$ taking a vector in $[0,1]^n$ to its partial sums is volume preserving, as it corresponds to a triangular matrix of determinant 1 (or by ...
Yuval Peres's user avatar
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14 votes
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How to show the following process is a local martingale but not a martingale?

We have to show that $(X_t^{S_n})_{t \geq 0}$ is a martingale. Since $$X_t^{S_n} = W_{t/(1-t)}^{T \wedge n} 1_{\{t<1\}} + W_{T \wedge n} 1_{\{t \geq 1\}}$$ this follows if we can prove the ...
saz's user avatar
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13 votes
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"Converse" of optional stopping theorem

Neat question! It is true, if I'm not mistaken. Let $n$ be arbitrary and set $A = \{E[X_{n+1} \mid F_n] > X_n\} \in F_n$. Set $\tau = (n+1) 1_A + n 1_{A^c}$; you may verify that $\tau$ is a ...
Nate Eldredge's user avatar
12 votes

Understanding the $\sigma$-algebra of a sum of random variables

$\sigma\{X+Y\}$ is contained in $\sigma\{X,Y\}$. This because measurability of $X$ and $Y$ implies measuribility of $g(X,Y)$ for any Borel-measurable $g:\mathbb R^2\to\mathbb R$. Involved here is the ...
drhab's user avatar
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12 votes
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Some version of Itô isometry with conditional expectations

Let $G \in \mathcal{F}_r$. Using that $$1_G \int_r^t u_s \, dB_s = \int_r^t 1_G u_s \, dB_s \quad \text{a.s.} $$ (see the lemma below) it follows from Itô's isometry that $$\begin{align*} \int_G \...
saz's user avatar
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12 votes
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Martingale oscillating between three values

Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of independent random variables such that $$\mathbb{P}(Y_n = 1) = \mathbb{P}(Y_n=-1) = \frac{1}{2n} \qquad \mathbb{P}(Y_n=0) = 1- \frac{1}{n}.$$ If we ...
saz's user avatar
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12 votes
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Stochastic processes - Why do we need filtration?

You know the probability of every event in every $\mathcal{F}_t$, but the idea is that at time $t$ you know specifically which event you are in. For an easy example, you can think of flipping two ...
user6247850's user avatar
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11 votes

Is every Markov Process a Martingale Process?

For a simple counterexample, let $X_t=t$ and $\mathcal F_t$ be the natural filtration. Then for $s<t$ and nonnegative measurable $f$, $$\mathbb E[f(X_t)\mid\mathcal F_s] = f(X_s+t-s)=:g(X_s) $$ so ...
Math1000's user avatar
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11 votes
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Why is $\int_0^t t \, dW_s$ not a martingale?

The process $(t W_t)_{t \geq 0}$ is indeed not a martingale. The reason is simply that you cannot use the martingale property of stochastic integrals to deduce the martingale property of the process. ...
saz's user avatar
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11 votes

Prove that a martingale is uniformly integrable.

The martingale $M_t$ is not uniformly integrable. If it were, then by standard martingale facts, it would converge a.s. and in $L^1$ to some $M_\infty$. (An $L^1$-bounded martingale always converges ...
Nate Eldredge's user avatar
11 votes
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Can a martingale converge in $L^1$ but not almost surely?

Yes, for a martingale convergence in $L^1$ implies almost sure convergence. Suppose $(M_n) \rightarrow M_\infty$ in $L^1$ where $(M_n)$ is a martingale. Then $\sup_n \mathbb{E}[|M_n|] < \infty$ ...
user6247850's user avatar
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10 votes
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Understanding the $\sigma$-algebra of a sum of random variables

In general, there's no connection between $\sigma(X+Y)$ and $\sigma(X)$ and $\sigma(Y)$. If $Y=-X$, for instance, then $X+Y$ is constant so it has a trivial sigma-field, whereas $\sigma(X)$ typically ...
grand_chat's user avatar
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10 votes

"Converse" of optional stopping theorem

Nate Eldredge's answer provides some nice intuition on why the result holds true. Here's a different approach. The task at hand is to prove that $\forall n, E(X_{n+1}|F_n)=X_n$. Let $n$ be fixed. ...
Gabriel Romon's user avatar
10 votes
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Show that the stochastic exponential is a true martingale

Itô's formula shows that $$M_t = 1+ \int_0^t f(s) M_s \, dW_s \tag{1}$$ and this implies, in particular, that $(M_t)_{t \geq 0}$ is a local martingale. (Note that $(M_t)_{t \geq 0}$ has continuous ...
saz's user avatar
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9 votes
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Radon-Nikodym-derivative as a martingale

Let $(\Omega, \mathcal{F}_{\infty}, \mathbb{P})$ be a probability space and let $(B_t)$ be a standard Brownian motion with its natural filtration $(\mathcal{F}_t)_{t\geq 0}$. Let $\mathbb{Q}$ be a ...
m_gnacik's user avatar
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9 votes

How long until a random word with letters "A", "B", "C" ends in the pattern "ABC"?

Expectation is easy enough (Variance seems like more work). We have four states, according to how much of $ABC$ is complete. Thus the states are $\emptyset, A, AB, ABC$. Of course, Start is $\...
lulu's user avatar
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8 votes
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Trying to understand $\mathbb E[X\mid \mathcal F]$ and Martingale concept

(Q1) I think it's helpful to think of it as an integral. For outcome $\omega\in\Omega$ we have that $\mathbb E [X \mid \mathcal G](\omega)=\frac{1}{P(A)}\int_A X(x) P(dx)$, where $A$ is the "...
jdods's user avatar
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8 votes
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Proving Wald's identity for Brownian motion

Without any additional information on the stopping time it will be difficult to find an integrable dominating function, I think. In order to avoid the problem of finding a dominating function you can ...
saz's user avatar
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8 votes
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Uniformly integrable local martingale

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $...
Rhys Steele's user avatar
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8 votes
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The distribution of a stopping time

As Mike pointed out, $\tau$ is a geometric random variable. To see this, let $n \geq 0$ then $$\begin{align}P(\tau = n) &= P(X_0 \notin S, X_1 \notin S, \ldots, X_{n-1} \notin S, X_n \in S) \\ &...
Michh's user avatar
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8 votes
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If $(X_n)$ is a martingale, prove that $(X_{n\wedge N})_n$ is a martingale, where $N$ is a stoping time

Note that $$ \begin{aligned} E(X_{N\wedge n+1}| \mathcal F_n) &= E(X_{N} 1_{N\leq n}+X_{n+1} 1_{N\geq n+1}| \mathcal F_n)\\ &= 1_{N\leq n} E(X_{N} | \mathcal F_n) + 1_{N\geq n+1}E(X_{n+1}| \...
Gabriel Romon's user avatar
8 votes
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General version of the fundamental theorem of asset pricing in continuous time

Let's begin with a preliminary definition. Definition No free lunch with vanishing risk (NFLVR). We say that a process $S$ satisfies NFLVR if there does not exist a sequence of admissible integrands $...
Jose Avilez's user avatar
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8 votes
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How to apply Ito's Formula to show that this is a martingale?

Note $f(x)=(1+\lambda |x|)e^{-\lambda |x|}$ is twice continuously differentiable. Indeed: $f'(x)=-\lambda^2xe^{-\lambda |x|}$ and $f''(x)=\lambda^2e^{-\lambda |x|}(\lambda |x|-1)$. Furthermore, if $...
Snoop's user avatar
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7 votes

Law of Large Numbers for Martingales

The claim in question is a corollary of a standard SLLN for martingale difference sequences (MDS). SLLN for MDS The statement of SLLN for MDS is as follows. Let $N_t$ be a martingale difference ...
Michael's user avatar
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