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
8 votes
Accepted

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
  • 19.7k
7 votes
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Differentiation under the integral sign and uniform integrability

Thinking a little more, I believe this is true. Let's adopt the strong sense of uniform integrability, and further assume that $\partial_t f$ is jointly measurable. Set $G(t) = \int_X \partial_t f(t,x)...
Nate Eldredge's user avatar
6 votes
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Why $\lim_{n\rightarrow \infty}\int_0^1 e^{x^n} = 1$

To show $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{e^{x^n}\,dx}} = 1$, it suffices to show that $\lim\limits_{n\rightarrow\infty}{\int\limits_{0}^{1}{(e^{x^n}-1)\,dx}} = 0$. Clearly, we ...
Joey Zou's user avatar
  • 8,466
6 votes
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Is $Y_n := \prod_1^n \xi_i$ for $\xi_i$ i.i.d. $\text{Unif}(0,2)$ a sequence of uniformly integrable random variables?

Note that since $(Y_n,\mathcal F_n^Y)$ is a martingale, then uniform integrability is equivalent with convergence in $L_1$. Moreover, if the limit exists (call it $Y$), then it must be the case that $\...
Presage's user avatar
  • 8,196
6 votes

Showing $\{X_n\}$ is uniformly integrable when $\sup _{n} \mathbb{E}\left[X_{n}^{2}\right]<\infty$

Let $M^2=\sup_n E[|X_n|^2]$. Then $E[|X_n|\mathbb{1}(|X_n|>a)\leq \sqrt{P(|X_n|\geq a)}M\leq\frac{1}{a}M^2 $
Mittens's user avatar
  • 39.2k
5 votes

Why $\lim_{n\rightarrow \infty}\int_0^1 e^{x^n} = 1$

Clearly $e^{x^n} \ge 1$, hence the integral is $\ge 1$ for all $n$. On the other hand, we have for each $\delta>0$ $$ \int_{0}^1e^{x^n}\mathrm{d}x\le \int_{0}^{1-\delta}e^{x^n}\mathrm{d}x+\delta e \...
Paolo Leonetti's user avatar
5 votes
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Uniform integrability and stopping times

This is not true. Moreover, it seems pretty obvious to me that the fact like this cannot be true without any further assumption like the martingale property; however, the requirement that $N$ should ...
zhoraster's user avatar
  • 25.5k
5 votes
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Uniform Integrability implies boundedness of $\sup_i\int|f_i|dP$?

For the second part in more general terms. Note that if we assume $(f_i)_{i \in I}$ is uniformly integrable, then for all $M>0$ we have for all $i \in I$ that $$ \int_\Omega |f_i| dP = \int_{(|f_i|...
K. Brix's user avatar
  • 555
5 votes
Accepted

Compactness in $L^1$ implies equi-integrability

I don't see how the shift operators come into play. You don't need them. As you said, there exist $f_1,\ldots,f_n$ such that the open balls $B(f_j,\epsilon/2)$, $j=1,\ldots,n$, cover the compact set $...
saz's user avatar
  • 120k
5 votes

If $\mathbb{E}(\sup_n |X_n|)< \infty$ then $(X_n)_n$ is uniformly integrable

Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of random variables. If there exists an integrable random variable $Y$ such that $|X_n| \leq Y$ for all $n \in \mathbb{N}$, then $(X_n)_{n \in \mathbb{N}}$ ...
saz's user avatar
  • 120k
5 votes
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Show that $\text{ sup}_n \int{|f_n|^pdm<C}$ implies $\int_{E}{|f_n|dm}<\epsilon \text{ for large enough } n$.

That comes for instance from Hölder's inequality: $$\int_E |f| \ d \mu = \int_\Omega \mathbf{1}_E |f| \ d \mu \leq \|f\|_{\mathbb{L}^p (\Omega)} \|\mathbf{1}_E \|_{\mathbb{L}^{p^*} (\Omega)} = \|f\|_{...
D. Thomine's user avatar
  • 10.9k
5 votes
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Almost-sure convergence of a series of log-normal random variables

By taking logarithms, dividing by $n$, and re-arranging, we obtain: $$\frac{1}{n} \sum_{i=1}^n X_i = \frac{1}{n} \log W_n + \frac{1}{2}$$ By the strong law of large numbers, $$\frac{1}{n} \sum_{i=1}^n ...
Jose Avilez's user avatar
  • 12.7k
5 votes
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Limit and conditional expectation commute in a uniformly integrable sequence

The proposition in the post does not hold. The problem arises from the implicit assumption that the a.s. limit of the conditional expectations exists. The example described in the link [1] is correct, ...
Yuval Peres's user avatar
4 votes

Counterexample for uniform integrability of an $\mathbb{L}^1$-bounded sequence

Let $\Omega=[0,1]$ with Lebesgue measure, and $X_n=n1_{[0,\frac{1}{n}]}$. Then the $X_n$ are bounded in $L^1$ but not uniformly integrable.
carmichael561's user avatar
4 votes

Uniform Integrability of Random Variables

You meant $\{X_n\}$ are uniformly integrable if $$ \lim_{M\to\infty} \sup_n E[ |X_n|\chi_{\{|X_n|>M\}}]=0$$ (e.g. https://en.wikipedia.org/wiki/Uniform_integrability) Now for your question. ...
Fnacool's user avatar
  • 7,519
4 votes
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Limit of an integral: Possible issue with uniform integrability?

Let $y = nx$. Then $$\int_0^1 f_n(x) \, dx = \int_0^n \frac{e^{y/n} \cos(y/n)}{1 + y^2}\, dy = \int_0^\infty g_n(y)\,dy$$ with $$g_n(y)=\frac{e^{y/n} \cos(y/n)}{1 + y^2} 1_{[0,n]}(y).$$ We have for ...
RRL's user avatar
  • 90.8k
4 votes

Is a sequence of identical distributed integrable random variables even uniformly integrable?

Since the random variables are identically distributed, we have $$\mathbb{E}(1_{\{|X_n| > c\}} |X_n|) = \mathbb{E}(1_{\{|X_1|>c\}} |X_1|)$$ for all $n \geq 1$ and therefore $$\sup_{n \geq 1} \...
saz's user avatar
  • 120k
4 votes
Accepted

Non uniformly integrable sequence

Consider $$X_n := n 1_{(0,1/n)}-n 1_{(1/n,2/n)}$$
saz's user avatar
  • 120k
4 votes
Accepted

Stopped random walk is not uniformly integrable

It follows from the optional stopping theorem that $(X_{n \wedge \tau})_{n \in \mathbb{N}}$ is a martingale; in particular, $\mathbb{E}(X_{n \wedge \tau}) = \mathbb{E}(X_0)=0$. Since \begin{align*} 0 =...
saz's user avatar
  • 120k
4 votes
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Difficult example of Uniformly absolute continuous sequence but not $L^1$ bounded

If $\mu$ is counting measure on a finite set then uniform absolute continuity trivially holds for any sequence $(f_n)$: take $\delta <1$. Obviously, $(f_n)$ need not be $L^{1}$ bounded.
Kavi Rama Murthy's user avatar
4 votes
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Convergence in measure implies $L^1$ norm-(Vitali Convergence Theorem)

It is something related to uniform integrability. The following theorems and the notion of uniform integrability are due to Vitali, which play an important role in probability theory. For your problem,...
Danny Pak-Keung Chan's user avatar
4 votes
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Showing $\{X_n\}$ is uniformly integrable when $\sup _{n} \mathbb{E}\left[X_{n}^{2}\right]<\infty$

$E|X_n|1_{|X_N| \geq \alpha} \leq \sqrt {EX_n^{2}} \sqrt {P(|X_n|\geq \alpha)}$. Hence it is enough to show that $\sup_n {P(|X_n|\geq \alpha)} \to 0$ as $\alpha \to \infty$. But ${P(|X_n|\geq \alpha)=...
Kavi Rama Murthy's user avatar
4 votes
Accepted

If $f$ us periodic and even, what I can conclude about of $\int f \;dx$?

Take $f(x)=\sin^{2}x$ for a counter-example. Here $L=\pi$ and the integral is strictly positive.
Kavi Rama Murthy's user avatar
4 votes
Accepted

stopped UI discrete time martingale is UI

Since $(X_n)$ is UI, we know there exists $X_\infty \in L^1(\mathcal F_\infty)$ such that $(X_n) \rightarrow X_\infty$ a.s. and in $L^1$. Furthermore, the uniform integrability and optional stopping ...
user6247850's user avatar
  • 13.5k
4 votes
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Uniformly integrability of inverse of sample moments

Cannot be true in general: consider $k=1$ (or any odd $k$), take $X_i$'s to be i.i.d normal with mean $0$ and variance $1$, we show below that $Y_n$ is not UI. Note that $$ \mathbb{E}\left(|Y_n| 1\{|...
Suman Chakraborty's user avatar
4 votes
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Stopped version of UI martingale is UI

Any uniformly integrable martingale can be written in the form $(E(X|\mathcal F_n))$ with $E|X| <\infty$. Optional Sampling Theorem (SeeTheorem 9.3.3, p.324 of K L Chung's "A Course in ...
Kavi Rama Murthy's user avatar
4 votes
Accepted

Proof of a martingale not being uniformly integrable

If the martingale $\left(\varphi\left(S_n\right)\right)_{n\geqslant 0}$ was uniformly integrable, then the optional stopping theorem for uniformly integrable martingale would give that $\mathbb E\left[...
Davide Giraudo's user avatar
3 votes

Limit of an integral: Possible issue with uniform integrability?

Consider $g_n(x) = \frac{n}{1+n^2x^2}$. Then $c g_n \leq f_n \leq Cg_n$ on $[0,1]$ and $\int_0^1 g_n(x)\,dx = \arctan(n)$. Also notice that $g_n \to 0$ but $arctan(n) \to \frac{\pi}{2}$, Since the $...
nullUser's user avatar
  • 27.9k
3 votes
Accepted

$\lim X_n = 0$ iff $b > 0$

First, rewrite the exponenent as \begin{align*} \exp \left( n\left( a\frac{S_n}{n} - b \right) \right) \end{align*} By the Law of Large Numbers, we have $a S_n/n- b \to a \mathbb{E}(\xi_1) - b$ a.s.. ...
Hetebrij's user avatar
  • 3,981

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