4

I consider a family of sub sigma algebra $(\mathcal{F}_s)_{s\in S}$ on ($\Omega,\mathcal{A}, \mathbb{P}$) and $X\in L^1(\Omega,\mathcal{A}, \mathbb{P})$. I want to show that $Y_s =\mathbb{E}(X | \mathcal{F}_s)$ is uniformly integrable.

My attempt is the following : I use the caracterization in terms of boundedness of a uniformly integrable family.

First we notice that there exists $M\geq 0$ such that $\mathbb{E}(\lvert X\rvert)\leq M$.

Now we notice that for all $s\in S$ we have

$$ \lVert Y_s \rVert_{L^1} = \mathbb{E}\left\lvert[\mathbb{E}(X | \mathcal{F}_s)\right\rvert] \leq \mathbb{E}[\mathbb{E}(\lvert X \rvert | \mathcal{F}_s)] = \mathbb{E}(\lvert X\rvert)\leq M $$

Which shows that the family is bounded in $L^1$ and thus uniformly integrable. Is this seems correct ?

Edit : This is false, my memory was totally wrong as my intuition. To solve the problem unfortunately I have not found other solutions than strenghten hypothesis by considering that $X\in L^p$ in order to use a characterization in terms of epsilon delta of the uniform integrability.

First we notice that there exists $M$ such that for all $s\in S$ $\lVert Y_s\rVert_{L^p}\leq M$

Let $\epsilon>0$. Take $\delta = \frac{\epsilon^q}{M^q}$. We have

$$ \mathbb{E}(\lvert Y_s\rvert 1_{A})\leq \lVert Y_s\rVert_{L^p}(\mathbb{P}(A))^{1/q}\leq M(\mathbb{P}(A))^{1/q}\leq M\frac{\epsilon}{M} $$

I am almost sure the hypothesis I have made is superficial

G2MWF
  • 1,615
  • 2
    No: bounded in $L^1$ does not imply UI. See this for example. – Snoop Oct 01 '23 at 18:40
  • @Snoop You are right. I have tried something else. – G2MWF Oct 01 '23 at 19:15
  • 1
    @copper.hat Made the necessary modifications ! – G2MWF Oct 01 '23 at 19:28
  • 1
    Yes, it works. Without this hypothesis, you can consider instead some $C>0$ and note that for any $s \in S$; $\mathbb E|Y_s 1_{|Y_s| > C}| = \mathbb E| \mathbb E[X1_{|Y_s|>C} | \mathcal F_s] | ] \le \mathbb E|X1_{|Y_s| > C}|$. Now, $\mathbb P(|Y_s| > C) \le \frac{\mathbb E|Y_s|}{C} \le \frac{M}{C}$, so by integrability of $X$ it follows that $\lim_{C \to \infty} \sup_{s \in S} \mathbb E|Y_s 1_{|Y_s|>C}| = 0$, which gives Uniform Integrability (one of the equivalent definitions). – Dominik Kutek Oct 01 '23 at 19:33
  • @DominikKutek The proof I should have done ! Thank you a lot. – G2MWF Oct 01 '23 at 19:45
  • 1
    @DominikKutek I don't see how your argument concludes uniformly in $s$. – Snoop Oct 01 '23 at 20:25
  • 1
    It's rather standard approximation argument. Fix $\varepsilon > 0$. Due to integrability of $X$, there is some $K>0$ such that $\mathbb E|X1_{|X|>K}| < \varepsilon$. Take $C > \frac{MK}{\varepsilon}$. Then, $ \mathbb E|Y_s1_{|Y_s| > C}| \le \mathbb E|X1_{|Y_s|>C}| \le \mathbb E|X1_{|X| > K}| + \mathbb E|X1_{|X| \le K}1_{|Y_s| >C}| \le \varepsilon + K\mathbb P(|Y_s| > C) \le \varepsilon + K \frac{M}{C} = 2 \varepsilon$. The crucial part is that the bound $\mathbb P(|Y_s| > C) \le \frac{M}{C}$ does not depend on $s$. – Dominik Kutek Oct 01 '23 at 22:33

2 Answers2

1

Using Jensen's inequality and the fact that $X$ is in $L^1(\Omega, \mathcal{A},\mathbb{P})$, we have $$|Y_s|\leq \mathbb{E}(|X| | \mathcal{F}_s),$$ for all $s\in S$. Then, using the maximal inequality, $$\mathbb{P}(\sup_s|Y_s|>M)\leq \frac{\mathbb{E}|Y_s|}{M}\leq \frac{\mathbb{E}|X|}{M}\to0, $$ as $M\to\infty$, uniformly in $s$.

Now using measurability, independence and conditional expectation properties, write $$\mathbb{E}(|Y_s|1_{\{|Y_s|>M\}}) \leq \mathbb{E}(|X|1_{\{\sup_s|Y_s|>M\}}), $$ and conclude that the RHS goes to zero as $M\to\infty$.

user
  • 408
0

I think there is an other elegant way to prove the result using the fact $X$ is uniformly integrable. I post it as an answer for the moment hoping this is correct.

First we notice that for all $s\in S$ we have using the tower property :

$$ \mathbb{E}[\lvert Y_s\rvert 1_{\lvert Y_s\rvert> C}]\leq\mathbb{E}[\lvert X\rvert 1_{\lvert Y_s\rvert> C}] $$

Let $\epsilon>0$. By uniform integrability of $X$, there exists $\delta>0$ such that $\mathbb{P}(A)\leq\delta$ implies $\mathbb{E}[\lvert X\rvert 1_{A}]\leq\epsilon$.

We also have $\mathbb{P}(\lvert Y_s\rvert > C)\leq\frac{M}{C}$, taking $C\geq\frac{\delta}{M}$ gives $\mathbb{P}(\lvert Y_s\rvert > C)\leq\delta$.

Thus, for all $\epsilon>0$, we have found a treshold (namely $\frac{\delta}{M}$) such that for all $C$ greater than this treshold we have for all $s\in S$

$$ \mathbb{E}[\lvert Y_s\rvert 1_{\lvert Y_s\rvert> C}]\leq\mathbb{E}[\lvert X\rvert 1_{\lvert Y_s\rvert> C}]\leq\epsilon $$

G2MWF
  • 1,615