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Let $(X^n),(Y^n)$ be a random variables with $E[X^n]\leq x \, \forall n $ for some fixed $x\in \mathbb{R}$ and $X^n\geq Y^n\, \forall n$.

Also assume that $Y^n\rightarrow Y$ in $L^1(P)$ as $n\rightarrow \infty$ for some random variable $Y$.

Does that imply that $\big((X^n)^-\big)_{n\in \mathbb{N}}$ is uniformly integrable?

DrShredz
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    What do you mean by $Y^n \nearrow Y$ in $L^1(P)$? I think I have only seen the notation $Y^n \nearrow Y$ used for pointwise convergence. – Jordan Green Jan 23 '19 at 14:27
  • Yeah sorry, I shouldn't have written it that way, it was due to the fact that it is a maximizing sequence in the source, where the problem is taken from. I don't think that matters here though, hopefully. – DrShredz Jan 23 '19 at 14:33
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    What are your thoughts on the problem? What does the inequality $X^n \geq Y^n$ tell you about the negative parts of the randomv variables? What is the relation between $(X^{n})^-$ and $(Y^n)^-$...? – saz Jan 23 '19 at 14:39
  • Thanks for the kind advice, I didn't understand that the somehow "pointwise" (in terms of n) upper bound, translated to a sufficient upper bound of uniform integrability. – DrShredz Jan 23 '19 at 17:30

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Yes, it does. To prove so, let $\varepsilon > 0$. We must show that there is $\delta > 0$ so that if $\Omega$ is our sample space and $A \subset \Omega$ is measurable with $P(A) < \delta$, then $\int_A (X^n)^- dP< \varepsilon$. For this, we begin by noticing that $(X^n)^- \leq (Y^n)^- \leq |Y^n|$. (This can be proved by casework.) Thus, it suffices to prove that the family $\{ Y^n \}$ is uniformly integrable.

A proof that the family $Y^n$ is uniformly integrable is as follows. (There may be a more elegant solution, but this is what seemed most immediate.) Fix $\varepsilon > 0$. For any measurable $A$ we have $$ \int_{A} |Y^n|\, dP \leq \int_{A} |Y^n - Y|\, dP + \int_{A} |Y| \,dP. \, \, \, (\star) $$ We can make the integral of $|Y|$ less than $\frac{\varepsilon}{2}$ by taking $P(A)$ sufficiently small - say, less than $\delta_1$. This uses absolute continuity of the Lebesgue integral (see this stack exchange answer: Absolute continuity of the Lebesgue integral).

Meanwhile, we have $$ \int_{A} |Y^n - Y|\, dP \leq \int_{\Omega} |Y^n - Y|\, dP. $$ Since $Y^n \to Y$ in $L^1$, there is $N$ so that if $n \geq N$, then $\int_{\Omega} |Y^n - Y|\, dP < \frac{\varepsilon}{2}$. It follows by $(\star)$ that if $P(A) < \delta_1$, then $\int_A |Y^n| \, dP < \varepsilon$ for all $n \geq N$.

To deal with the case that $n < N$, let $S = |Y_1| + \dots |Y_N|$. Then for each $n < N$, we have $$ \int_{A} |Y^n|\, dP + \int_A |Y| \, dP \leq \int_{A} S + \int_{A} |Y|. $$ By the absolute continuity property mentioned above, we can make $\int_{A} S \, dP$ less than $\varepsilon$ by picking an appropriate $\delta_2$. Thus, for $A$ with $P(A) < \delta_2$, we’ll have $\int_A |Y^n| \, dP < \varepsilon$ for all $n < N$.

We can then take $\delta$ to be the minimum of $\delta_1$ and $\delta_2$.

Jordan Green
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