$E(f(|X_n|))$ property implies uniform integrability?

Question

This is exercise 6.10 in Resnick's book "A Probability Path".

We're given a sequence of random variables $(X_n)$ and an increasing function $f: [0, \infty) \rightarrow [0, \infty)$ such that

$$ \frac{f(x)}{x} \rightarrow \infty, \qquad x \rightarrow \infty, $$

and

$$ \sup_{n \geq 1} E(f(|X_n|)) < \infty, $$

and the goal is to conclude that $\{X_n\}$ is uniformly integrable. I know I'm supposed to show my work so far but honestly I find the problem hard to even get a start on, mostly because the existence of $f$ is abstract and I just don't see how to get from A to B. No, I am not a student. Any help is appreciated.

Answer 1

Hmm. Writing integrals as integrals: Say $\epsilon>0$. You need to show there exists $\delta>0$ so $P(A)<\delta$ implies $\int_A|X_n|\,dP<\epsilon$. You know that $\int f(|X_n|)\,dP\le K$ for some fixed $K$. Hmm.

Ok. Say $L>0$ is a large number. Choose $M$ so $$\frac{f(t)}t>L\quad(t>M).$$ Now write $A=B\cup C$, where $B=\{t\in A:|X_n(t)|>M\}$ and $C=A\setminus B$. Now $$\int_C|X_n|\,dP\le MP(C)\le MP(A).$$So $\int_C|X_n|<\epsilon/2$ if $M\delta<\epsilon/2$. Progress...

Ah. On the other hand, on $B$ we have $|X_n|>M$, hence $f(|X_n|)/|X_n|>L$, or $|X_n|<f(|X_n|)/L$. So $$\int_B|X_n|\,dP\le\frac1L\int_Bf(|X_n|)\,dP\le\frac KL.$$

There we are. First choose $L$ so that $K/L<\epsilon/2$. Now $L$ determines $M$; choose $\delta$ so $M\delta<\epsilon/2$ and you're done.

Answer 2

I just started using the math stack exchange community to help with my self study of probability theory, and I have a comment in general about this problem. It should really be a comment on the above post by David, but I don't have the reputation points for that, so I am posting it here. Also, at the end of this post I am posting my own solution to the problem, so in that way I suppose it is appropriate as an answer to the original question.

In regard to David's post, where have we used that $f$ is increasing? I also did the problem myself -- my proof is below -- and I have a slightly different argument, but I didn't use that $f$ is increasing. And I don't see in David's post the use of this assumption either. But I could be missing something, so please correct me if I'm wrong.

On a slightly more pedantic note, according to the definition of uniform integrability given in Resnick, David's argument is missing the check that $sup_{n \geq 1} E(|X_n|) < \infty$. This is not hard to provide, but it is worth noting.

So as promised, here is my solution to the problem -- it is based directly on the definition of uniform integrability given in Resnick. I'm not suggesting it is any better or worse than David's approach. Just giving my approach:

For $a > 0$,

$$\int_{[|X_n| > a]} |X_n| dP \leq \int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} |X_n| dP + \int_{[|X_n| > a] \cap [|X_n| > f(|X_n|)]} |X_n| dP$$. Thus

$$sup_{n \geq 1} \int_{|X_n| > a]} |X_n| dP \leq sup_{n \geq 1} \int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} |X_n| dP+sup_{n\geq1}\int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} |X_n| dP$$

So it is sufficient to show the two terms on the left $\to 0$ as $a \to \infty$

For the first term ( $sup_{n \geq 1} \int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} |X_n| dP$), Let $\epsilon > 0$. There exists $M > 0 $ such that if $a > M$ then $\frac{a}{f(a)} < \frac{\epsilon}{sup_{n \geq 1} E(f(|X_n|))}$. Note this is possible since $\frac{f(a)}{a} \to \infty$ as $a \to \infty$. So we have that for $a > M$

$$sup_{n \geq 1} \int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} |X_n| dP = sup_{n \geq 1} \int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} \frac{|X_n|}{f(|X_n|)} f(|X_n|) dP \leq \frac{\epsilon}{sup_{n \geq 1} E(f|X_n|)} sup_{n \geq 1} E(f|X_n|) = \epsilon$$

So the first term $\to 0$ as $n \to \infty$. For the second term ($sup_{n\geq1}\int_{[|X_n|>a] \cap [|X_n| \leq f(|X_n|)]} |X_n| dP$), there exists $c > 0$ such that if $a > c$, then $\frac{f(a)}{a} > 1$. Thus

$[|X_n| > f(|X_n|)] \subseteq [|X_n| \leq c]$. Thus for $a > c$, $[|X_n| > a] \cap [|X_n| > f(|X_n|)] = \emptyset$. Thus for $a > c$, the second term is $0$. Thus the second term $\to 0$ as $n \to \infty$