Does $E(X_{n} 1_{\vert X_{n} \vert \geq 1 })=\frac{1}{n}$ imply uniform integrability

Question

Say $E(X_{n} 1_{\vert X_{n} \vert \geq 1 })=\frac{1}{n}$ for all $n \in \mathbb N$.

Can I immediately deduce that $(X_{n})_{n}$ is uniformly integrable?

My idea:

$E(\vert X_{n}\vert)= E(\vert X_{n}\vert 1_{\vert X_{n} \vert \geq 1 })+E(\vert X_{n}\vert 1_{\vert X_{n} \vert < 1 })\leq\frac{1}{n}+1$

Thus $\sup\limits_{n} E(\vert X_{n}\vert)\leq 2$

Let $\epsilon > 0$, then choose $N \in \mathbb N>1$ so that $\frac{1}{N} < \epsilon$ so that for all $n \geq N$:

$E(\vert X_{n}\vert 1_{\vert X_{n} \vert \geq n })=E(\vert X_{n}\vert 1_{\vert X_{n} \vert \geq N })\leq E(\vert X_{n}\vert 1_{\vert X_{n} \vert \geq 1 })\leq \frac{1}{n}\xrightarrow{n \to \infty} 0$

I still have not shown uniform integrability. Any ideas?

Answer 1

For given $\epsilon>0$ choose $N \in \mathbb{N}$ sufficiently large such that $1/N \leq \epsilon$. Then

$$\int_{|X_n| \geq R} |X_n| \, d\mathbb{P} \leq \int_{|X_n| \geq 1} |X_n| \, d\mathbb{P} \leq \frac{1}{n} \leq \epsilon$$

for all $n \geq N$ and $R \geq 1$. On the other hand, the finite family $\{X_1,\ldots,X_{N-1}\}$ is uniformly integrable, and therefore there exists $R \gg 1$ sufficiently large such that

$$\int_{|X_n| \geq R} |X_n| \, d\mathbb{P} \leq \epsilon$$

for all $n \in \{1,\ldots,N-1\}$. Hence,

$$\sup_{n \geq 1} \int_{|X_n| \geq R} |X_n| \,d \mathbb{P} \leq \epsilon$$

which proves that $(X_n)_{n \geq 1}$ is uniformly integrable.

Answer 2

If the assumption was $$ E\left(\lvert X_{n}\rvert 1_{\vert X_{n} \vert \geq 1 }\right))=\frac{1}{n}, $$ then for all $R\geqslant 1$, $\limsup_{n\to+\infty}E\left(\lvert X_{n}\rvert 1_{\vert X_{n} \vert \geq R }\right))=0$ hence the uniform integrability follows.

However, with the assumption $$E\left( X_{n} 1_{\vert X_{n} \vert \geq 1 }\right))=\frac{1}{n},$$ the conclusion may not hold: let $X_n$ be a random variable taking the values $n$, $-n$, $1$ and $0$ with respective probabilities $1/n$, $1/n$, $1/n$ and $1-3/n$.