Let $(X_n)$ be a sequence of random variables, and $Y$ a integrable random variable with
$$\sup P(|X_n| \ge a) \le P(Y \ge a),$$
for all $a \in \mathbb{R}$. Show that $(X_n)$ is uniformly ntegrable.
This may be a stupid question, but I am having doubts if my solution is correct.
Since $P(|X_n| \ge a) \le P(Y \ge a)$ for all $n$, we have
$$\sup E(|X_n| ; |X_n|>a)=\sup \int_a^{\infty}xdF_{|X|}\le \int_a^{\infty}xdF_{Y}(x).$$
Since $Y$ is integrable, the limit when $a\rightarrow \infty$ is $0$.
Is my reasoning correct? If the dominance was pontual and not stochastic, I am certain of how to prove the result, but I'm not sure in this case.
Best Answer
Not quite. The best way to go about this is using the Darth Vader Rule. Applied here, it gives us that $$E(|X_n|1_{|X_n| \geq a})= \int_a^\infty P(|X_n|\geq a)dx \leq \int_a^\infty P(|Y|\geq a)dx = E(|Y|1_{|Y|\geq a}).$$