Bounded probability implies convergence in probability

Question

Let $(X_n)$ be a sequence of random variables and $(a_n),(b_n)$ be two sequences of non-negative real numbers such that $a_n\downarrow 0$ and $b_n\downarrow 0$ when $n\to\infty$.

If for any $t>0$, $$ P(|X_n|\geq a_n+t)\leq b_n, $$ can we conclude that $X_n\overset{P}{\to}0$ as $n\to\infty$? From the hypothesis I know that for any $t>0$ $$ \lim_{n\to\infty}P(|X_n|\geq a_n+t)=0. $$ But I do not see why the fact that $a_n\downarrow 0$ implies that $$ \lim_{n\to\infty}P(|X_n|\geq t)=0. $$ I was trying to use Slutsky's theorem, but I don't know anything of the convergence of $|X_n|-a_n$. Other thing is that for $n$ sufficiently large $a_n\leq \epsilon$ for any $\epsilon>0$, so $$ P(|X_n|\geq a_n+t)\leq P(|X_n|\geq \epsilon+t), $$ but that doesn't help either.

Any suggestions?

Answer 1

Fix $s > 0$. We want to show that $\mathbb{P}(|X_n| \geq s) \to 0$ so fix $\varepsilon > 0$ and we aim to show that for large enough $n$, $\mathbb{P}(|X_n| \geq s) < \varepsilon$.

By the assumption applied with $t = \frac{s}{2}$, $$\mathbb{P}(|X_n| \geq a_n + \frac{s}{2}) \to 0$$ Combining this fact with the assumption that $a_n \to 0$, there is an $N$ such that $n \geq N$ implies that $a_n < \frac{s}{2}$ and $\mathbb{P}(|X_n| \geq a_n + \frac{s}{2}) < \varepsilon$. This means that if $n \geq N$, $|X_n| \geq s$ implies that $|X_n| \geq a_n + \frac{s}{2}$. Therefore, for $n \geq N$, $$\mathbb{P}(|X_n| \geq s) \leq \mathbb{P}(|X_n| \geq a_n + \frac{s}{2}) < \varepsilon$$ which gives the desired convergence.

Answer 2

Let $\epsilon>0$ be fixed and let $t=\frac{1}{2}\epsilon$.

Then for $n$ large enough we have $a_{n}+t<\epsilon$ so that $P\left(\left|X_{n}\right|\geq\epsilon\right)\leq\left(\left|X_{n}\right|\geq a_{n}+t\right)\leq b_{n}$.

This leads to $\lim_{n\to\infty}P\left(\left|X_{n}\right|\geq\epsilon\right)=0$.

Then the fact that this works for every $\epsilon>0$ allows the conclusion that $X_{n}\stackrel{P}{\to}0$.