A random variable, $X_n$, is defined to be bounded in probability if there exists an $M$ and $N$ for which \begin{align*} \mathbb{P}\big(|X_n| < M\big) > 1 - \epsilon,\ \forall n > N,\ \forall \epsilon > 0, \end{align*} and it is defined to converge in probability to another random variable $X$ if \begin{align*} \mathbb{P}\big(|X_n - X| < \epsilon\big) \to 1, \text{ as } n \to \infty,\ \forall \epsilon > 0. \end{align*}
My question is as follows: Is saying that "$X_n$ is bounded in probability" equivalent to saying that "$X_n$ converges in probability to 0"?
Choose an $N$ such that $M = \epsilon$. $X_n$ can be then re-expressed as \begin{align*} \mathbb{P}\big(|X_n| < \epsilon\big) > 1 - \epsilon,\ \forall n > N_{\epsilon},\ \forall \epsilon > 0. \end{align*} This leads to \begin{align*} 1 - \epsilon < \mathbb{P}\big(|X_n| < \epsilon\big) \leq 1 \end{align*} for all $\epsilon > 0$. Since the set $\{|X_n| < \epsilon_1\} \subset \{|X_n| < \epsilon_2\}$ for all $\epsilon_1 < \epsilon_2$, by continuity of probability, \begin{align*} \lim_{\epsilon \to 0} \mathbb{P}\big(|X_n| < \epsilon\big) = \mathbb{P}\!\left(\lim_{\epsilon \to 0} |X_n| < \epsilon\right) = \mathbb{P}\big(|X_n| \leq 0\big), \end{align*} so \begin{align*} \lim_{\epsilon \to 0} 1 - \epsilon \leq \lim_{\epsilon \to 0} \mathbb{P}\big(|X_n| < \epsilon\big) \leq 1, \text{ or }\ \ 1 \leq \mathbb{P}\big(|X_n| \leq 0\big) \leq 1, \end{align*} which, by the Sandwich Theorem, gives $\mathbb{P}\big(|X_n| \leq 0\big) = 1$ for all $n \geq N_{\epsilon}$. Then $\mathbb{P}\big(|X_n| < 0\big) \to 1$ as $n \to \infty$, since $N_{\epsilon} < \infty$. Since for all $\epsilon > 0$, \begin{align*} \mathbb{P}\big(|X_n| < \epsilon\big) \geq \mathbb{P}\big(|X_n| \leq 0\big) = 1, \end{align*} $\mathbb{P}\big(|X_n| < \epsilon\big) \to 1$ and $X_n$ converges to 0 in probability.
Remark: The "$<$" inequalities are converted into "$\leq$" as the limit $\epsilon \to 0$ is approached from the right.
As I am new to the topic, can I check with the seasoned professionals here on whether the above argument is flawed? Thanks in advance.
$\endgroup$2 Answers
$\begingroup$If $X$ is a random variable, then $\mathbb P(|X|\geqslant M)\to 0$ as $M$ goes to infinity. A sequence is bounded in probability if this convergence is uniform with respect to the collection, that is, $\lim_{M\to \infty}\sup_n\mathbb P(|X_N|\geqslant M)=0$.
But it tells nothing about the convergence in probability to $0$, for example, if $X_n=X\neq 0$ for each $n$, we have a sequence which is bounded in probability but which does not converge to $0$ in probability.
What is true is the following: if $X_n\to 0$ in probability, then $\mathbb P(|X_n|>1)\to 0$ as $n$ goes to infinity. Fix $\varepsilon$. Pick $n_0$ such that $P(|X_n|>1)\lt \varepsilon$ if $n\geqslant n_0+1$, and conclude using the fact that the finite sequence $X_1,\dots,X_{n_0}$ is bounded in probability.
$\endgroup$ $\begingroup$As I see it - no, it is not equivalent.
Counterexample
Define sequence of trivial random variables $X_n(\omega) = (-1)^n$ $\forall \omega \in \Omega$. Then $ \forall n \in \mathbb{N} $ we have $|X_n| \leq 1$, so $X_n$ are bounded in probability, but they do not converge to anything.
Mistake
You made a mistake in the line:
"Choose an $N$ such that $M=\epsilon$"
The definition of being bounded in probability (which you've quoted) says that there exists some constant $M$, specific one. It doesn't say that for each constant $C$ you can find a number $N$ large enough to have:
$\forall n>N$ $\mathbb{P}(|X_n| < C)$.
You can see it clearly in our counterexample. $X_n$'s are bounded in probability by $M=2$, but none of them is bounded by $M \in (0,1)$.
Hope it helps,
Mateusz
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