2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.