$X_n\to0$ in probability and $\{c_n\}$ is a bdd seq of real numbers, then $c_nX_n\to0$ in probability.

Question

Suppose $X_n\to0$ in probability and $\{c_n\}$ is a bounded sequence of real numbers. Prove that $c_nX_n\to0$ in probability.

I believe I need to use a theorem, which states that:

If $\{X_n\}_{n=1}^\infty$ and $X$ are random variables on a probability space, $(\Omega,\mathcal{F},P)$. Then, $X_n\to X$ in probability as $n\to\infty$ iff every subsequence $\{X_{n_{m}}\}$ has a further subsequence $\{X_{n_{m_{k}}}\}$ such that $X_{n_{m_{k}}}\to X$ a.s. as $k\to\infty$.

I am having a difficult time formulating my thoughts though. Any help would be appreciated.

Answer 1

You can use that theorem to prove the result, but I think that makes it more complicated than it needs to be.

Hint: The typical condition of convergence in probability is: $$X_n \stackrel{p}{\to} X \iff \forall \epsilon > 0, \lim_{n \to \infty} \mathbb P(|X_n - X| > \epsilon) = 0.$$

Here, that means that for any $\epsilon > 0$, $\lim_{n \to \infty} \mathbb P(|X_n| > \epsilon) = 0$. What can you say about $\mathbb P(|c_n X_n| > \epsilon)$?

Answer 2

Abridged solution using the theorem you wrote. There is $|c_n| \leq \alpha$ for all $n,$ then the relation $X_{n_{m_k}} \to 0$ (being equivalent to $|X_{n_{m_k}}| \to 0$), implies $0 \leq |c_{n_{m_k}} X_{n_{m_k}}| \leq \alpha |X_{n_{m_k}}| \to 0,$ hence the result. Q.E.D.

Answer 3

Note that $$ C|X_n|\ge |c_n X_n| $$ for some $C>0$ (if $c_n$ is identically zero the result is trivial) since the sequence $(c_n)$ is bounded. In particular, given $\epsilon>0$, $$ P(|c_nX_n|>\epsilon)\leq P(C|X_n|>\epsilon)\to 0 $$ since one event is a subset of the other and $X_n\to 0$ in probability.