If a product is bounded in probability, are the terms also bounded in probability?

Question

Suppose we have $a_n$ and $b_n$ random variables such that $a_n \times b_n = Op(n^{k})$ for $n \to \infty$.

We have $a_n = C + op(1)$, with $C \ne 0$.

Intuition tells me $b_n = Op(n^{k})$ , but how do I prove this?

Answer 1

Assume, without loss of generality, that C > 0.

For any $\epsilon > 0$ and $\delta = C/2$ there is a $N_a$ such that for $n > N_a$,

$P( |a_n|> C - \delta ) < \epsilon/2$.

Then, for $n > N_a$

$P( n^{-k}|a_n b_n| > M ) > P( n^{-k} |b_n| > 2 M / C ) = P( n^{-k} |b_n| > M_* )$

with $M_* = 2 M /C $

hence if $b_n$ isn't $Op(n^k)$ the right side doesn't go less than an $\epsilon/2$ no matter the $M$, which would implies that $a_n \times b_n \ne Op(n^{k})$