3
$\begingroup$

I have found the following very useful theorem and I would appreciate some help comprehending it fully.

Theorem Let $\{X_n \} $ be a sequence of random variables bounded in probability and let $ \{Y_n \} $ be a sequence of RVs which converge to $0$ in probability. Then

$$X_n Y_n \rightarrow 0 \quad \text{in probability} $$

And its proof:

For a given $\epsilon$ since $X_n$ is bounded in probability we can choose $N$ and a constant $B$ such that

$$n \geq N \Rightarrow P \left[ |X_n| \leq B \right] \geq 1-\epsilon$$

Then

$$ \overline{\lim_{n \to \infty}}P \left[|X_n Y_n| \geq \epsilon \right] \leq \overline{\lim_{n \to \infty}} P \left[ |X_n Y_n| \geq \epsilon, |X_n| \leq B \right]+ \overline{\lim_{n \to \infty}}P \left[|X_n Y_n| \geq \epsilon, |X_n| > B \right] \leq \overline{\lim_{n \to \infty}} P\left[|Y_n| \geq \epsilon/B \right] +\epsilon=\epsilon $$

If someone could help me understand how the last inequality is derived, given the information we have, I would be grateful. I understand the second to last inequality comes from the subadditivity property of limitsuperior so I merely need to comprehend the last one. Thank you in advance.

EDIT: I think it is because $$P(A) \times P(B|A) \leq P(A) $$ and the complement of the event that is implied by the bound in probability.

$\endgroup$
3
  • $\begingroup$ Could you provide a reference for this theorem? $\endgroup$
    – TrungDung
    Jan 17, 2022 at 15:29
  • $\begingroup$ I think it is still correct if we replace "in probability" by "a.s." in those to places. $\endgroup$
    – TrungDung
    Jan 20, 2022 at 15:07
  • $\begingroup$ The theorem appears to state $$O_p(1)\cdot o_p(1) = o_p(1)$$. This is a rather old result, for example in White's Asymptotic Theory for Econometricians (1984), it is stated as an exercise. $\endgroup$ Sep 28 at 1:18

1 Answer 1

4
$\begingroup$

First you should show that: $$P[|X_n Y_n| \geq \epsilon,|X_n|\leq B]\leq P[|Y_n|\geq \epsilon/B]$$ Hint: If $A\subset B$ then $P(A)\leq P(B)$.
For the 2nd part, again by using above property, you can say that: $$P \left[|X_n Y_n| \geq \epsilon, |X_n| > B \right]\leq P[|X_n|>B]$$ Now try to use $P[|X_n|\leq B]\geq 1-\epsilon$.

$\endgroup$
3
  • $\begingroup$ For the second case, that is easy to see so no problems there. The problem is for the first case, because we are also replacing $X_n$ with its upper bound. I find that somewhat hard to digest. I'll think about it some more. Thank you. $\endgroup$
    – JohnK
    Dec 18, 2013 at 14:35
  • $\begingroup$ Could you provide a reference for this theorem? $\endgroup$
    – TrungDung
    Jan 17, 2022 at 15:29
  • $\begingroup$ Is it correct if we assume that $Y_n$ converges a.s. to 0 then the $X_nY_n$ converges a.s. to 0? $\endgroup$
    – TrungDung
    Jan 20, 2022 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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