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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.

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  • $\begingroup$ Could you provide a reference for this theorem? $\endgroup$
    – TrungDung
    Jan 17 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 at 15:07

1 Answer 1

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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 $P[|X_n|\leq B]\geq 1-\epsilon$.

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  • $\begingroup$ 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 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 at 13:08

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