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I'm self-studying probabilistic order notation, and I wanted to show some properties to get used to it. But now I'm having trouble showing that the sum of two random variables that are bounded in probability is bounded in probability (also known as tight, or $O_p(1)$).

Let $X_n = O_p(1)$ and $Y_n = O_p(1)$, then for any $\epsilon > 0$, there exists $M$ such that $\sup_n P(|X_n| > M) < \epsilon/2$ and $\sup_n P(|Y_n| > M) < \epsilon/2$. Now I'd like to show that $\sup_n P(|X_n + Y_n| > M) < \epsilon$, but I haven't succeeded so far. I couldn't go further than $P(|X_n + Y_n| > M) \le P(|X_n|+|Y_n| > M)$.

Thanks!

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1 Answer 1

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The secret is to observe the following

$$\{|X_n| + | Y_n| > M \} \subset \{ |X_n| \ge M/2 \} \cup \{|Y_n| \ge M/2 \}$$

otherwise, $|X_n| + | Y_n| $ would be strictly less than $M$.

Then

$$P(|X_n| + | Y_n| > M ) \le P(|X_n| \ge M/2) + P(|Y_n| \ge M/2)$$

now you use properties of sup for a set of positive numbers and the inequality that you suggested.

Hope this can help.

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