若$X_n$为$O_P(1)$且$Y_n$为$O_P(1)$，$X_nY_n$是否也为$O_P(1)$？

Great question! You're absolutely correct that if ${X_n}$ and ${Y_n}$ are both bounded in probability (i.e., $X_n = O_P(1)$ and $Y_n = O_P(1)$), then their product ${X_nY_n}$ is also bounded in probability ($X_nY_n = O_P(1)$). Let's break this down using formal definitions and a straightforward proof.

Formal Definition of $O_P(1)$ #

First, let's recap the core definition we'll use:

A sequence of random variables ${Z_n}$ is bounded in probability (written $Z_n = O_P(1)$) if for every $\epsilon > 0$, there exists a constant $M > 0$ and an integer $N$ such that for all $n \geq N$,
$$P(|Z_n| > M) < \epsilon.$$

Step-by-Step Proof #

To show $X_nY_n = O_P(1)$ given $X_n = O_P(1)$ and $Y_n = O_P(1)$, follow these steps:

• Pick any arbitrary $\epsilon > 0$. Our goal is to find a constant $K > 0$ and integer $N$ such that $P(|X_nY_n| > K) < \epsilon$ for all $n \geq N$.
• Since $X_n = O_P(1)$, by definition there exists some $M_1 > 0$ and integer $N_1$ where for all $n \geq N_1$, $P(|X_n| > M_1) < \epsilon/2$.
• Similarly, for $Y_n = O_P(1)$, there exists some $M_2 > 0$ and integer $N_2$ where for all $n \geq N_2$, $P(|Y_n| > M_2) < \epsilon/2$.
• Let $N = \max(N_1, N_2)$ (so both conditions hold for $n \geq N$) and set $K = M_1 \times M_2$.
• Now, note that the event $|X_nY_n| > K$ can only occur if either $|X_n| > M_1$ or $|Y_n| > M_2$ (if both were bounded by their respective $M$s, their product would be bounded by $K$).
• Apply the union bound to the probabilities:
  $$P(|X_nY_n| > K) \leq P(|X_n| > M_1) + P(|Y_n| > M_2)$$
• For $n \geq N$, this sum is less than $\epsilon/2 + \epsilon/2 = \epsilon$.
• Since $\epsilon$ was chosen arbitrarily, we've satisfied the definition of $X_nY_n = O_P(1)$.

Quick Context #

This result is indeed a standard (though sometimes unstated) asymptotic probability property. It's often taken for granted because it follows directly from the definition of boundedness in probability, which is why you might not find a dedicated proof in every textbook. That said, it's a useful building block for more complex asymptotic arguments in probability and statistics.

内容的提问来源于stack exchange，提问作者cgmil