If $Y_n$ is limited in probability and $X_n \to 0$ in probability then $X_nY_n \to 0$ in probability.

Attempt: I would go like:

for every $\epsilon$, $\mathbb P(|X_nY_n|\leq \epsilon) = \mathbb P(|Y_n|\leq \frac{ \epsilon}{ |X_n|})\geq\mathbb P(|Y_n|\leq C_{\delta}) \geq 1-\delta \to 1$ but the truth is that $\frac{ \epsilon}{ |X_n|}\geq C_{\delta}$ is not always true but almost certainly, so this strategy should be wrong!



For any fixed positive $\epsilon$:

$\mathbb{P}[|X_nY_n| \ge \epsilon] \leq$

$\mathbb{P}[|X_n| \ge \epsilon^2]$ $+$ $\mathbb{P}[|Y_n| \ge \frac{1}{\epsilon}]$

But (for any such $\epsilon$) the above 2 terms on the rhs of the above inequality go to 0 as $n$ goes to $\infty$.

  • $\begingroup$ Why is the first inequality correct? Imo $\mathbb P(|X_n| \ge \epsilon^2,|Y_n| \ge \frac{1}{\epsilon}) \leq \mathbb P(|X_nY_n| \ge \epsilon)$ and not the other way around. $\endgroup$ – Maffred Jun 28 '18 at 20:29
  • $\begingroup$ The only way $|X_nY_n|$ can be at least $\epsilon$ is if either $|X_n|$ is at least $\epsilon^2$ OR $|Y_n|$ is at least $\frac{1}{\epsilon}$. If both inequalities $|X_n| < \epsilon^2$ and $|Y_n| \le \frac{1}{\epsilon}$ hold then the inequality $|X_nY_n| < \epsilon$ also holds. $\endgroup$ – Mike Jun 28 '18 at 20:35
  • $\begingroup$ So, as you wrote above, $\mathbb{P}[|X_n| \le \epsilon^2$ AND $|Y_n| \le \frac{1}{\epsilon}]$ $\le$ $\mathbb{P}[|X_nY_n| \le \epsilon]$. But we also have $\mathbb{P}[|X_n| \le \epsilon^2$ OR $|Y_n| \le \frac{1}{\epsilon}]$ goes to 0, and this is at least $\mathbb{P}[|X_nY_n| \le \epsilon]$. $\endgroup$ – Mike Jun 29 '18 at 3:18

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