Given that $$\forall \epsilon_1\ \exists \delta_{\epsilon_1}>0, N_{\epsilon_1}>0,\text{ s.t. } \Pr\{ n^\alpha |X_n| \ge \delta_{\epsilon_1} \}<\epsilon_1\ \forall n>N_{\epsilon_1}$$ $$\forall \epsilon_2\ \exists \delta_{\epsilon_2}>0, N_{\epsilon_2}>0,\text{ s.t. }\Pr\{ n^\beta |Y_n| \ge \delta_{\epsilon_2} \}<\epsilon_2\ \forall n>N_{\epsilon_2} $$
How to prove that $$\forall \epsilon\ \exists \delta_\epsilon>0, N_\epsilon>0,\text{ s.t. } \Pr\{ n^{\alpha+\beta} |X_nY_n| \ge \delta_\epsilon \}<\epsilon\ \forall n>N_\epsilon $$
In other word, can we get $X_nY_n=O_p(n^{-(\alpha+\beta)})$ based on $X_n=O_p(n^{-\alpha})$ and $Y_n=O_p(n^{-\beta})$? what conditions should be added? Thanks!
Thank you @Michael. I guess the proof shuold be: $\forall \epsilon_1<\frac{\epsilon}{2},\ \epsilon_2<\frac{\epsilon}{2},\ \exists\ \delta_{1}\ge\delta_{\epsilon_1}\ge\sqrt{\delta},\ \delta_{2}\ge\delta_{\epsilon_2}\ge\sqrt{\delta},\ N_{\epsilon_1,\delta_1}\ge 0,\ N_{\epsilon_2,\delta_2}\ge 0$ s.t.
\begin{align*} P\{ n^{\alpha+\beta}\| X_n Y_n \| \ge \delta \} &\le P\{ n^{\alpha}\| X_n \| \ge \delta_1 \} + P\{ n^{\beta}\| Y_n \| \ge \delta_2 \} \\ &\le \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon \ \text{for} \ \forall n>\max\{N_{\epsilon_1,\delta_1}, N_{\epsilon_2,\delta_2}\} \end{align*}
Am I right?