Question

in convergence of probability $\lim_{n\to\infty} P(|X_n - X| + |Y_n - Y| > \varepsilon) \le \lim_{n\to\infty} \{P(|X_n - X| > \frac{\varepsilon}{2}) + P(|Y_n - Y| > \frac{\varepsilon}{2})\}$

          in convergence of probability
$\lim_{n\to\infty} P(|X_n - X| + |Y_n - Y| > \varepsilon) \le \lim_{n\to\infty} \{P(|X_n - X| > \frac{\varepsilon}{2}) + P(|Y_n - Y| > \frac{\varepsilon}{2})\}$

in convergence of probability
limn→∞ P(|Xn - X| + |Yn - Y| > ε) ≤limn→∞{P(|Xn - X| > (ε)/(2)) + P(|Yn - Y| > (ε)/(2))}

Added by Yolanda P.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Sri K

09/28/2023

Step 1

The absolute value function is defined as follows: \[ |a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases} \] Show more…

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In convergence of probability: [lim_{n o infty} P(|x_{n}-x|+|y_{n}-y|>epsilon) leq lim_{n o infty} left( P(|x_{n}-y|>frac{epsilon}{2}) + P(|mu_{n}-y|>frac{epsilon}{2}) ight)] How? In convergence of probability: [P(|x_{n}-x|+|y_{n}-y|>epsilon) leq P(|x_{n}-y|) + P(|mu_{n}-y|) ] How?