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If $(x_{n}) \to x$ in probability convergence

Prove $(\sqrt{x_n}) \to \sqrt{x}$ in probability convergence

My attempt:

$$\text{Pr}(|\sqrt {x_n} - \sqrt{x}| \geq \epsilon) \leq \text{Pr}(|x_n- x| \geq \epsilon)$$

Is this true?

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  • $\begingroup$ Not if $x_n, x < 1$. $\endgroup$ – Pedro M. Dec 23 '17 at 17:51
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$$\text{Pr}(|\sqrt {x_n} - \sqrt{x}| \geq \epsilon) \leq \text{Pr}(\sqrt {|x_n -x|}\geq \epsilon)=\text{Pr}(|x_n- x| \geq \epsilon^2)$$

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  • $\begingroup$ what eq you reffer it ? $\endgroup$ – Doāa Y. El-borai Dec 23 '17 at 17:57
  • $\begingroup$ @DoāaY.El-borai He is using that $\{ w \in \omega : \sqrt{x_n} - \sqrt{x} \geq \epsilon \} \subset \{ w \in \omega : \sqrt{x_n - x} \}$ $\endgroup$ – Olba12 Dec 23 '17 at 18:07

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