Convergence in probability

Question

Can anyone tell me how they got the regions $0<\epsilon<\theta$ and $\epsilon >0 $.

Also to clarify, is the last step where it says $\lim_{n \to \infty} P(|Y_n-\theta|>\epsilon)=0$

Answer 1

\begin{align} \mathbb{P}(|Y_n - \theta | \geq \epsilon) &= 1- \mathbb{P}(-\epsilon \leq Y_n - \theta \leq \epsilon) \\ &= 1 - \mathbb{P}(Y_n \leq \theta + \epsilon) + \mathbb{P}(Y_n \leq \theta - \epsilon) \\ &= 1 - F_{Y_n}(\theta + \epsilon) + F_{Y_n}(\theta-\epsilon). \\ &= F_{Y_n}(\theta- \epsilon). \end{align} Note that the last equality follows from the fact that $\theta + \epsilon \geq \theta$ for any $\epsilon>0$ and consequently $F_{Y_n}(\theta+ \epsilon) = 1$. According to the definition of the distribution function of $Y_n$, there are two cases to be considered:

(1) Assume $0<\epsilon<\theta$, then $0 < \theta-\epsilon < \theta$ and consequently $$\mathbb{P}(|Y_n - \theta| \geq \epsilon) = \left(\frac{\theta-\epsilon}{\theta}\right)^n \to 0,$$ as $n \to \infty$.

(2) Assume $\epsilon > \theta$, then $\theta - \epsilon < 0$ and consequently $F_{Y_n}(\theta- \epsilon) = 0, \forall n \geq 1$. Therefore, the result holds in the limit.