Convergence in probability 
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 $\lim_{n \to \infty} P(|Y_n-\theta|>\epsilon)=0$
 A: \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. 
A: Imagine we have $X$ representing the toss of a fair die.
Then $\text{Range}(X) = \{1,2,\cdots,6\}$
$$P(X = x) = \frac16$$
So:
$$P(X \le 1) = \frac16$$
$$P(X \le 2) = \frac26$$
$$P(X \le 3) = \frac36$$
$$P(X \le 4) = \frac46$$
$$P(X \le 5) = \frac56$$
$$P(X \le 6) = \frac66 = 1$$
Also:
$$P(X \le 7) = \frac66 = 1$$
$$P(X \le 0) = \frac06$$
$$P(X \le 0.5) = \frac06$$
$$P(X \le -2000) = \frac06$$
$$P(X \le 4.5) = \frac46$$
$$P(X > 7) = \frac06$$
Thus:
The '$\varepsilon > \theta$' (I'll assume '$\varepsilon > 0$' is a typo) is like '$X > 7$'
The '$0 < \varepsilon < \theta$' (I'll assume '$\varepsilon > 0$' is a typo) is like '$1 < X < 6$'

Also
Show that $Y_n$ converges in probability
